4.1 Introduction
Chapter 4 — Laws of Motion
On September 5, 1977, NASA launched a spacecraft called Voyager 1 from Cape Canaveral. Its original mission was to fly past Jupiter and Saturn and send back photographs. That primary mission ended in 1980. Yet nearly five decades later, Voyager 1 is still moving — travelling at about 61,000 kilometres per hour, carrying it further from Earth every second. It has now crossed into interstellar space, the most distant human-made object in existence.
But here is the puzzle. Voyager’s main engines have been switched off for decades. Nothing pushes it. Nothing pulls it forward. So what keeps it going?
The answer took humanity nearly two thousand years to work out. From Aristotle in ancient Greece, to Galileo in Renaissance Italy, to Isaac Newton in seventeenth-century England, thinkers struggled with a question that sounds deceptively simple: what does a force actually do to a body? Does moving require force? Does stopping? Does turning?
The answers Newton finally wrote down in 1687 — three short laws — turned out to govern almost everything mechanical around us. They explain why we lurch forward when a bus brakes, why a rocket rises when its hot exhaust rushes down, why a cyclist leans into a sharp turn, why a cricketer draws his hands back while catching a fast ball, and yes, why Voyager keeps drifting silently through the empty dark of space.
Figure to come
Fig. 4.0 – Illustration of Voyager 1 spacecraft coasting through interstellar space with Earth as a distant blue dot in the background; inset panels showing (a) a bus passenger jerking backward, (b) a rocket rising with exhaust below, (c) a car banking on a curved road — all captioned with “one chapter, one set of laws.”
This chapter will build up the answers step by step, beginning with a mistake Aristotle made and ending with the three laws that guide spacecraft, engineers, cyclists, and cricketers alike.
In the previous chapter, we spent our effort on describing motion — where a particle is at each instant, how fast it moves, and how its velocity changes. We saw that motion at a constant velocity can be described using velocity alone, while motion that speeds up, slows down, or changes direction needs the extra concept of acceleration.
But describing motion is not the same as explaining it. So far we have never once asked the deeper question: what actually causes a body to move the way it does? What decides whether it stays still, speeds up, slows down, or turns? That is the question this chapter takes up.
Let us begin the way scientists often do — by trusting common experience.
To make a football at rest start rolling, someone has to kick it. To throw a stone upward, your hand must give it an upward push. A gentle breeze can set the branches of a tree swinging, and a strong wind can even shift heavy objects that a person would struggle to move. A boat drifts down a flowing river without anyone rowing it, because the water carries it along. In each of these examples, something outside the body must act on it before its state of motion changes.
The same is true when we want to stop something. A ball rolling down an inclined plane can be brought to rest by pushing against its motion. If you did nothing, it would keep going. Something outside the ball has to act on it to slow it down or halt it.
This “something” is what we call a force, and the body providing it is called an external agency. In everyday physics language, a force is simply a push or a pull applied to a body by some agent outside it.
In all the examples above — hands, wind, stream, another player kicking a ball — the external agency is in direct contact with the body it acts on. Contact seems natural: to push something, you usually have to touch it.
But contact is not the only way a force can act. A stone released from the top of a building accelerates downward, and no visible hand pushes it. The Earth pulls it toward itself through the invisible force we call gravity. A bar magnet can lift an iron nail from a small distance away without ever touching it. These are examples of forces acting across empty space, without contact.
Figure to come
Fig. 4.1 – Two panels side by side: (a) a foot kicking a football and a breeze bending tree branches, labelled “Contact forces”; (b) a stone falling from a rooftop with a downward arrow labelled “gravity” and a bar magnet attracting a nearby iron nail, labelled “Non-contact forces.”
So, gathering what common experience tells us so far: a force is needed to put a stationary body in motion, and a force is also needed to stop a moving one. The agency providing this force may or may not be in contact with the body.
That covers starting and stopping. But now comes a subtler question — one whose answer took nearly two thousand years to work out.
Imagine a skater gliding in a straight line across a perfectly smooth sheet of ice at a steady speed. She is not speeding up, not slowing down, not turning. She is simply moving.
Does she need a force to keep going?
Common sense whispers “yes” — after all, when we push a book across a table and let go, it soon stops, so surely something must keep pushing to keep it moving. But is this really a law of nature, or is it just a habit of thought we have formed by living in a world full of friction?
That is the puzzle we opened this chapter with — the same puzzle that connects a skater on ice to the Voyager spacecraft drifting through interstellar space. The answer, worked out first by Galileo and then sharpened into precise laws by Newton, will turn out to be very different from what everyday experience suggests. Working it out will open the door to almost all of modern mechanics.
4.2 Aristotle’s Fallacy
At the end of Section 4.1 we left ourselves with a sharp question: does a body need a continuous external force to keep moving at constant velocity? The answer took nearly two thousand years to work out. The person who finally cracked it was Galileo Galilei in the seventeenth century, and his answer became the foundation of Newtonian mechanics — the framework that marks the birth of modern science.
To appreciate what Galileo saw, we first have to understand the answer that had been accepted before him — an answer given by one of the most influential thinkers in the history of Western thought.
Almost all of Aristotle’s specific ideas about motion are now known to be wrong. But one of them, stated simply, was this:
Why the Aristotelian view feels correct
The Aristotelian view is not silly — in fact, it is the view almost anyone would arrive at just by looking around. Consider a small child playing with a simple toy car on the floor, dragging it with a string. As long as the child pulls, the car rolls; the moment the string is released, the car slows and stops. What could be more obvious? To keep it moving, something must be applied. To make it stop, do nothing.
The same pattern is everywhere in daily life. Push a book across a table — let go, and it stops within seconds. Kick a football across dry ground — however hard the kick, the ball eventually rolls to a halt. Cycle without pedalling — you gradually slow down. Left to themselves, terrestrial bodies always seem to come to rest.
Spotting the flaw
So where exactly does Aristotle’s reasoning go wrong?
Look again at the toy car. When the child pulls it at constant speed across the floor, the child assumes the pull is what “keeps it moving.” But there is another force in the picture that the eye does not see — the force of friction between the wheels and the floor, always pointing opposite to the motion.
When the car moves at constant velocity, the child’s pull and the floor’s friction are equal in size and opposite in direction. Their sum is zero. In other words, the net external force on the car is already zero — even though the child is pulling.
The moment the string is released, the child’s pull disappears but friction does not. Now the only remaining external force is friction, and it acts against the motion, slowing the car until it stops.
Figure to come
Fig. 4.2 – Toy car being pulled across a floor. Two horizontal arrows on the car, drawn to equal length: one forward labelled “child’s pull,” one backward labelled “friction.” Caption: “At constant velocity, the two forces cancel; the net external force is zero.”
The correct picture, then, is not that motion needs a force — it is that motion at constant velocity needs zero net force. The child’s applied force is required only to cancel friction. Take friction away, and no applied force would be needed at all.
Why the flaw took so long to spot
The reason Aristotle’s mistake stood for so long is that opposing forces are always present in the natural world we live in. Friction acts wherever solids slide against solids; a related resisting force called viscous drag acts on bodies moving through fluids like air or water. You cannot walk down a street, push a trolley, or throw a stone through the air without one of these forces acting somewhere.
In such a world, it is extremely easy to mistake “the force needed to cancel friction” for “the force needed to keep a body moving.” Aristotle, in effect, coded common terrestrial experience into a law of nature — but he coded the wrong law, because the friction was invisible in his picture.
To dig down to the true relationship between force and motion, one has to strip the picture bare — to imagine a world in which friction and drag are absent, so that only the applied forces show. That imaginative leap is exactly what Galileo took, and it is the subject of the next section.
4.3 The Law of Inertia
At the end of Section 4.2 we saw that Aristotle’s mistake was rooted in the ever-present presence of friction. To find the true law of motion, someone would have to strip friction away — at least in imagination — and see what remained.
That someone was Galileo Galilei. And the way he stripped friction away was not by pure thought, but by a beautifully simple set of experiments with balls rolling on inclined planes.
Galileo’s single inclined plane
Galileo noticed three distinct kinds of behaviour on an inclined plane, illustrated in Fig. 4.3(a):
A ball rolling down an inclined plane speeds up — it accelerates.
A ball rolling up an inclined plane slows down — it retards.
A ball rolling on a horizontal plane is somewhere in between — neither pulled downhill nor pushed uphill.
Case (iii) is the interesting one. If moving down a slope causes speeding up, and moving up a slope causes slowing down, then on a truly horizontal plane — where the ball is neither climbing nor descending — there is no reason for it to speed up or slow down at all. Galileo concluded that a ball moving on a perfectly smooth (frictionless) horizontal plane must move with constant velocity, indefinitely.
Figure to come
Fig. 4.3(a) – Three sketches of a ball on inclined planes: (i) rolling down a slope with velocity arrow lengthening, (ii) rolling up a slope with velocity arrow shortening, (iii) rolling along a horizontal plane with constant-length velocity arrow.
Galileo’s double inclined plane
To test this idea more sharply, Galileo designed a second, ingenious experiment involving two inclined planes joined at the bottom, forming a V-shape (Fig. 4.3(b)).
He released a ball from rest at some height on one plane, and let it roll down and climb up the other. He observed that if the planes were smooth, the ball climbed to nearly the same height it had started from — a little less because of friction, but never more.
In the ideal case, with no friction at all, Galileo argued that the ball should climb back to exactly the same height. This was his first idealisation.
Now came the clever variation. He gradually decreased the slope of the second plane. Each time, the ball still climbed back to the same original height — but to do so, it had to travel a longer horizontal distance along the gentler slope.
What if the second plane were made completely horizontal? The ball, still “chasing” its original height, would never reach it. In this idealised frictionless picture, the ball would then roll along the horizontal plane forever — its motion would never cease.
Figure to come
Fig. 4.3(b) – Three stacked sketches of a double inclined plane: (i) equal slopes with ball reaching the same height on the other side; (ii) gentler right slope with ball travelling farther to reach same height; (iii) right plane fully horizontal, with dashed arrow showing ball continuing indefinitely.
In practice, of course, the ball does stop after some finite distance, because friction can never be totally eliminated. But Galileo had cleanly separated the behaviour of the ball itself from the effects of friction. Without friction, the ball’s uniform motion would simply continue.
The equivalence of rest and uniform motion
This was Galileo’s breakthrough insight — one that had eluded Aristotle and everyone who followed him for two thousand years. The state of rest and the state of uniform linear motion (motion in a straight line at constant velocity) are not two different situations. They are physically equivalent. In both cases, the net external force on the body is zero.
It is incorrect to assume that a net force is needed to keep a body moving at constant velocity. The applied force we usually need in everyday life is only there to cancel friction, so that the two forces add up to zero net external force. If friction were absent, no applied force would be required at all.
This is exactly why Voyager 1, out in interstellar space where friction and drag are effectively absent, needs no engine to keep moving. Its state of uniform motion is preserved as long as the net force on it is (nearly) zero.
Inertia — resistance to change
Galileo’s conclusion can be stated in one clean sentence: if the net external force on a body is zero, a body at rest stays at rest, and a body in motion continues to move with the same velocity in a straight line. This tendency of a body to resist any change in its state of rest or uniform motion is called inertia.
The word literally means “sluggishness” or “resistance to change” — and that captures the physical idea perfectly. A body left to itself does not spontaneously start moving, stop moving, speed up, slow down, or turn. Something outside it has to compel any such change.
This law was Galileo’s gift to physics. It cleared the ground for Isaac Newton, who took it as his starting point and built the whole edifice of classical mechanics upon it — as we shall see in the next section.
Long before Galileo, ancient Indian thinkers had developed an elaborate system of ideas about motion. Force, the cause of motion, was classified into several kinds — nodan (force due to continuous pressure, like wind on a sail), abhighat (impact force, like a potter’s rod striking a wheel), sanskara (the persistent tendency of a body to keep moving in a straight line, called vega), and forces transmitted through strings or rods. Of these, the concept of vega in the Vaisesika school comes remarkably close to the modern idea of inertia — the natural tendency of a body to continue in a straight line unless opposed by contact with air or other objects.
Indian thinkers also correctly recognised that the various motions of an extended body — translational, rotational, and vibrational — all arise from the translational motions of its constituent particles. A falling leaf, for example, may show downward motion (patan) as well as rotational and vibrational motion (bhraman, spandan) as a whole, but each individual particle at any instant undergoes only a small, definite displacement. The 12th-century mathematician Bhaskara (Bhaskara II) introduced the concept of tatkaliki gati — “instantaneous motion” — anticipating by several centuries the modern notion of instantaneous velocity used in differential calculus. The distinction between a wave and a current of water was also clearly understood: a current is a bulk motion of water particles under gravity, while a wave is the transmission of vibrations through them.
4.4 Newton’s First Law of Motion
Galileo’s simple but revolutionary ideas dethroned Aristotelian mechanics — but they did not yet form a complete science of motion. Somebody had to take Galileo’s law of inertia, turn it into a precise starting point, and build the rest of mechanics on top of it.
That task was accomplished almost single-handedly by Isaac Newton (1642–1727), one of the greatest scientists in history. Working roughly a generation after Galileo, Newton wrote down three short laws of motion that together explain almost every mechanical phenomenon around us, from a falling stone to the orbits of the planets. Newton did not begin from scratch — he built directly on Galileo’s law of inertia and made it the starting point of his mechanics.
Newton’s first law, in his own crisp language, states:
Rewriting the first law using acceleration
Both “at rest” and “moving at constant velocity in a straight line” have one thing in common — the velocity is not changing with time. So the acceleration is zero in both cases.
This lets us restate Newton’s first law more compactly, using the language of acceleration:
Written this way, the law works in both directions:
- If we know the net force is zero, we can conclude the acceleration is zero.
- If we know the acceleration is not zero, we can conclude that some net external force must be acting on the body.
Both directions are useful in practice, and they correspond to two rather different problem-solving situations.
Situation 1 — We know the net force is zero
Sometimes we can be reasonably sure, from the physical setup itself, that no significant external force acts on a body.
For instance, an astronaut floating far out in deep interstellar space, with no nearby stars or planets and with all thrusters switched off, has essentially no forces acting on him. By the first law, his acceleration must be zero. If he happens to be moving at the moment he shuts off his thrusters, he will simply keep drifting at that velocity forever — precisely the situation we described for Voyager 1 in Section 4.3.
Because the setup tells us that the force is zero, the first law gives us the motion for free: zero acceleration, constant velocity.
Situation 2 — We know the body is unaccelerated
Much more often, we do not have a clear list of all the forces on a body to begin with. But if we can observe that the body is at rest or moving with uniform velocity, we can turn the argument around and infer that the net external force on it must be zero.
This is a surprisingly useful trick. It does not require us to know each individual force — it only requires knowing the state of motion. If the body is not accelerating, whatever forces are acting on it must cancel out overall.
Consider a book at rest on a horizontal table (Fig. 4.4(a)). Two external forces act on it: its weight \(W\) pulling it down (the pull of the Earth), and the normal force \(R\) pushing up on it (the contact push from the table).
Figure to come
Fig. 4.4(a) – A book resting flat on a table. Two vertical arrows on the book of equal length: one pointing down labelled \(W\), one pointing up labelled \(R\).
Now, we do not know from first principles what \(R\) actually is. Unlike gravity, which we can calculate independently, the normal force is a self-adjusting force: the table pushes up with whatever value is needed to prevent the book from sinking into it.
But we can pin \(R\) down using the first law. The book is observed to be at rest, so its acceleration is zero, so the net external force on it must be zero. That in turn forces the normal force to equal the weight in magnitude and oppose it in direction:
\[R = W\]
Note carefully the direction of the reasoning here. It is tempting to say, “Since \(R\) equals \(W\), the two forces cancel, and so the book stays at rest.” This gets cause and effect backwards. The correct chain of reasoning is:
Because the book is observed to be at rest, the net force on it must be zero (by the first law), which forces the table’s normal force \(R\) to adjust to be exactly equal and opposite to the weight \(W\).
This distinction is not just philosophical fussiness — it matters in problems where the body is accelerating vertically (say, a book on the floor of a lift going up with acceleration). There, \(R\) will not equal \(W\), precisely because the acceleration is not zero. Confusing the two directions of reasoning leads to wrong answers on exactly such problems.
A subtler example: a car picking up speed
Now consider a car that starts from rest, picks up speed, and then cruises on a smooth straight road at constant velocity (Fig. 4.4(b)).
Figure to come
Fig. 4.4(b) – A car shown in three snapshots along a straight road: (i) stationary with zero-velocity marker, (ii) accelerating with a growing velocity arrow and a friction-force arrow pointing forward on the driven tyres, (iii) moving at constant velocity with no net horizontal force.
Let us walk through the three stages using the first law:
- When the car is stationary, the net horizontal force on it is zero. The first law is trivially satisfied.
- During pick-up, the car accelerates in the forward direction. By the first law, a net external force must be acting on it in the direction of motion.
- Once the car cruises at constant velocity, the net force on it is once again zero.
Here comes the subtle bit. What is the external force that accelerates the car during pick-up?
It cannot be anything internal — no internal engine push or driver muscle can, by itself, accelerate the whole car as a rigid body. (We shall see the deep reason for this when we discuss Newton’s third law.) The only conceivable external horizontal force available to push the car along the road is the friction between the tyres and the road surface.
Friction, which we usually think of as opposing motion, is here acting in the direction of motion on the driven tyres — because the engine tries to spin the tyres backwards against the road, and the road pushes the tyres forward in reaction. This forward friction is what accelerates the whole car. We will study friction properly in Section 4.9.
Inertia in daily life — the bus jerk
The inertia buried inside the first law shows up dramatically whenever a vehicle changes its motion suddenly.
Suppose you are standing in a stationary bus, holding no support, and the driver suddenly starts the bus forward. You get thrown backward with a jerk. Why? Nobody has pushed you backward.
Your feet are in contact with the floor of the bus. Friction between your shoes and the floor is (usually) enough to accelerate your feet along with the bus. But your body is not a rigid block — it is deformable, and different parts of it can shift slightly relative to each other. So while your feet go forward with the bus, the upper part of your body, by inertia, tries to stay where it was. Relative to the bus, then, you get thrown backward.
The instant after, muscular forces in your legs and torso pull the rest of your body forward, bringing everything up to the bus’s velocity again. That is why the jerk is momentary rather than permanent.
The reverse happens when the driver brakes suddenly. Now your feet stop along with the bus (again because of friction between shoe and floor), but the upper part of your body continues forward by inertia. You are thrown forward. Muscular forces then act to bring the body back to rest.
So the “invisible hand” that seems to throw you backward or forward in a bus is really just inertia — your body’s resistance to any change in its state of motion. Nobody pushed you; the bus pushed your feet, and the rest of you had to catch up (or slow down) a moment later.
4.5 Newton’s Second Law of Motion
The first law told us what happens when the net external force on a body is zero — the body stays at rest, or continues moving with a constant velocity. But that is just a special case. The far more common situation is one in which the net force is not zero and the body’s velocity is changing.
For that general situation, we need a law that tells us how much the velocity changes for a given force. That is what Newton’s second law does. It relates the net external force on a body to its acceleration — but in a subtler way than you might first guess. To see the subtlety, we first need a new quantity: momentum.
Momentum
Momentum is defined as the product of a body’s mass and its velocity. It is written as \(\vec{p}\), and
\[\vec{p} = m\,\vec{v} \qquad (4.1)\]
where \(m\) is the body’s mass (SI unit: kg) and \(\vec{v}\) is its velocity (SI unit: m s⁻¹). Momentum is a vector, pointing in the same direction as the velocity. Its SI unit is kg m s⁻¹ (also written N s).
Why introduce momentum at all? Because a whole cluster of everyday observations shows that when we ask “how much force is needed to affect a body’s motion?”, the answer depends on both the body’s mass and its velocity — not on either alone. Momentum ties them into a single quantity.
Consider four familiar observations.
Mass matters. Imagine a small car and a loaded truck parked on a horizontal road. To push each up to the same speed in the same time, you need a much greater force on the truck than on the car. Likewise, if the two are moving at the same speed, a much greater force is needed to bring the truck to rest in the same time. For the same change in velocity over the same time, more massive bodies need more force. Mass is clearly a key parameter.
Speed matters. A bullet fired from a gun at very high speed can pierce human tissue; the same bullet lobbed by hand at moderate speed barely dents the skin. For a body of given mass, greater speed means a greater force is needed to bring it to rest in the same time. So velocity is also a key parameter.
Falling stones. If two stones — one light, one heavy — are dropped from the top of a building, the light stone is much easier to catch than the heavy one. Both reach the ground at nearly the same speed (gravity gives them the same acceleration), but the heavier stone needs more force to stop.
Taken together, these observations point to a single conclusion: the product of mass and velocity — momentum — is the quantity that captures “how hard it is to change a body’s motion.”
Rate of change of momentum
Now for the crucial next step. It is not momentum by itself, but the rate of change of momentum, that matters for force.
A cricketer’s catch. A seasoned cricketer catches a fast-moving ball far more easily than a novice, who often hurts his hands. Why? Both stop the same ball with the same initial momentum — so both change the ball’s momentum by the same amount. The difference is how fast they do it. The seasoned cricketer draws his hands backwards as the ball arrives, so the ball takes a longer time to come to rest. The novice keeps his hands fixed and stops the ball almost instantly. Same change in momentum, longer time in one case → smaller force. Same change in momentum, shorter time in the other case → larger force. That is what hurts.
Figure to come
Fig. 4.5 – A cricketer catching a ball, drawing his hands backwards during the catch to lengthen the impact time. Labels: “ball’s momentum before catch,” “ball’s momentum after catch (zero),” “time of contact — longer when hands are drawn back.”
The physics is clear: force depends not just on how much the momentum changes, but on how fast the change happens. The greater the rate of change of momentum, the greater the applied force.
Same force, same time → same change in momentum. Another observation seals the point. Apply a fixed force for a fixed interval of time to two bodies of different masses, both starting at rest. The lighter body picks up a greater speed than the heavier one — but at the end, both have gained the same momentum. So the same force acting for the same time causes the same change in momentum, regardless of mass. This is a direct clue to what the second law should look like.
Direction matters too. In all the cases above, the change in momentum is in the same direction as the force. But momentum is a vector, and its direction can change even when its magnitude does not. Consider a stone whirled in a horizontal circle at uniform speed by a string. The magnitude of the momentum is fixed, but its direction keeps rotating. A force is needed to cause this change of direction — and we feel this force as the tension in the string. If we whirl the stone faster, or on a smaller circle, we must pull harder. Again, the greater the rate of change of the momentum vector, the greater the force required.
Figure to come
Fig. 4.6 – A stone tied to a string being whirled in a horizontal circle above a person’s head. Two snapshots of the stone at different positions show equal-length momentum vectors pointing in different directions.
Stating the Second Law
Newton pulled these observations together into a single elegant statement:
In symbols: if a force \(\vec{F}\) acts on a body for time \(\Delta t\), and the body’s momentum changes from \(\vec{p}\) to \(\vec{p} + \Delta \vec{p}\), then
\[\vec{F} \propto \frac{\Delta \vec{p}}{\Delta t} \quad \text{or} \quad \vec{F} = k\,\frac{\Delta \vec{p}}{\Delta t}\]
where \(k\) is a constant of proportionality. Taking the limit \(\Delta t \to 0\), the ratio \(\Delta \vec{p}/\Delta t\) becomes the derivative \(\mathrm{d}\vec{p}/\mathrm{d}t\):
\[\vec{F} = k\,\frac{\mathrm{d}\vec{p}}{\mathrm{d}t} \qquad (4.2)\]
From momentum-form to \(\vec{F} = m\vec{a}\)
For a body of fixed mass \(m\), we can pull the mass out of the derivative:
\[\frac{\mathrm{d}\vec{p}}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}(m\vec{v}) = m\,\frac{\mathrm{d}\vec{v}}{\mathrm{d}t} = m\,\vec{a} \qquad (4.3)\]
So Eq. (4.2) becomes
\[\vec{F} = k\,m\,\vec{a} \qquad (4.4)\]
Force is proportional to the product of mass and acceleration.
Now, the unit of force has not been fixed anywhere yet. We are free to choose the constant \(k\) to give ourselves a convenient unit. The simplest choice is \(k = 1\). With this choice, the second law takes its most famous form:
\[\vec{F} = \frac{\mathrm{d}\vec{p}}{\mathrm{d}t} = m\,\vec{a} \qquad (4.5)\]
The SI unit of force — the newton
Setting \(k = 1\) fixes the unit of force. In SI, a unit force is defined as the force that gives a mass of 1 kg an acceleration of 1 m s⁻². This unit is named the newton (symbol N):
\[1 \text{ N} = 1 \text{ kg} \cdot 1 \text{ m s}^{-2} = 1 \text{ kg m s}^{-2}\]
For a rough feel: the weight of a 100-gram apple on Earth is about 1 N.
Four important points about the second law
The second law is deceptively simple. Four features are worth spelling out.
(i) Consistent with the first law. In Eq. (4.5), if \(\vec{F} = 0\) then \(\vec{a} = 0\), and the body moves at constant velocity — exactly what the first law says. So the first law is not really a separate rule; it is the special case of the second law when the net force is zero.
(ii) It is a vector law — three equations in one. The equation \(\vec{F} = m\vec{a}\) is a vector equation. In three dimensions, it splits into one scalar equation for each direction:
\[F_x = \frac{\mathrm{d}p_x}{\mathrm{d}t} = m a_x, \quad F_y = \frac{\mathrm{d}p_y}{\mathrm{d}t} = m a_y, \quad F_z = \frac{\mathrm{d}p_z}{\mathrm{d}t} = m a_z \qquad (4.6)\]
A key consequence: if a force is not parallel to the velocity, it only changes the component of velocity along its own direction. The perpendicular component of velocity is left unchanged. This is exactly what happens in projectile motion: gravity acts vertically downward, so only the vertical component of velocity changes, while the horizontal component stays constant.
Figure to come
Fig. 4.7 – A person standing at the door of an accelerating train drops a stone. Two panels: (a) the moment of release, with the stone carrying the train’s horizontal velocity; (b) a moment later, with the stone still moving with the same horizontal component but a growing downward vertical velocity — no horizontal force, no horizontal acceleration.
(iii) Applicable to a single point particle — and to systems. Strictly, Eq. (4.5) is stated for a single point particle: \(\vec{F}\) is the net external force on the particle, and \(\vec{a}\) is its acceleration. It turns out that the same equation applies to a rigid body — or, more generally, to any system of particles — provided we understand \(\vec{F}\) as the total external force on the system and \(\vec{a}\) as the acceleration of the system’s centre of mass. Internal forces between particles of the system are not to be included in \(\vec{F}\); they cancel out in pairs (a direct consequence of the third law, which we take up in the next section). The precise treatment of extended bodies will come in Chapter 6.
(iv) It is a local relation. The force \(\vec{F}\) acting on a body at a certain point in space at a certain instant of time determines the acceleration at that point at that instant — nothing more. Acceleration does not carry any memory of past motion. The moment after a stone is released from an accelerating train, for example, the stone no longer accelerates horizontally — it “carries no memory” of the train’s acceleration a moment ago. (Air resistance is neglected here.) The only force on the free-falling stone is gravity, and the only acceleration it has is \(g\), downwards.
Impulse
There are many situations in which a very large force acts on a body for a very short time — and yet produces a finite change in the body’s momentum. A hammer striking a nail, a bat hitting a ball, a bullet striking a wall, a hand catching a fast-moving ball: in each case, the force during the tiny contact time is huge and difficult to measure directly. But the product of that force and the contact time is much easier to obtain, because by the second law it equals the change in momentum:
\[\text{Impulse} = \vec{F} \times \Delta t = \Delta \vec{p} \qquad (4.7)\]
A large force acting for a short time to produce a finite change in momentum is called an impulsive force. Historically, impulsive forces were often put in a separate conceptual category from “ordinary” forces, but in Newtonian mechanics they are exactly the same kind of thing — they simply happen to be large and brief.
The concept of impulse turns two hard-to-measure quantities (a large force and a tiny contact time) into one easily calculable quantity (change in momentum). This is what makes it so useful in problems involving collisions and impacts.
4.6 Newton’s Third Law of Motion
The second law tells us how a force changes a body’s motion. But it leaves one obvious question unanswered: where does that external force come from in the first place? What agency actually provides it?
The answer in Newtonian mechanics is beautifully simple: the external force on any body always arises from some other body. Forces are not properties of a single object floating around on its own — they are the mechanism by which two bodies interact with each other.
This raises a natural question. Consider two bodies \(A\) and \(B\), where \(B\) exerts a force on \(A\). Does \(A\), in turn, exert a force on \(B\)? And if so, how are the two forces related?
Two suggestive examples
Sometimes the two-way nature of force is obvious. Press a coiled spring with your hand — the spring compresses under your push, but you can also feel the spring pushing back on your hand. Both bodies (you and the spring) are exerting forces on each other simultaneously.
Sometimes it is less obvious. The Earth pulls a stone downwards through gravity. Does the stone pull the Earth upwards? At first glance the answer seems to be “no” — we never actually see the Earth jump up to meet a falling stone. But Newton insisted the answer is yes. The stone exerts an equal and opposite gravitational pull on the Earth. We do not see the Earth move because the Earth is enormously more massive: the same magnitude of force, applied to a body about \(10^{25}\) times more massive than the stone, produces an acceleration far too small to observe.
Stating the Third Law
Newton captured this idea in one crisp sentence — arguably the most quoted sentence in all of physics:
The wording is so memorable that it has passed into common English. But partly because it is so memorable, it is also widely misunderstood. In modern physics language, a clearer statement is:
In symbols, if \(\vec{F}_{AB}\) is the force on \(A\) by \(B\), and \(\vec{F}_{BA}\) is the force on \(B\) by \(A\), then
\[\vec{F}_{AB} = -\vec{F}_{BA} \qquad (4.8)\]
Force, then, is never a one-body property. It is always a mutual interaction between two bodies.
Three common misconceptions cleared up
Because the action–reaction language is so old, several confusions have grown around it. Let us settle them one by one.
(i) “Action” and “reaction” both simply mean force. There is no special physical concept called “action” and a separate one called “reaction.” Both are just forces. Using two names for the same physical concept is only a linguistic habit, and it can mislead. The cleaner statement — “the force on \(A\) by \(B\) equals the negative of the force on \(B\) by \(A\)” — avoids the trap entirely.
(ii) Action does not precede reaction in time. The third law does not mean that \(A\) first pushes \(B\), and then \(B\) — as a delayed effect — pushes back. There is no cause–effect relationship implied. The two forces \(\vec{F}_{AB}\) and \(\vec{F}_{BA}\) arise together and act at the very same instant. Since neither comes first, either one may be called “the action” and the other “the reaction” — the choice is arbitrary. They are two aspects of a single mutual interaction.
(iii) Action and reaction act on different bodies. This is the most important point. The two forces of a third-law pair act on two different bodies — one on \(A\) and one on \(B\). They never act on the same body. Two consequences follow:
- If you are analysing the motion of \(A\) alone, only \(\vec{F}_{AB}\) enters \(A\)’s equation of motion — the other force in the pair acts on \(B\), not on \(A\).
- It is a serious error to add up \(\vec{F}_{AB}\) and \(\vec{F}_{BA}\) and conclude “the net force is zero, so nothing accelerates.” The two forces are indeed equal and opposite, but they do not cancel each other on any single body because they don’t act on the same body.
However — and this matters — if you consider the two bodies \(A\) and \(B\) together as a single system, then \(\vec{F}_{AB}\) and \(\vec{F}_{BA}\) are both internal forces of that system. Now they do cancel out, and their contribution to the total external force on the system is zero.
This is a deep fact: all internal forces in a body or a system of particles cancel out in pairs. It is exactly what allows Newton’s second law to be applied to extended bodies (a whole car, a whole planet), not just to point particles. We will make heavy use of this in Chapter 6 when we study systems of particles.
Real-world manifestations
The third law is not an abstract idea — it is at work every time anything mechanical happens.
Rockets — going up by pushing something down. A rocket engine burns fuel and expels hot exhaust gases downwards through its nozzle at very high speed. The rocket pushes those gases down; by the third law, the gases push the rocket up with an equal and opposite force. That upward push (called the thrust) is what lifts the rocket. Notice something important — the rocket does not need to push against the ground, or against the air below it. It only needs its own exhaust to push against. This is why rockets can operate perfectly well in the vacuum of outer space, where there is no air to react against.
Applying the Third Law to a Collision
The third law is often the last step in solving collision problems: use the second law to find the force (or impulse) on one body, then use the third law to flip the direction and get the force on the other.
4.7 Conservation of Momentum
The second and third laws of motion, taken together, lead to one of the most powerful and widely used principles in all of physics: the law of conservation of momentum.
Deriving conservation from Newton’s laws
Let us derive the law from a familiar example: a bullet fired from a gun.
At the moment of firing, the gun exerts a force \(\vec{F}\) on the bullet, driving it forward. By Newton’s third law, the bullet exerts an equal and opposite force \(-\vec{F}\) on the gun. Both forces act for the same brief interval of time \(\Delta t\) — the time during which the bullet is being pushed out by the barrel.
By Newton’s second law in impulse form, the change in the bullet’s momentum is \(\vec{F}\,\Delta t\), and the change in the gun’s momentum is \(-\vec{F}\,\Delta t\). Before firing, both were at rest, so the initial momentum of each was zero — the change of each therefore equals its final momentum:
\[\vec{p}_b = \vec{F}\,\Delta t, \qquad \vec{p}_g = -\vec{F}\,\Delta t\]
Adding these,
\[\vec{p}_b + \vec{p}_g = 0\]
The total momentum of the (bullet + gun) system after firing is exactly what it was before — namely, zero. The forward momentum of the bullet is exactly balanced by the equal and opposite backward (recoil) momentum of the gun. Momentum has simply been redistributed within the system, not created or destroyed.
Isolated systems and the general principle
The bullet–gun case is a special case of something much more general. The only forces that changed the two momenta were mutual forces between the two bodies of the system. There was no external agent acting on the pair from outside. A system on which the net external force is zero is called an isolated system.
For any isolated system of particles, internal forces can shuffle momentum around between individual particles, but by Newton’s third law these internal forces come in equal and opposite pairs. The individual changes cancel in pairs, and the total momentum of the system stays exactly the same.
Application: collisions
The commonest application of conservation of momentum is to collisions — two bodies briefly interacting with each other while no significant outside force acts on the pair.
Consider two bodies \(A\) and \(B\) with initial momenta \(\vec{p}_A\) and \(\vec{p}_B\). They collide, interact for a short time \(\Delta t\), then separate with final momenta \(\vec{p}'_A\) and \(\vec{p}'_B\).
Let \(\vec{F}_{AB}\) be the force on \(A\) by \(B\) during the collision, and \(\vec{F}_{BA}\) the force on \(B\) by \(A\). Applying Newton’s second law in impulse form to each body:
\[\vec{F}_{AB}\,\Delta t = \vec{p}'_A - \vec{p}_A\] \[\vec{F}_{BA}\,\Delta t = \vec{p}'_B - \vec{p}_B\]
(We have used a common time interval \(\Delta t\) because that is the time for which the two bodies are in contact.)
By Newton’s third law, \(\vec{F}_{AB} = -\vec{F}_{BA}\), so adding the two impulses gives zero on the left:
\[(\vec{p}'_A - \vec{p}_A) + (\vec{p}'_B - \vec{p}_B) = 0\]
Rearranging,
\[\vec{p}'_A + \vec{p}'_B = \vec{p}_A + \vec{p}_B \qquad (4.9)\]
The total momentum of the two-body system after the collision equals its total momentum before.
Notice what we have not assumed. Nothing about how the collision happens — whether the bodies stick together, bounce apart cleanly, or deform temporarily. Nothing about kinetic energy. Momentum conservation therefore holds for every collision in an isolated system, regardless of the internal details. That is what makes it such a powerful tool: it works even for messy, complicated collisions where the interior dynamics are hopeless to analyse in detail.
A brief note on elastic vs inelastic collisions
Collisions are often classified as elastic or inelastic depending on whether kinetic energy is also conserved:
- In an elastic collision, the total kinetic energy of the system is conserved as well as the total momentum. Two hard billiard balls colliding is close to elastic.
- In an inelastic collision, some kinetic energy is converted into heat, sound, or permanent deformation, so kinetic energy is not conserved. Total momentum, however, still is. Two bodies that stick together after impact (a completely inelastic collision) is a common example.
The crucial point: momentum conservation applies to both types equally. Kinetic energy conservation is an additional condition that only elastic collisions satisfy. We will study elastic and inelastic collisions in detail in Chapter 5 (Work, Energy, and Power).
4.8 Equilibrium of a Particle
The three laws and the conservation principle we have built up so far are enough to handle a wide range of problems. Before moving on to the specific kinds of forces we meet in mechanics, let us isolate one particularly useful special case: the situation in which a particle is in equilibrium.
What “equilibrium” means
In mechanics, a particle is said to be in equilibrium when the net external force on it is zero.
Notice at once what the first law tells us about such a particle: since the net force is zero, its acceleration is zero, so the particle is either at rest or moving with a uniform velocity in a straight line. Both possibilities count as “equilibrium.”
Two forces on a particle
The simplest equilibrium case is a particle acted on by just two forces \(\vec{F}_1\) and \(\vec{F}_2\). For zero net force,
\[\vec{F}_1 = -\vec{F}_2 \qquad (4.10)\]
The two forces must be equal in magnitude and opposite in direction. This is exactly the situation of the book resting on a table that we studied in Section 4.4: the table’s normal reaction is equal and opposite to the book’s weight.
Three concurrent forces on a particle
The next-simplest case is three forces acting at the same point on the particle. Forces that all act at a common point are called concurrent forces. For equilibrium under three concurrent forces \(\vec{F}_1\), \(\vec{F}_2\) and \(\vec{F}_3\), the vector sum must vanish:
\[\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \vec{0} \qquad (4.11)\]
Rearranging this equation gives a very useful geometric picture: the resultant of any two of the forces, obtained by the parallelogram law, must be equal in magnitude and opposite in direction to the third. Equivalently, if we draw the three force vectors head-to-tail, they close up to form a triangle (Fig. 4.9). The three forces in equilibrium can therefore be represented by the sides of a triangle taken in the same sense — all arrows going the same way around.
Figure to come
Fig. 4.9 – Three panels: (a) three vectors \(\vec{F}_1, \vec{F}_2, \vec{F}_3\) radiating from a common point, showing concurrent forces at a particle; (b) parallelogram construction showing that \(\vec{F}_1 + \vec{F}_2\) is equal and opposite to \(\vec{F}_3\); (c) the same three vectors drawn head-to-tail, closing into a triangle. Caption: “Three concurrent forces in equilibrium form a closed triangle.”
This result generalises easily. A particle acted on by \(n\) concurrent forces \(\vec{F}_1, \vec{F}_2, \ldots, \vec{F}_n\) is in equilibrium if and only if these vectors, drawn head-to-tail, close into an \(n\)-sided polygon with all arrows going the same way around.
Component form
For actual calculations, the vector equation \(\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = 0\) is usually split into three scalar equations — one for each coordinate direction:
\[F_{1x} + F_{2x} + F_{3x} = 0\] \[F_{1y} + F_{2y} + F_{3y} = 0\] \[F_{1z} + F_{2z} + F_{3z} = 0 \qquad (4.12)\]
Here \(F_{1x}, F_{1y}, F_{1z}\) are the components of \(\vec{F}_1\) along the \(x\), \(y\), and \(z\) axes, and similarly for the other forces. In two-dimensional problems (by far the most common case in this chapter), only the \(x\) and \(y\) equations are needed.
Real-world equilibrium
A worked example
4.9 Common Forces in Mechanics
Now that we have Newton’s three laws in hand, and the special case of equilibrium settled, we need to know what forces to plug in when we apply the laws to real situations. In this section we survey the forces that appear again and again in mechanics problems: gravity, normal reaction, friction, tension, spring force, and a few others.
Gravity — the ever-present non-contact force
Every object on Earth experiences a downward pull from the Earth called the force of gravity (or the object’s weight). Gravity is not a contact force: it acts across empty space, without the two bodies needing to touch. It is also universal — the same force that pulls an apple to the ground governs the motion of the Moon around the Earth, and of the Earth around the Sun. Because gravity acts at a distance, it is called a non-contact force.
We will study gravity in depth in Chapter 7. For present purposes, all we need is that gravity gives every body of mass \(m\) near the Earth’s surface a downward weight of
\[W = mg\]
where \(g \approx 9.8 \text{ m s}^{-2}\) is the acceleration due to gravity.
Contact forces — from one body touching another
Apart from gravity, almost every force we meet in this chapter is a contact force — it arises because two bodies touch each other. When two bodies are in contact (a book on a table, two blocks pressed together, a hand on a spring, a foot on the ground), they exert forces on each other at every point where they touch. These forces obey Newton’s third law: the force on body A by B is equal and opposite to the force on B by A.
The contact force at any point can be decomposed into two components:
- The component perpendicular to the surfaces in contact is called the normal reaction (\(N\) or \(R\)).
- The component parallel to the surfaces in contact is called friction (\(f\)).
We already met the normal reaction in Section 4.4, where it kept a book from falling through a table. Friction will be the subject of subsection 4.9.1.
Contact forces also arise when a solid touches a fluid. A body immersed in a liquid experiences an upward buoyant force equal to the weight of the fluid it displaces (Archimedes’ principle). Fluids also exert a viscous force on solid bodies moving through them, and gases exert air resistance or drag. All of these are contact forces because they need the solid and the fluid to be in physical contact.
Figure to come
Fig. 4.11 – Four small sketches illustrating contact forces in mechanics: (a) a block on a table with normal reaction \(N\) upward and weight \(W\) downward; (b) a stretched spring pulling a block, with spring force \(F = -kx\) marked; (c) a stick stirring a liquid, showing viscous drag opposing the stick’s motion; (d) a small object floating on water, with buoyant upthrust and weight marked.
Tension and spring force
Two other contact forces show up so often that they deserve special attention: tension in a string, and force by a spring.
Spring force. When you push or pull on a spring, the spring pushes or pulls back with a restoring force — a force that tries to return the spring to its natural (unstretched) length. For small displacements from the natural length, the restoring force is proportional to the displacement:
\[F = -kx\]
where \(x\) is the displacement from the unstretched state and \(k\) is a positive constant called the spring constant or force constant (SI unit: N m⁻¹). The negative sign says that the force always points opposite to the displacement: pull the spring rightward, and it pulls back leftward; compress it leftward, and it pushes back rightward.
Tension. For an ideal, inextensible (unable-to-stretch) string, the “spring constant” is effectively infinite — the string does not stretch even under large forces. The restoring force in such a string is called tension (\(T\)). For a massless string passing around a smooth (frictionless) pulley, the tension is the same throughout the length of the string. This is a very useful modelling assumption in mechanics problems: unless told otherwise, assume the string is inextensible and massless, and the pulley is smooth, so that a single symbol \(T\) describes the tension everywhere.
Where do contact forces come from?
Contact forces feel like a category distinct from gravity, but at a deeper level they aren’t. We know from modern physics that there are four fundamental forces in nature: gravitational, electromagnetic, weak, and strong. The weak and strong forces act only inside atomic nuclei — they don’t concern us in mechanics. That leaves gravity and electromagnetic forces.
Here comes the surprise: every contact force in mechanics — normal reaction, friction, tension, spring force, buoyancy, viscous drag — is fundamentally electromagnetic in origin. At the microscopic level, all bodies are made of electrons and nuclei, and when two bodies are pushed together, it is the electrostatic repulsion between the electron clouds at the surface of one body and the electron clouds at the surface of the other that stops them from passing through each other. Contact forces are the macroscopic bulk effect of trillions of these tiny electrical interactions.
Why, then, do we treat contact forces as separate types rather than derive them from electromagnetism? Because the microscopic details are hopelessly complicated. It is far more useful to describe contact forces empirically — measure their characteristic properties from experiment (like the coefficient of friction or the spring constant) and use these in problems — than to try to compute them from first principles.
4.9.1 Friction
Let us now examine the friction force in detail.
Return to the setup of a body of mass \(m\) resting on a horizontal table (as in Section 4.4). The weight \(mg\) is cancelled by the table’s normal reaction \(N\); there is no motion. Now apply a small horizontal force \(F\) to try to push the body sideways.
Common experience tells us that a small applied force may not move the body at all — a small push on a heavy wooden crate does nothing. But according to Newton’s second law, if \(F\) were the only horizontal force on the body, the body would have to accelerate at \(F/m\) no matter how small \(F\) is. Since it doesn’t, there must be another horizontal force present, one that cancels \(F\). That force is called static friction \(f_s\), and it points opposite to \(F\) (Fig. 4.12(a)).
Figure to come
Fig. 4.12 – Two panels: (a) A block on a horizontal surface with an applied force \(F\) pushing it rightward; a static friction force \(f_s\) acts leftward at the base, equal in magnitude to \(F\); block does not move. (b) The same block sliding rightward with velocity \(v\); kinetic friction \(f_k\) acts leftward at the base, opposing the motion.
Static friction — the self-adjusting force
Static friction is unusual. Unlike gravity, whose value is fixed, static friction adjusts itself to whatever is needed to keep the body at rest:
- If you don’t push, static friction is zero (there is no need for it).
- Push a little, and static friction pushes back the same little amount — the body doesn’t move.
- Push a bit more, and static friction increases to match — the body still doesn’t move.
This can’t go on forever. There is a maximum value beyond which static friction can no longer match the applied force. Experiments show that this limiting value depends on the normal force \(N\) pressing the two surfaces together, and on the nature of the surfaces themselves, but not (to a good approximation) on the area of contact:
\[(f_s)_{\text{max}} = \mu_s N \qquad (4.13)\]
Here \(\mu_s\) is a dimensionless constant called the coefficient of static friction, depending only on the pair of surfaces in contact. Typical values: rubber on dry concrete, \(\mu_s \approx 1\); steel on ice, \(\mu_s \approx 0.03\).
So the general law of static friction reads:
\[f_s \le \mu_s N \qquad (4.14)\]
with equality only at the point of impending motion — that is, the instant when the body is on the verge of sliding.
Static friction is said to oppose impending motion — that is, the motion that would have taken place if friction were absent.
Kinetic (sliding) friction
Once the applied force \(F\) exceeds the maximum static friction, the body begins to slide. The friction force does not vanish; a resistive force still opposes the sliding motion. This is called kinetic friction or sliding friction \(f_k\). Experiments give a similar-looking law:
\[f_k = \mu_k N \qquad (4.15)\]
where \(\mu_k\) is the coefficient of kinetic friction. Three important experimental facts:
- \(\mu_k\) is (approximately) independent of the sliding speed.
- \(\mu_k\) is (approximately) independent of the area of contact.
- \(\mu_k\) is less than \(\mu_s\) for the same pair of surfaces.
That last point is important. Because \(\mu_k < \mu_s\), once you overcome static friction and start something sliding, it takes less force to keep it moving than it did to get it started — as anyone who has pushed a heavy crate has felt.
Once the body is sliding, its acceleration from Newton’s second law is
\[a = \frac{F - f_k}{m}\]
If the applied force just equals \(f_k\), the body slides at constant velocity. If the applied force is removed, the body decelerates at \(-f_k/m\) and eventually comes to a stop.
Friction is empirical, not fundamental
The laws of friction (Eqs. 4.13, 4.14, 4.15) are empirical — that is, approximate rules found by experiment, not exact fundamental laws like those of gravitation or electromagnetism. They break down for very small or very large loads, for extremely rough or highly polished surfaces, for very high sliding speeds, and so on. But over a wide range of everyday conditions they work well enough to be extremely useful in engineering and physics calculations.
It is relative motion, not motion, that friction opposes
An important subtlety: friction opposes relative motion between two surfaces in contact — not motion in general.
Consider a wooden box sitting on the floor of a train that suddenly accelerates. If the box is stationary with respect to the train, it must be accelerating in the same direction as the train (relative to the ground). What force accelerates it? The only horizontal force on the box is from its contact with the floor — and if there were no friction, the floor would slide forward under the box, leaving the box behind (by inertia, hitting the rear wall of the compartment). Static friction between box and floor prevents this: it acts forward on the box (in the direction of the train’s acceleration), providing exactly the force needed to accelerate the box along with the train.
So here, static friction is causing the box’s motion (relative to the ground), not opposing it. What it opposes is the relative sliding between box and floor.
Rolling friction — the reason wheels exist
Consider a ring or a sphere rolling without slipping over a horizontal surface. At any instant, only a single point of the wheel is in contact with the surface, and that point is momentarily at rest relative to the surface. Since there is no relative sliding, in principle there should be no kinetic friction. And since the body is neither at rest nor being pushed (in this idealised picture), there is no impending slide either — hence no static friction. In this ideal situation, the body should roll on forever at constant velocity.
In practice, this does not happen. A rolling body eventually slows down and stops, so some resistive force must be present. This force is called rolling friction, and it comes from a very different mechanism than sliding friction.
During real rolling, both the wheel and the surface deform very slightly under the contact load. The contact is no longer a mathematical point but a small area, and the wheel effectively has to “climb out” of the tiny depression it makes ahead of itself at every instant. This costs energy, showing up as a small resistive force. The net result is that some applied force is needed to keep a real wheel rolling at constant velocity.
The critical point is that rolling friction is typically two or three orders of magnitude smaller than sliding friction for the same load. A crate that requires a 30 N push to slide across a floor might need only 0.3 N to roll across on wheels. This is why the invention of the wheel — allowing rolling instead of sliding — has been called one of the greatest technological milestones in human history.
Managing friction — reducing it and using it
Friction is often a nuisance. In an engine or gearbox, sliding friction between metal parts wastes energy as heat and wears the parts down over time. Two standard tricks are used to reduce it.
Figure to come
Fig. 4.15 – Two panels: (a) Ball bearings between a rotating inner shaft and a stationary outer housing; small balls arranged around the shaft roll as the shaft turns. (b) A puck floating on a thin cushion of compressed air over a smooth surface — the working principle of an air hockey table.
But not all friction is bad. In many situations, we need it:
- Walking and running. As we saw in Section 4.6, static friction between shoe and ground is what pushes us forward. On slippery ice, walking becomes almost impossible.
- Braking. When we press the brakes of a car or bicycle, brake pads are pressed against a rotating disc or drum; kinetic friction between them slows the wheel and dissipates the vehicle’s kinetic energy as heat. Without friction, we could not stop.
- Driving. On a dry road, static friction between the tyre and the road pushes the car forward (as we discussed for the accelerating car in Section 4.4). On a very slippery road — wet ice, an oil spill — this friction drops sharply, and neither accelerating nor stopping is possible in the usual way.
So friction is a double-edged tool: undesirable when we want efficient motion, and indispensable when we want to control it.
4.10 Circular Motion
We now use the three laws to understand a very common kind of motion — motion in a circle.
In Chapter 3 we established that a body moving in a circle of radius \(R\) with uniform speed \(v\) has an acceleration of magnitude \(v^2/R\) directed towards the centre of the circle. This is the centripetal acceleration. It exists even when the speed is constant, because the direction of motion keeps changing.
By Newton’s second law, an acceleration requires a force. So a body moving in a circle at uniform speed must experience a net force directed towards the centre, of magnitude
\[f_c = \frac{mv^2}{R} \qquad (4.16)\]
where \(m\) is the body’s mass. This inward force is called the centripetal force.
Where does the centripetal force come from?
It is important to realise that “centripetal force” is not a new kind of force like gravity or tension. It is a label for the net radial force that a body in circular motion needs, whatever its physical origin. In different situations, different physical forces play the centripetal role:
- For a stone whirled at the end of a string, the centripetal force is the tension in the string.
- For a planet orbiting the Sun, it is the gravitational pull of the Sun.
- For a car turning on a flat road, it is the friction between the tyres and the road.
Let us now examine two very common cases in detail: a car turning on a flat road, and a car turning on a banked road. Both are direct applications of Newton’s second law, and both build heavily on what we learned about friction in the previous section.
Motion of a car on a level (flat) road
Consider a car of mass \(m\) moving at speed \(v\) on a horizontal circular track of radius \(R\) (Fig. 4.16(a)). Three forces act on it:
- its weight \(mg\), acting vertically downwards;
- the normal reaction \(N\) from the road, acting vertically upwards;
- the friction \(f\) between the tyres and the road, acting horizontally.
Figure to come
Fig. 4.16 – Two panels: (a) A car viewed from behind, on a horizontal circular track; weight \(mg\) downward, normal \(N\) upward, and friction \(f\) acting horizontally along the road surface toward the centre of the circle. A dashed circle of radius \(R\) shows the path. (b) The same car on a banked road tilted at angle \(\theta\); the normal \(N\) is now perpendicular to the tilted road surface (leaning slightly toward the centre) and friction \(f\) acts along the tilted surface.
Since the car has no acceleration in the vertical direction, the vertical components must balance:
\[N - mg = 0 \quad \Longrightarrow \quad N = mg \qquad (4.17)\]
The centripetal force needed for circular motion is horizontal, pointing toward the centre of the circle. The only horizontal force available is the friction between the tyres and the road — specifically, static friction. (Kinetic friction would only appear if the tyres were sliding across the road, which we assume they are not.) Static friction here opposes the impending slide of the car radially outward, away from the centre.
Setting the friction equal to the required centripetal force,
\[f = \frac{mv^2}{R}\]
But static friction cannot exceed its maximum value \(\mu_s N\):
\[f \le \mu_s N\]
Combining, and using \(N = mg\):
\[\frac{mv^2}{R} \le \mu_s mg \quad \Longrightarrow \quad v^2 \le \mu_s R g\]
Notice the mass has cancelled — the maximum turning speed does not depend on how heavy the car is. This gives the maximum safe speed for taking the turn without slipping:
\[v_{\text{max}} = \sqrt{\mu_s R g} \qquad (4.18)\]
If the actual speed of the car exceeds this, static friction is not enough to provide the required centripetal force, and the car skids outward.
Motion of a car on a banked road
Highways and racetracks that involve high-speed turns are usually banked — the outer edge of the curve is raised so that the road tilts inward at some angle \(\theta\) to the horizontal (Fig. 4.16(b)). The idea is to let some part of the normal force itself contribute to the centripetal force, reducing the reliance on friction. This lets the car take the turn faster without slipping.
On a banked road, the normal reaction \(N\) is perpendicular to the tilted road surface, so it has both a vertical component \(N\cos\theta\) and a horizontal component \(N\sin\theta\) pointing toward the centre of the circle. Friction \(f\) acts along the road surface. At the maximum permissible speed the impending slide is outward and upward along the slope, so friction acts down the slope.
Since there is no vertical acceleration, the net vertical force must be zero:
\[N \cos\theta = mg + f \sin\theta \qquad (4.19a)\]
The horizontal components of \(N\) and \(f\) together provide the centripetal force:
\[N \sin\theta + f \cos\theta = \frac{mv^2}{R} \qquad (4.19b)\]
To find \(v_{\text{max}}\), static friction is at its maximum: \(f = \mu_s N\). Substituting into Eqs. (4.19a) and (4.19b):
\[N \cos\theta = mg + \mu_s N \sin\theta \qquad (4.20a)\]
\[N \sin\theta + \mu_s N \cos\theta = \frac{mv_{\text{max}}^2}{R} \qquad (4.20b)\]
From Eq. (4.20a),
\[N = \frac{mg}{\cos\theta - \mu_s \sin\theta}\]
Substituting into Eq. (4.20b),
\[\frac{mg(\sin\theta + \mu_s \cos\theta)}{\cos\theta - \mu_s \sin\theta} = \frac{mv_{\text{max}}^2}{R}\]
Solving for \(v_{\text{max}}\) (dividing numerator and denominator by \(\cos\theta\) to bring out \(\tan\theta\)):
\[v_{\text{max}} = \left[R g\,\frac{\mu_s + \tan\theta}{1 - \mu_s \tan\theta}\right]^{1/2} \qquad (4.21)\]
Comparing this with Eq. (4.18) for the flat road, the maximum speed on a banked road is greater than on a flat road of the same radius. This is exactly why we bank roads.
The optimum speed on a banked road
There is one special speed at which the banking is just right — friction is not needed at all. Set \(\mu_s = 0\) in Eq. (4.21):
\[v_0 = \sqrt{R g \tan\theta} \qquad (4.22)\]
At this optimum speed \(v_0\), the horizontal component \(N \sin\theta\) of the normal force by itself provides exactly the required centripetal force \(mv_0^2/R\), and no friction is called into play. Driving at \(v_0\) on a banked turn causes essentially no wear on the tyres.
At speeds below \(v_0\), the car tends to slide down the slope, so friction acts up the slope to hold it. At speeds above \(v_0\) (but below \(v_{\text{max}}\)), the car tends to slide up and outward, and friction acts down the slope. A stationary car (\(v = 0\)) on a banked road has the strongest tendency to slide down; for such a car to be able to park at all, one needs \(\tan\theta \le \mu_s\).
4.11 Solving Problems in Mechanics
The three laws of motion you have learnt in this chapter are the foundation of the whole of mechanics. Combined with the principle of conservation of momentum and the common force laws we surveyed in Section 4.9, they give you everything you need to attack a very wide range of problems.
But real-world problems have a complication we have mostly sidestepped so far. In each of Newton’s laws, we talked about a single body — one particle, one book, one car. Most interesting problems in mechanics, however, involve several bodies interacting with each other: a block on a trolley, a load hanging from a pulley system, a set of connected railway carriages, a person standing in an elevator, a car pulling a trailer. How do we apply the three laws to such an assembly?
Choosing your system
The key trick is this: you can pick any part of the assembly you like, and apply Newton’s laws to that part alone — provided you include every force acting on it from everything else.
The part you choose is called the system, and everything else — the rest of the assembly plus all outside agencies (Earth’s gravity, walls, ropes attached from outside, the air) — is called the environment.
You are free to choose the system in whatever way makes the problem simplest. Sometimes it is a single body; sometimes it is a group of bodies treated as a single unit; sometimes the same problem is solved twice, with two different choices of system, to extract two different unknowns.
The five-step procedure
Here is a systematic method that works for essentially any mechanics problem:
(i) Draw a schematic of the whole assembly. Show the different bodies, the connecting links (strings, rods, pulleys), the supports (floors, tables, hinges), and the given quantities (masses, applied forces, angles, coefficients of friction).
(ii) Choose one part of the assembly as your system. Often the object whose motion is asked about, or an object connected by a single string to it.
(iii) Draw a separate diagram of just that system, with only the forces acting on it. Include forces from the environment (like gravity), from bodies in contact with the system (the tension of a rope attached to it, the normal reaction from a surface), and any applied forces given in the problem. Do not include forces exerted by the system on other bodies — those are third-law pairs of the environment’s forces on the system, and by our choice of “system,” they are not forces on the system itself. Such a diagram is called a free-body diagram (or FBD). The name does not imply zero net force — a “free” body may well be accelerating.
(iv) Label every force with what you know. For each force in the FBD, mark its direction (which is often given — for example, gravity always points down, and the tension in an inextensible string acts along its length), and its magnitude if known. Any unknowns — an unknown tension, an unknown friction, an unknown normal reaction — are the quantities you will solve for using the laws of motion.
(v) Apply Newton’s second law to the system. Write \(\vec{F}_{\text{net}} = m\vec{a}\) for each direction (usually two: horizontal and vertical, or along-slope and perpendicular-to-slope). If you have more unknowns than equations, repeat steps (ii)–(v) with a different choice of system. When the same interaction reappears from both bodies’ viewpoints, remember Newton’s third law: if in one FBD you label the force on body \(A\) by body \(B\) as \(\vec{F}\), then in the FBD of body \(B\) you must label the force by \(A\) as \(-\vec{F}\).
We have already used this method silently in most of the solved examples in this chapter — for the book on the table (Section 4.4), the suspended mass (Example 4.6 in Section 4.8), the block on the incline (Example 4.8 in Section 4.9), and the block–trolley system (Example 4.9). Free-body diagrams are the single most important habit for physics problem-solving.
A worked example combining all the ideas
Identifying action–reaction pairs
For case (a):
- The force of gravity (\(20 \text{ N}\) downward) on the block by the Earth is one member of an action–reaction pair. Its partner is the gravitational pull on the Earth by the block (\(20 \text{ N}\) upward on the Earth), which is not drawn in the figure.
- The force on the floor by the block (\(20 \text{ N}\) downward) is one member of another action–reaction pair. Its partner is the normal force on the block by the floor (\(R = 20 \text{ N}\) upward).
For case (b):
- The gravitational pull (\(270 \text{ N}\) down) on the system by the Earth pairs with the gravitational pull (\(270 \text{ N}\) up) on the Earth by the system.
- The force on the floor by the system (\(267.3 \text{ N}\) down) pairs with the force on the system by the floor (\(R' = 267.3 \text{ N}\) up).
- In addition, the force on the block by the cylinder pairs with the force on the cylinder by the block. These are the “internal forces” we did not have to compute — but they exist and form a valid third-law pair.
A common trap: two forces on the same body ≠ action–reaction pair
Notice something crucial about case (a). The block’s weight (\(20 \text{ N}\) down) and the normal reaction from the floor (\(R = 20 \text{ N}\) up) are equal in magnitude and opposite in direction. It is very tempting to call these an action–reaction pair. They are not.
An action–reaction pair consists of mutual forces between two different bodies. Weight and normal reaction both act on the same body (the block) — so they cannot form a third-law pair, no matter how equal and opposite they happen to be. The equality here comes from the first law (the block is in equilibrium), not from the third law.
The trap is easier to see in case (b), where the two forces are not equal any more: the weight is \(270 \text{ N}\) but \(R'\) is only \(267.3 \text{ N}\). If they had been a third-law pair, they would still have been equal and opposite. Their inequality here is a direct signal that they never formed a third-law pair to begin with.
The correct third-law partner of the weight (\(270 \text{ N}\) down on the system by the Earth) is the pull on the Earth by the system (\(270 \text{ N}\) up on the Earth). The correct third-law partner of \(R'\) (\(267.3 \text{ N}\) up on the system by the floor) is the force on the floor by the system (\(267.3 \text{ N}\) down on the floor).
Why free-body diagrams matter
Every mechanics problem you will encounter in this chapter’s exercises, in the next chapter’s exercises, and in your board and entrance examinations boils down to the same core skill: pick a system, draw its free-body diagram, and apply Newton’s second law. The forces get more varied as the physics gets richer — springs, tensions, frictions on inclined planes, buoyant forces, and later on electric and magnetic forces — but the method never changes. Cultivate the habit now, and mechanics stops being a memory task and becomes a matter of clear drawing.
4.12 Summary
1. Aristotle was wrong about force and motion. The view held for two thousand years — that a force is necessary to keep a body in uniform motion — is not correct. In practice, we do need an applied force, but only to counter the ever-present opposing force of friction. Take friction away, and no applied force is needed to maintain uniform motion.
2. The law of inertia (Galileo → Newton’s First Law). Galileo extrapolated observations of bodies on inclined planes to conclude that motion at constant velocity requires no net force. Newton restated this as the First Law of Motion:
Every body continues to be in its state of rest or of uniform motion in a straight line, unless compelled by some external force to act otherwise.
Equivalently: if the net external force on a body is zero, its acceleration is zero.
3. Momentum. The momentum \(\vec{p}\) of a body is the product of its mass \(m\) and its velocity \(\vec{v}\):
\[\vec{p} = m\vec{v}\]
Momentum is a vector. Its SI unit is kg m s⁻¹ (or N s).
4. Newton’s Second Law of Motion. The rate of change of momentum of a body is proportional to the applied external force and takes place in the direction in which the force acts:
\[\vec{F} = k\,\frac{\mathrm{d}\vec{p}}{\mathrm{d}t} = k\,m\,\vec{a}\]
where \(\vec{F}\) is the net external force and \(\vec{a}\) is the acceleration. Choosing \(k = 1\) in SI units,
\[\vec{F} = \frac{\mathrm{d}\vec{p}}{\mathrm{d}t} = m\vec{a}\]
The SI unit of force is the newton: \(1 \text{ N} = 1 \text{ kg m s}^{-2}\).
Key features of the Second Law:
It is consistent with the First Law: \(\vec{F} = 0\) implies \(\vec{a} = 0\).
It is a vector equation (splitting into three scalar component equations).
It applies to a particle and, more generally, to any body or system of particles, provided \(\vec{F}\) is the total external force on the system and \(\vec{a}\) is the acceleration of the system’s centre of mass.
It is a local law: \(\vec{F}\) at a point at a certain instant determines \(\vec{a}\) at that same point at that instant. Acceleration does not depend on the past history of motion.
5. Impulse. Impulse is the product of a force and the time interval for which it acts, and it equals the change in momentum:
\[\text{Impulse} = \vec{F} \times \Delta t = \Delta \vec{p}\]
The idea of impulse is especially useful when a large force acts for a very short time (an impact or a collision) — during this short interval the position of the body barely changes, but its momentum changes measurably.
6. Newton’s Third Law of Motion. To every action, there is always an equal and opposite reaction. Restated in modern language:
Forces in nature always occur between pairs of bodies. The force on body A by body B is equal in magnitude and opposite in direction to the force on body B by A.
Action and reaction are simultaneous — there is no cause-effect ordering. They act on two different bodies, and therefore cannot cancel each other on any single body. The internal action–reaction forces between different parts of an extended body, however, sum to zero.
7. Law of Conservation of Momentum. The total momentum of an isolated system of particles (net external force zero) is conserved. This law follows directly from Newton’s Second and Third Laws.
8. Friction. The frictional force opposes (impending or actual) relative motion between two surfaces in contact. It is the component of the contact force along the common tangent to the surfaces.
- Static friction \(f_s\) opposes impending relative motion, and is self-adjusting up to a limit:
\[f_s \le (f_s)_{\text{max}} = \mu_s N\]
- Kinetic friction \(f_k\) opposes actual relative motion:
\[f_k = \mu_k N\]
Here \(N\) is the normal reaction between the two surfaces, \(\mu_s\) is the coefficient of static friction, and \(\mu_k\) is the coefficient of kinetic friction. Both coefficients are constants characteristic of the pair of surfaces in contact. Both frictional forces are (approximately) independent of the area of contact. Experimentally, \(\mu_k < \mu_s\) for the same pair of surfaces.
4.13 Physical Quantities in this Chapter
| Quantity | Symbol | SI Unit | Dimensions | Remarks |
|---|---|---|---|---|
| Momentum | \(\vec{p}\) | kg m s⁻¹ (or N s) | \([MLT^{-1}]\) | Vector; \(\vec{p} = m\vec{v}\) |
| Force | \(\vec{F}\) | N | \([MLT^{-2}]\) | Vector; \(\vec{F} = m\vec{a}\) (Second Law) |
| Impulse | \(\vec{J}\) | kg m s⁻¹ (or N s) | \([MLT^{-1}]\) | Impulse = force × time = change in momentum |
| Static friction | \(f_s\) | N | \([MLT^{-2}]\) | \(f_s \le \mu_s N\) |
| Kinetic friction | \(f_k\) | N | \([MLT^{-2}]\) | \(f_k = \mu_k N\) |
| Coefficient of friction | \(\mu_s, \mu_k\) | dimensionless | — | \(\mu_k < \mu_s\) |
| Spring constant | \(k\) | N m⁻¹ | \([MT^{-2}]\) | \(F = -kx\) (Hooke’s law) |
| Centripetal force | \(f_c\) | N | \([MLT^{-2}]\) | \(f_c = mv^2/R\); directed towards centre |
4.14 Points to Ponder
1. Force is not always in the direction of motion. Depending on the situation, \(\vec{F}\) may be along \(\vec{v}\), opposite to \(\vec{v}\), normal to \(\vec{v}\), or make some other angle with \(\vec{v}\). But in every case, force is parallel to acceleration — not to velocity. Confusing velocity and acceleration directions is a common exam error.
2. Zero velocity does not mean zero force. If \(\vec{v} = 0\) at an instant — that is, if a body is momentarily at rest — it does not follow that the force or acceleration on it is zero at that instant. For example, when a ball thrown upward reaches its maximum height, \(\vec{v} = 0\), but the force on it is still its weight \(mg\), and its acceleration is still \(g\) (downward).
3. Force has no memory. The force on a body at a given time is determined only by the situation at the location of the body at that time. Force is not “carried” by the body from its earlier history of motion. The instant a stone is released from an accelerating train, no horizontal force (or acceleration) acts on it, even though the train is still accelerating — the stone remembers no push. The only force on it is gravity, straight down (air resistance neglected).
4. \(m\vec{a}\) is not a force. In Newton’s second law, \(\vec{F} = m\vec{a}\), the left side \(\vec{F}\) is the net force due to all material agencies external to the body, and \(\vec{a}\) is the effect of that force. The product \(m\vec{a}\) should not be regarded as yet another separate force — it is simply what happens when the net force acts on a body of mass \(m\).
5. Centripetal force is not a new kind of force. It is simply the name given to the net radial force that provides inward acceleration to a body in circular motion. We should always look for some material force — tension, gravitational pull, friction, an electrical force — as the centripetal force. If no such force is present, no circular motion is possible.
6. Static friction is self-adjusting. Static friction takes whatever value is needed to prevent relative motion, up to a maximum of \(\mu_s N\). Do not automatically write \(f_s = \mu_s N\) in a problem without first checking that the body is on the verge of slipping — otherwise you may drastically overestimate the friction force that is actually acting.
7. Normal reaction equals weight only in equilibrium. The familiar \(R = mg\) for a body on a table is true only if the body is in equilibrium. In an accelerating lift, for example, \(R \ne mg\). And the equality (when it holds) has nothing to do with the third law — it is a consequence of the first law.
8. Action and reaction are simultaneous and act on different bodies. In Newton’s third law, “action” and “reaction” simply stand for mutual forces between a pair of bodies. Unlike the everyday meaning of the words, action does not precede or cause reaction — they arise together and act on two different bodies. This is why they cannot cancel each other.
9. There is really only one thing called “force.” The many words we use in mechanics — friction, normal reaction, tension, air resistance, viscous drag, thrust, buoyancy, weight, centripetal force — all stand for “force” in different contexts. For clarity in problem-solving, every force should be reducible to the phrase “force on A by B” (naming the two bodies involved).
10. Living bodies obey Newton’s laws too. In applying the second law, there is no conceptual distinction between inanimate and animate objects. A human is also subject to \(\vec{F} = m\vec{a}\): to accelerate, you need an external force. Without the external friction of the ground on your shoes, you could not walk.
11. The physical concept of force is not the same as the feeling of force. On a merry-go-round, every part of your body is being pulled inward (toward the axis) by real centripetal forces. But you feel pushed outward — because that is the direction your body would fly off in if the ropes let go. Physics answers the question “what force acts?”; sensation answers a different question — “what motion is impending?” Do not confuse them.
4.15 NCERT Questions
(For simplicity in numerical calculations, take \(g = 10 \text{ m s}^{-2}\) throughout.)
- Give the magnitude and direction of the net force acting on
- a drop of rain falling down with a constant speed,
- a cork of mass 10 g floating on water,
- a kite skilfully held stationary in the sky,
- a car moving with a constant velocity of 30 km/h on a rough road,
- a high-speed electron in space far from all material objects, and free of electric and magnetic fields.
- A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,
- during its upward motion,
- during its downward motion,
- at the highest point where it is momentarily at rest.
Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction? Ignore air resistance.
- Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg,
- just after it is dropped from the window of a stationary train,
- just after it is dropped from the window of a train running at a constant velocity of 36 km/h,
- just after it is dropped from the window of a train accelerating with \(1 \text{ m s}^{-2}\),
- lying on the floor of a train which is accelerating with \(1 \text{ m s}^{-2}\), the stone being at rest relative to the train.
Neglect air resistance throughout.
- One end of a string of length \(l\) is connected to a particle of mass \(m\) and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed \(v\), the net force on the particle (directed towards the centre) is:
- \(T\), (ii) \(T - \dfrac{mv^2}{l}\), (iii) \(T + \dfrac{mv^2}{l}\), (iv) \(0\)
Here \(T\) is the tension in the string. [Choose the correct alternative.]
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of \(15 \text{ m s}^{-1}\). How long does the body take to stop?
A constant force acting on a body of mass 3.0 kg changes its speed from \(2.0 \text{ m s}^{-1}\) to \(3.5 \text{ m s}^{-1}\) in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?
A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.
The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg.
A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of \(5.0 \text{ m s}^{-2}\). Calculate the initial thrust (force) of the blast.
A body of mass 0.40 kg moving initially with a constant speed of \(10 \text{ m s}^{-1}\) to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be \(t = 0\), the position of the body at that time to be \(x = 0\), and predict its position at \(t = -5 \text{ s}\), \(25 \text{ s}\), \(100 \text{ s}\).
A truck starts from rest and accelerates uniformly at \(2.0 \text{ m s}^{-2}\). At \(t = 10 \text{ s}\), a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at \(t = 11 \text{ s}\)? (Neglect air resistance.)
A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is \(1 \text{ m s}^{-1}\). What is the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean position?
A man of mass 70 kg stands on a weighing scale in a lift which is moving
- upwards with a uniform speed of \(10 \text{ m s}^{-1}\),
- downwards with a uniform acceleration of \(5 \text{ m s}^{-2}\),
- upwards with a uniform acceleration of \(5 \text{ m s}^{-2}\).
What would be the readings on the scale in each case?
d. What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?- The figure below (Fig. 4.18) shows the position-time graph of a particle of mass 4 kg. What is the (a) force on the particle for \(t < 0\), \(t > 4 \text{ s}\), \(0 < t < 4 \text{ s}\)? (b) impulse at \(t = 0\) and \(t = 4 \text{ s}\)? (Consider one-dimensional motion only.)
Figure to come
Fig. 4.18 – A position-time (\(x\) vs \(t\)) graph. The particle is at \(x = 0\) for all \(t < 0\); between \(t = 0\) and \(t = 4 \text{ s}\), the position increases linearly from 0 to 3 m (a straight line ending at point A); for \(t > 4 \text{ s}\), the position remains constant at \(x = 3 \text{ m}\).
Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force \(F = 600 \text{ N}\) is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case?
Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.
A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.
Two billiard balls each of mass 0.05 kg moving in opposite directions with speed \(6 \text{ m s}^{-1}\) collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?
A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is \(80 \text{ m s}^{-1}\), what is the recoil speed of the gun?
A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)
A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev/min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?
If, in Exercise A-Q21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks:
- the stone moves radially outwards,
- the stone flies off tangentially from the instant the string breaks,
- the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle?
Explain why
- a horse cannot pull a cart and run in empty space,
- passengers are thrown forward from their seats when a speeding bus stops suddenly,
- it is easier to pull a lawn mower than to push it,
- a cricketer moves his hands backwards while holding a catch.
4.16 Check Your Concepts
A student says: “Newton’s first law is unnecessary because it is contained in the second law — just set \(\vec{F} = 0\) in \(\vec{F} = m\vec{a}\) and you get \(\vec{a} = 0\).” Explain why, historically and conceptually, Newton chose to state the first law separately, even though it is mathematically a special case of the second.
Two students argue about a book at rest on a table.
- Student A says: “The book stays at rest because the normal reaction \(R\) from the table and the weight \(W\) of the book are equal and opposite — they are an action–reaction pair.”
- Student B says: “\(R\) and \(W\) are equal and opposite because the book is in equilibrium, but they are not an action–reaction pair.”
Who is correct, and why? Identify the correct action–reaction partners of \(R\) and \(W\) separately.
A block of mass 5 kg lies on a horizontal surface with \(\mu_s = 0.4\). A horizontal force of 10 N is applied. Without doing any full calculation, state (a) whether the block moves, and (b) the magnitude of the frictional force acting on it. Justify your answer in one or two sentences using the correct law of static friction.
Rockets can accelerate through the vacuum of outer space where there is no air. A friend argues that this is impossible, since “there is nothing for the rocket to push against.” Using Newton’s third law and the principle of conservation of momentum, explain what the rocket actually pushes against, and why the vacuum is no obstacle.
“Centripetal force is not a new kind of force.” Explain this statement, giving three different physical situations in which the centripetal force is provided by three different kinds of real forces. In each case, identify which physical force plays the centripetal role.
A car travels around a horizontal circular track at a constant speed. Its passenger claims to feel “thrown outward” and blames a mysterious “centrifugal force.” From the point of view of an observer standing on the ground, explain what force is actually acting on the passenger and why the passenger feels pushed outward.
A book of mass \(m\) is at rest on the floor of a lift.
- Draw the free-body diagram of the book when the lift is (i) stationary, (ii) moving up with constant velocity, (iii) moving up with acceleration \(a\), and (iv) in free fall (cable snapped).
- In which of the four cases is the normal reaction on the book equal to \(mg\)? In which case is it zero?
- Explain, using the concept of impulse (\(\vec{F}\,\Delta t = \Delta \vec{p}\)), why:
- modern cars are designed with “crumple zones” that deliberately crush during a crash,
- high jumpers land on thick foam mattresses rather than bare ground,
- a person jumping from height instinctively bends the knees on landing.
Which physical quantity is the same in each case, and which is being controlled by the design or the action?
A ball is whirled in a vertical circle by a string. State how the tension in the string changes as the ball moves from the lowest point to the highest point of the circle. At which position is the string most likely to break, and why? (Assume the ball moves at constant speed for this qualitative reasoning.)
A body slides down a rough inclined plane at constant velocity. Draw its free-body diagram, and use it to write the relationship between the angle of the incline \(\theta\) and the coefficient of kinetic friction \(\mu_k\) between the body and the plane. What does your result say about the angle of the incline in this special situation?
4.17 Practice with Numericals
A force of \(6 \text{ N}\) acts on a body of mass 2 kg initially at rest. Find (a) the acceleration produced, (b) the velocity of the body after 4 s, and (c) the distance covered by the body in these 4 s.
A cricket ball of mass 160 g moving horizontally at \(20 \text{ m s}^{-1}\) is struck by a bat and returns along the same line with a speed of \(30 \text{ m s}^{-1}\). If the contact between bat and ball lasts for \(0.01 \text{ s}\), find (a) the impulse imparted to the ball, and (b) the average force exerted by the bat on the ball.
A person of mass 50 kg is standing on a frictionless frozen lake. She holds a bag of mass 5 kg. She throws the bag horizontally in one direction with a speed of \(4 \text{ m s}^{-1}\) (measured relative to the ground). Using conservation of momentum, find her recoil velocity.
A block of mass 10 kg is suspended in equilibrium by two ropes attached to the ceiling. One rope makes an angle of 30° with the vertical and the other makes an angle of 60° with the vertical. Find the tensions in the two ropes.
A block of mass 4 kg is placed on a rough inclined plane of inclination 30°. The coefficient of kinetic friction between the block and the plane is 0.3. Find (a) the frictional force acting on the block, and (b) the acceleration with which the block slides down the plane.
Two blocks of masses 3 kg and 5 kg are placed in contact with each other on a smooth horizontal surface. A horizontal force of 40 N is applied to the 3 kg block, pushing both blocks together. Find (a) the acceleration of the two-block system, and (b) the contact force between the two blocks.
A stone of mass 500 g tied to the end of a string is whirled in a horizontal circle of radius 1 m at a constant angular speed of 4 revolutions per second. Find (a) the linear speed of the stone, (b) the centripetal acceleration, and (c) the tension in the string.
A car of mass 800 kg is moving at \(20 \text{ m s}^{-1}\) on a circular horizontal track of radius 40 m. Find (a) the centripetal force required, and (b) the minimum coefficient of static friction between the tyres and the road for the car not to skid.
A circular road of radius 50 m is banked at an angle of 30° to the horizontal. Assuming friction is negligible, find the optimum speed at which a car can safely negotiate the curve.
A block of mass 2 kg rests on a horizontal surface. The coefficient of static friction between the block and the surface is 0.25 and the coefficient of kinetic friction is 0.2. A horizontal force is applied to the block that increases uniformly from 0 to 8 N over 10 s. Find (a) the applied force at the instant the block just begins to move, (b) the acceleration of the block at the instant when the applied force reaches its maximum value of 8 N.
Two blocks of masses 5 kg and 3 kg are connected by a light inextensible string that passes over a smooth pulley fixed at the edge of a table. The 5 kg block lies on the horizontal surface of the table, and the 3 kg block hangs vertically off the edge. The coefficient of kinetic friction between the 5 kg block and the table is 0.2. When the system is released, find (a) the acceleration of the blocks, and (b) the tension in the string.
A bullet of mass 50 g moving at \(500 \text{ m s}^{-1}\) strikes a stationary wooden block of mass 4.95 kg and gets embedded in it. Using conservation of momentum, find the velocity of the (bullet + block) system immediately after the collision. Is this collision elastic or inelastic? Give a one-line justification.