Chapter 1: Force
In the year 1173, builders in the Italian city of Pisa began work on a bell tower beside the cathedral. By the time the third storey was finished, something had gone wrong. The ground beneath one side of the tower was soft, and that side had sunk. The tower had begun to lean.
Work stopped for almost a hundred years. When later builders returned, they tried to correct the tilt by making the upper storeys slightly taller on the lower side, which is why the tower is faintly curved rather than straight. It was finished in the fourteenth century, and it kept leaning, a little more with every passing decade.
By 1990 the lean had grown worrying enough for the tower to be closed to visitors. Engineers did not try to push the tower upright. Instead they worked quietly on the ground, removing small amounts of soil from beneath the higher side so that the tower settled gently back towards the vertical by a fraction of a degree. The tower reopened in 2001, still leaning by about four degrees, and still standing.
The question the engineers had to answer was not how heavy the tower was. It was where its weight acted, and how far that line of action lay from the edge it might turn about. That single idea — a force acting at a distance from a point — runs through this whole chapter.
Diagram
Fig. 1.0-1 — The Leaning Tower And Its Line Of Weight Simple side view of a tall cylindrical tower tilted to one side on level ground. Draw a dashed vertical line from a point marked G near the middle of the tower straight down to the ground; label G as ‘centre of gravity’ and the dashed line as ‘line along which the weight acts’. Mark the base of the tower with a short horizontal line and label its two edges as ‘edge the tower could turn about’. Show with an arrow that the vertical line still meets the ground inside the base.
1.1 Translational And Rotational Motions
In Class IX you learnt that a force is a push or a pull that can change the state of rest or of motion of a body. What a force actually does to a body depends on the kind of body it acts on.
A rigid body is one whose shape and size do not change when a force acts on it — a steel rod, a wooden plank, a metal disc. A force applied to a rigid body can only set it moving. On a non-rigid body, such as a rubber ball or a lump of dough, the same force changes the shape or the size of the body as well as moving it.
Force can also be described in a quantitative way. When a force acts on a body, it changes the body’s linear momentum, and the faster that momentum changes, the greater the force. In other words, force is the rate of change of linear momentum of a body. For a body whose mass stays constant, this statement takes the familiar form
\[F = ma\]
where \(F\) is the force in newton (N), \(m\) is the mass of the body in kilogram (kg), and \(a\) is the acceleration produced in metre per second squared (m s⁻²). The relation in this form holds only while the mass of the body remains constant.
Force is a vector quantity, so a force is described completely only when both its size and its direction are stated. Its SI unit is the newton (symbol N). Force is sometimes also measured in a gravitational unit called the kilogram-force (symbol kgf), where
\[1\\ \\text{kgf} = g\\ \\text{N} = 9.8\\ \\text{N}\]
taking \(g = 9.8\\ \\text{m s}^{-2}\), the average value of the acceleration due to gravity at the earth’s surface.
Two Ways A Body Can Move
A rigid body acted upon by a force can have two kinds of motion:
- linear or translational motion, and
- rotational motion.
Which of the two occurs depends on whether the body is free to move, or is held fixed at some point.
1.1.1 Linear Or Translational Motion
When a force acts on a stationary rigid body that is free to move, the body begins to move in a straight path along the direction of the force. This is called linear motion or translational motion. Every particle of the body moves through the same distance in the same direction.
A ball lying on the floor behaves in exactly this way. Push it, as in Fig. 1.1-1, and it moves off along the direction of your push.
Diagram
Fig. 1.1-1 — A Ball Set Into Translational Motion Side view of a ball resting on a horizontal floor. Draw a horizontal arrow labelled F pushing the ball from the left, and a second horizontal arrow to the right of the ball labelled ‘direction of motion’. Label the ball, and label the floor line.
1.1.2 Rotational Motion
Now hold the body down at one point so that it cannot move away — that is, pivot it. The point at which the body is held is called the pivot, and the fixed line about which the body then turns is called the axis of rotation.
If a force is now applied at a suitable point of the pivoted body, the body cannot travel along the direction of the force. Instead it turns about the axis passing through the pivot. This turning is the turning effect of the force, and the motion is called rotational motion.
A wheel pivoted at its centre shows this clearly. Apply a force along the rim, as in Fig. 1.1-2, and the wheel spins about the axis through its centre while the centre itself stays where it is.
Diagram
Fig. 1.1-2 — A Wheel In Rotational Motion Front view of a circular wheel with a dot at its centre labelled ‘pivot’. Draw a short vertical arrow labelled F touching the rim at the left, directed upwards along the rim. Draw a curved arrow along the rim showing the wheel turning. Label the wheel.
A door behaves in the same way. Its hinges hold one edge fixed, so a force applied normally on the handle cannot carry the door across the room. The door turns about the vertical axis passing through the hinges.
The two kinds of motion are set side by side in Table 1.1.
1.2 Moment Of A Force Or Torque
Consider a body pivoted at a point O. A force \(F\) is applied horizontally on the body, its line of action lying along the direction AP as shown in Fig. 1.2-1. The line of action of a force is simply the straight line drawn through the point where the force acts, along the direction in which the force acts.
Because the body is pivoted, it is not free to move away, so this force cannot produce linear motion along its own direction. What the force does instead is turn the body about the vertical axis passing through O. In Fig. 1.2-1 the force turns the body anticlockwise.
Diagram
Fig. 1.2-1 — The Turning Effect Of A Force About A Pivot An irregular flat body seen from above, with a dot near its left edge labelled O and marked ‘pivot’. A point A lies on the body towards the lower right; an arrow at A labelled F points downwards to the right, and its line of action is drawn as a dotted straight line through A extended to a point P above and right of O. From O draw a dotted perpendicular OP meeting this line of action at P, with a right-angle symbol marked at P and the segment labelled ‘perpendicular distance OP’. Draw a curved arrow showing the body turning anticlockwise about O.
1.2.1 Factors Affecting The Turning Of A Body
You already know from experience that a heavier push turns a door faster. You also know that a push near the hinges is almost useless, while the same push at the outer edge swings the door easily. So the turning effect depends on two things:
- the magnitude of the force applied, and
- the perpendicular distance of the line of action of the force from the axis of rotation (or pivoted point).
The turning effect depends on the product of these two factors, not on either one alone. Doubling the force doubles the turning effect, and so does doubling the perpendicular distance. This product is called the moment of force, or the torque. A pivoted body rotates because of the moment of force acting on it about the pivot.
In Fig. 1.2-1, the perpendicular drawn from the pivoted point O to the line of action AP is OP. Therefore the moment of the force about the axis passing through O is
\[\\text{moment of force} = F \\times \\text{OP} \\qquad \\dots (1.1)\]
where \(F\) is the magnitude of the applied force in newton (N), and OP is the perpendicular distance of the line of action of the force from the axis of rotation, in metre (m). The distance to be used is always the perpendicular distance, never the slanting distance from the pivot to the point where the force acts.
This relation also tells you how to get the strongest turn out of a given force. Since the force is fixed, the moment is largest when the perpendicular distance is largest. So the force should be applied at the point for which the perpendicular distance of its line of action from the axis of rotation is the greatest. In that position, the given force provides the maximum torque to turn the body.
1.2.2 Units Of Moment Of Force
The unit of the moment of a force follows directly from the relation that defines it:
\[\\text{unit of moment of force} = \\text{unit of force} \\times \\text{unit of distance}\]
In the SI system, force is measured in newton and distance in metre, so the SI unit of moment of force is newton metre, abbreviated as N m. In the C.G.S. system, force is measured in dyne and distance in centimetre, so the C.G.S. unit is dyne centimetre (dyne × cm).
If force is measured in gravitational units instead, the unit of moment of force in the SI system is kgf × m, and in the C.G.S. system it is gf × cm. These four units are related as follows:
\[1\\ \\text{N m} = 10^{5}\\ \\text{dyne} \\times 10^{2}\\ \\text{cm} = 10^{7}\\ \\text{dyne cm}\] \[1\\ \\text{kgf} \\times \\text{m} = 9.8\\ \\text{N m}\] \[1\\ \\text{gf} \\times \\text{cm} = 980\\ \\text{dyne cm} \\qquad \\dots (1.2)\]
The first of these follows from \(1\\ \\text{N} = 10^{5}\\ \\text{dyne}\) and \(1\\ \\text{m} = 10^{2}\\ \\text{cm}\); the other two follow from \(1\\ \\text{kgf} = 9.8\\ \\text{N}\) and \(1\\ \\text{gf} = 980\\ \\text{dyne}\).
One point of care: the unit N m of moment of force is never written as joule (J). The unit N m for work or energy is written as joule, because work and energy are scalar quantities, while torque is a vector quantity. Same-looking units, different physical quantities.
1.2.3 Clockwise And Anticlockwise Moments
A turning effect can act in one of two senses, and the two must be told apart. Conventionally, if the effect on the body is to turn it anticlockwise, the moment of force is called an anticlockwise moment and it is taken positive. If the effect on the body is to turn it clockwise, the moment of force is called a clockwise moment and it is taken negative.
The moment of force is a vector quantity, so it has a direction of its own. The direction of an anticlockwise moment is along the axis of rotation outwards, that is, towards the observer. The direction of a clockwise moment is along the axis of rotation inwards, that is, away from the observer.
1.2.4 Changing The Direction Of Rotation
When a force is applied on a pivoted body, the sense in which the body turns depends not only on the direction of the force, but also on the point at which the force is applied. So the direction of rotation of a body can be changed in two ways.
The first way is by changing the point of application of the force. Fig. 1.2-2 shows a disc pivoted at its centre. The same force \(F\), in the same direction, produces an anticlockwise moment when applied at point A and a clockwise moment when applied at point B on the opposite side of the pivot.
Diagram
Fig. 1.2-2 — Changing The Point Of Application Of The Force Two circular discs drawn side by side, each with a dot at the centre labelled ‘pivot’. In the left disc, mark point A on the lower rim and draw an arrow labelled F at A pointing horizontally to the right; draw a curved arrow showing anticlockwise turning and label the disc ‘anticlockwise (positive)’. In the right disc, mark point B on the upper rim and draw an identical arrow labelled F at B pointing horizontally to the right; draw a curved arrow showing clockwise turning and label the disc ‘clockwise (negative)’. Label both discs ‘disc’.
The second way is by changing the direction of the force. Fig. 1.2-3 shows a pivoted axle with a force applied at its free end. Reversing the direction of that force at the same point turns the anticlockwise moment into a clockwise one.
Diagram
Fig. 1.2-3 — Changing The Direction Of The Force Two identical sketches side by side, each showing a short rod fixed at one end to a pivot labelled ‘axle’ with a dot at the pivot. In the left sketch draw an arrow labelled F at the free end of the rod, pointing perpendicular to the rod and upwards; add a curved arrow showing anticlockwise turning and label it ‘anticlockwise (positive)’. In the right sketch draw an arrow labelled F at the same free end, perpendicular to the rod but pointing downwards; add a curved arrow showing clockwise turning and label it ‘clockwise (negative)’.
1.2.5 Common Examples Of Moment Of Force
The everyday world is full of arrangements built around this one relation. In each of the examples below, look for the same idea: the force is deliberately applied as far as possible from the axis of rotation.
Opening A Door
To open or shut a door, we apply a force (a push or a pull) normal to the door at its handle P, which is fixed at the maximum distance from the hinges, as shown in Fig. 1.2-4.
You must have noticed that applying the force at a point Q near the hinge R requires a much greater force to open the door. If the force is applied at the hinge R itself, the door will not open however large the force is, because for a force at R the torque is zero. This is why the handle P is provided near the free end of the door: a smaller force acting at a larger perpendicular distance from the hinges produces the large moment of force needed to swing the door.
Diagram
Fig. 1.2-4 — Opening Of A Door A door seen in perspective within its frame. Label the frame ‘frame (fixed in wall)’. Mark the two hinges on the right-hand vertical edge and label them ‘hinge’; mark the upper hinge region with the letter R. Mark the handle on the left-hand vertical edge with the letter P and label it ‘handle’. Mark a point Q on the door close to the hinge edge. Draw an arrow labelled F at P, normal to the door surface, and a second shorter arrow labelled F at Q, also normal to the door.
Handles, Wheels And Pedals
The upper movable circular stone of a hand flour grinder carries a handle near its rim, that is, at the maximum distance from the centre. A small force applied at that handle rotates the heavy stone about the iron pivot at its centre.
For turning a steering wheel, a force is applied tangentially on the rim of the wheel, as in Fig. 1.2-5. The sense of rotation of the wheel can be changed either by reversing the direction of the force, or by changing the point of application of the force without changing its direction. In Fig. 1.2-5(a), the force \(F\) applied at point A turns the wheel anticlockwise; in Fig. 1.2-5(b), the same force \(F\) in the same direction applied at point B turns it clockwise.
Diagram
Fig. 1.2-5 — Sense Of Rotation Of A Steering Wheel Two steering wheels drawn side by side, each a circular rim with three spokes and a hub. In (a), mark point A on the lower left of the rim and draw an arrow labelled F at A tangential to the rim; add curved arrows showing anticlockwise rotation and caption the sketch ‘anticlockwise rotation’. In (b), mark point B on the upper right of the rim and draw an arrow labelled F at B tangential to the rim in the same direction as in (a); add curved arrows showing clockwise rotation and caption it ‘clockwise rotation’.
In a bicycle, a small force applied on the foot pedal turns the rear wheel. The front toothed wheel carrying the pedal is bigger than the rear wheel, so the perpendicular distance of the point of application of the force from the axle is large, as shown in Fig. 1.2-6. The front toothed wheel is joined to the rear wheel by a chain passing over their teeth, and the chain pulls and drives the rear wheel when the pedal is pushed down.
Diagram
Fig. 1.2-6 — Turning The Toothed Wheel Of A Bicycle Side view of a bicycle chain drive. Draw a large toothed wheel on the left with a dot at its centre labelled ‘axle’, and a small toothed wheel on the right with a dot at its centre labelled ‘axle’; label the small wheel ‘rear wheel’. Join the two by a chain running over the teeth. Draw the crank arm on the large wheel with the foot pedal at its end, labelled ‘foot pedal’, and a downward arrow labelled F on the pedal.
Spanners And Screw Jacks
A spanner used to tighten or loosen a nut has a long handle, so that a small force applied normally at the end of the handle produces a large moment of force, as shown in Fig. 1.2-7. The spanner is turned anticlockwise to loosen the nut by applying the force in the direction shown, and clockwise to tighten the nut by applying the force in the opposite direction.
Diagram
Fig. 1.2-7 — A Spanner Used To Loosen A Nut Top view of a spanner gripping a hexagonal nut at its left end, with a long straight handle extending to the right. Label the nut ‘nut’ and the handle ‘handle’. Draw an arrow labelled ‘force’ at the far right end of the handle, perpendicular to the handle. Add a curved arrow near the nut showing anticlockwise turning.
A jack screw used to lift a heavy load such as a vehicle is provided with a long arm, so that less effort is needed to rotate it and raise or lower the load table.
Conclusion
From all these examples we conclude that the turning of a body about an axis depends not only on the magnitude of the force, but also on the perpendicular distance of the line of action of the applied force from the axis of rotation. The larger the perpendicular distance, the less is the force needed to produce the same turning effect, and the other way round.
1.3 Couple
A single force acting alone does not cause the rotation of a pivoted body. Rotation is always produced by a pair of forces. This may sound surprising, because in the examples of the previous section only one force seemed to be at work.
Look again at those examples. Rotation there occurs due to two forces: the force applied from outside, and the force of reaction at the pivoted point. The force of reaction produced at the pivot is equal in magnitude but opposite in direction to the applied force. Its moment about the pivot is zero, because its distance from the axis of rotation is zero, and that is why this reaction is not shown explicitly in the diagrams of the previous section.
This pair of forces — the external force together with the force of reaction — is called a couple.
When we open a door, for example, the rotation of the door is produced by a couple consisting of two forces: the force we exert at the handle of the door, and an equal and opposite force of reaction at the hinge.
Couples We Apply Deliberately
Sometimes we require a larger turning effect. Then two forces, equal in magnitude and opposite in direction, are applied on the body explicitly, in such a way that both the forces turn the body in the same direction.
To open the nut of a car wheel, we apply equal forces, each of magnitude \(F\), at the two ends of a wheel wrench’s arm in opposite directions, as shown in Fig. 1.3-1. Both hands work together, and the nut turns far more easily than it would under a single push.
Diagram
Fig. 1.3-1 — A Couple Applied To Open The Nut Of A Car Wheel Front view of a cross-shaped car wheel wrench, with the socket at the centre fitted over a nut; label the centre ‘nut’. Draw an arrow labelled F at the left arm pointing downwards, and an equal arrow labelled F at the right arm pointing upwards. Add a curved arrow around the centre showing the turning produced. Label a hand holding each of the two arms.
The same thing happens whenever two fingers, or two hands, work on opposite sides of an object: turning a water tap, tightening the cap of an inkpot, or turning the key in the hole of a lock, as shown in Fig. 1.3-2. Winding a clock or a watch with a key also needs a pair of forces applied in this way.
Diagram
Fig. 1.3-2 — Everyday Couples Three small sketches side by side. (a) A hand turning a water tap, with two arrows labelled F on opposite sides of the tap handle, pointing in opposite directions, and a curved arrow showing the turn. (b) A hand tightening the cap of an inkpot, with two arrows labelled F on opposite sides of the cap in opposite directions. (c) Fingers turning a key in the hole of a lock, with two arrows labelled F on opposite sides of the key head in opposite directions.
1.3.1 Moment Of A Couple
Fig. 1.3-3 illustrates the effect produced by a couple. AB is a bar which is pivoted at a point O. At the ends A and B, two equal and opposite forces, each of magnitude \(F\), are applied. The perpendicular distance between the two forces is AB, written as \(d\), which is called the couple arm.
Diagram
Fig. 1.3-3 — Action Of A Couple On A Bar A horizontal bar drawn as a long thin rectangle, labelled A at its left end and B at its right end, with a dot marked O at its middle labelled ‘pivot’. Draw an arrow labelled F at end A pointing vertically downwards, and an equal arrow labelled F at end B pointing vertically upwards. Mark the whole length AB with a double-headed arrow labelled d and the words ‘couple arm’. Add a curved arrow showing the bar turning anticlockwise about O.
These two forces cannot produce translational motion, because their resultant sum in any direction is zero. Yet each force is capable of producing a turning effect on the bar, and in Fig. 1.3-3 both turning effects are in the same sense. Thus the two forces together form a couple which rotates the bar about the point O, here in the anticlockwise direction.
The total turning effect is simply the sum of the two separate moments. Taking moments about O:
\[\\text{moment of force } F \\text{ at end A} = F \\times \\text{OA} \\quad \\text{(anticlockwise)}\] \[\\text{moment of force } F \\text{ at end B} = F \\times \\text{OB} \\quad \\text{(anticlockwise)}\]
Since both are anticlockwise, they add:
\[\\text{total moment of the couple} = F \\times \\text{OA} + F \\times \\text{OB} = F \\times (\\text{OA} + \\text{OB})\]
But OA + OB is the whole length AB, which is the couple arm \(d\). Therefore
\[\\text{moment of couple} = F \\times \\text{AB} = F \\times d \\qquad \\dots (1.3)\]
where \(F\) is the magnitude of either one of the two forces in newton (N), and \(d\) is the perpendicular distance between the two forces, that is, the couple arm, in metre (m). The SI unit of the moment of a couple is newton metre (N m), the same as that of the moment of a force.
Notice something useful in this result: the position of the pivot O never appears in the final answer. Wherever the pivot lies along the bar, the moment of the couple comes out as \(F \\times d\).
1.4 Equilibrium Of Bodies
We have read that when a single force acts on a body, it can produce translational motion if the body is free to move, or rotational motion if the body is pivoted or fixed at a point. But in certain circumstances a body can preserve its state of rest or of motion even when several forces are acting on it.
For this to happen, the forces must satisfy two conditions:
- the resultant of all the forces is zero, so they do not change either the state of rest or the state of linear motion of the body, and
- the algebraic sum of the moments of all the forces about the fixed point is zero, so they do not change the rotational state of the body.
A body preserving its state, static or dynamic, in the presence of two or more forces is said to be in equilibrium.
Equilibrium is of two kinds: static equilibrium and dynamic equilibrium.
1.4.1 Static Equilibrium
When a body remains in a state of rest under the influence of several forces, the body is said to be in static equilibrium.
Consider first the body lying on a table top in Fig. 1.4-1. It is pulled by a force \(F\) towards its left and by an equal force \(F'\) towards its right, both along the same line. The body does not move. The reason is that the applied forces are equal and opposite and also act along the same line, so they balance each other and there is no net horizontal force on the body. Hence the body remains at rest, that is, in static equilibrium.
Diagram
Fig. 1.4-1 — A Body In Static Equilibrium A rectangular block drawn in simple three-dimensional outline resting on a surface. Draw a horizontal arrow labelled F pointing away to the left from the left face of the block, and a horizontal arrow labelled F′ pointing away to the right from the right face, both arrows along the same horizontal dotted line drawn through the block. Mark the dotted line ‘same line of action’.
For a book lying on a table, the force exerted by the weight of the book on the table vertically downwards is balanced by an equal and opposite force of reaction exerted by the table on the book vertically upwards. Thus the book too is in static equilibrium.
A beam balance shows the second condition at work as well. When the beam is balanced in the horizontal position, as in Fig. 1.4-2, the clockwise moment of force due to the object on its right pan balances the anticlockwise moment of force due to the weights on its left pan. The beam has no rotational motion, so it is in static equilibrium.
Diagram
Fig. 1.4-2 — A Balanced Beam Balance A beam balance with a central vertical pillar and a horizontal beam pivoted at its middle; mark the pivot with a dot. Hang a pan from each end of the beam by threads. Show a set of standard weights in the left pan, labelled ‘weights’, and an object in the right pan, labelled ‘object’. Draw a curved arrow near the pivot on the left labelled ‘anticlockwise moment’ and one on the right labelled ‘clockwise moment’. Show the beam exactly horizontal.
1.4.2 Dynamic Equilibrium
When a body remains in the same state of motion, translational or rotational, under the influence of several forces, the body is said to be in dynamic equilibrium.
A rain drop reaching the earth’s surface with a constant velocity is one example. The weight of the falling drop is balanced by the sum of the buoyant force and the force due to friction, or viscosity, of the air. The net force on the drop is therefore zero, so it falls down with a constant velocity.
An aeroplane moving at a constant height is another. The upward lift on it balances its weight downwards. A car moving with a constant velocity on a straight road, and a ceiling fan turning at a constant angular velocity, are also in dynamic equilibrium — the first in steady translational motion, the second in steady rotational motion.
Steady motion along a circular path, such as that of a satellite going round its planet, is also treated as dynamic equilibrium. In such cases the force of attraction on the moving body provides the force needed to keep it in its circular path, a point we return to in Section 1.8.
The two kinds of equilibrium are compared in Table 1.2.
1.4.3 Conditions For Equilibrium
From all the above examples we find that two conditions must be satisfied for a body to be in equilibrium.
- The resultant of all the forces acting on the body should be zero. This keeps the body from starting to move, or from changing the motion it already has.
- The algebraic sum of the moments of all the forces acting on the body about the point of rotation should be zero. That is, the sum of the anticlockwise moments about the axis of rotation must be equal to the sum of the clockwise moments about the same axis.
The second condition is stated as an algebraic sum because anticlockwise moments are taken positive and clockwise moments negative. Their algebraic sum can be zero only when the two kinds of moment are equal in size.
Both conditions must hold together. Satisfying only the first leaves the body free to turn, as the next example shows.
1.5 Principle Of Moments
When several forces act on a pivoted body, they all tend to rotate it about the axis passing through the pivot. Some of them turn the body one way and some the other way, so their effects must be added with proper regard to sense.
The resultant moment of all the forces about the pivoted point is obtained by taking the algebraic sum of the moment of each force about that point. To find this algebraic sum, the anticlockwise moment is taken positive, while the clockwise moment is taken negative — the convention already fixed in Section 1.2.3.
If that algebraic sum comes out to be zero, the turning effects cancel exactly and the body does not begin to rotate. This is the principle of moments.
The beam balance met in Section 1.4.1 is a device that works on this principle: it reads a correct mass only when the moments on its two sides are equal.
1.5.1 Verification Of The Principle Of Moments
The principle can be checked directly on the laboratory bench, with a metre rule, two spring balances and some slotted weights. The arrangement is shown in Fig. 1.5-1.
Suspend a metre rule horizontally from a fixed support by means of a strong thread at a point O. Now suspend two spring balances A and B on the metre rule, one on either side of the thread. Suspend some slotted weights \(W\_1\) and \(W\_2\) from the spring balances A and B respectively. The metre rule may tilt to one side.
Now adjust either the slotted weights on the spring balances, or the positions of the spring balances on either side of the thread at O, in such a way that the metre rule becomes horizontal again. A spring balance is used at each point so that the pull acting there can simply be read off its scale.
Diagram
Fig. 1.5-1 — Verification Of The Principle Of Moments A horizontal metre rule with its scale markings shown, hung from a fixed support at the top by a strong vertical thread tied at a point O on the rule; label the support ‘support’ and the point O. On the right of O, hang a spring balance labelled A from the rule at a point marked A, with slotted weights labelled W₁ hanging from it. On the left of O, hang a spring balance labelled B from the rule at a point marked B, with slotted weights labelled W₂ hanging from it. Mark the distance from O to A with a horizontal double-headed arrow labelled l₁, and the distance from O to B with a horizontal double-headed arrow labelled l₂. Show the rule exactly horizontal. Add curved arrows showing that W₁ tends to turn the rule clockwise and W₂ anticlockwise.
Let the weight suspended from the spring balance A on the right side of the thread be \(W\_1\), at a distance \(\\text{OA} = l\_1\), while the weight suspended from the spring balance B on the left side of the thread is \(W\_2\), at a distance \(\\text{OB} = l\_2\).
The weight \(W\_1\) tends to turn the metre rule clockwise, while the weight \(W\_2\) tends to turn the metre rule anticlockwise. Taking moments about the point O:
\[\\text{clockwise moment of } W\_1 \\text{ about O} = W\_1 \\times l\_1\] \[\\text{anticlockwise moment of } W\_2 \\text{ about O} = W\_2 \\times l\_2\]
In equilibrium, when the metre rule is horizontal, it is found by measurement that
\[W\_1 l\_1 = W\_2 l\_2\]
that is, clockwise moment = anticlockwise moment. This verifies the principle of moments. Repeating the experiment with different weights and different distances gives the same result each time.
Using The Principle In Calculations
In problems on balancing, the working is always the same. Choose the pivot as the point about which moments are taken, list the anticlockwise moments and the clockwise moments separately, and set the two sums equal. One useful consequence: a force acting through the pivot itself has zero moment about it, so it drops out of the equation.
The arrangement of Example 1.5 is drawn in Fig. 1.5-2.
Diagram
Fig. 1.5-2 — A Metre Rule Balanced On A Knife Edge At Its Centre A horizontal metre rule with markings 0, 10, 20 … 100 cm shown along it, resting on a knife edge drawn as a small triangle under the 50 cm mark; label the knife edge ‘pivot’. Hang a weight labelled ‘20 gf’ by a short thread from the 20 cm mark, and a weight labelled ‘30 gf’ by a short thread from the 70 cm mark. Mark the distance from the 20 cm mark to the pivot with a double-headed arrow labelled 30 cm, and the distance from the pivot to the 70 cm mark with a double-headed arrow labelled 20 cm. Show the rule exactly horizontal.
1.6 Centre Of Gravity
We have read in Class IX that the gravitational force between two masses is always attractive. The earth attracts every particle of a body towards its centre by the force of gravity acting on that particle, and this force is called the weight of the particle.
Each body can be considered to be made up of a large number of particles of weight \(w\_1\), \(w\_2\), \(w\_3\), and so on. As the size of the body is quite small in comparison to the size of the earth, the forces of gravity acting on these particles can be assumed to be parallel to each other, as shown in Fig. 1.6-1.
Diagram
Fig. 1.6-1 — Weights Of The Particles Of A Body And Their Resultant An irregular closed outline representing a body. Inside it, draw eight small dots spread over the area, and from each dot draw a short vertical arrow pointing downwards; label these arrows w₁, w₂, w₃, w₄, w₅, w₆, w₇, w₈. Mark a point near the middle of the body with a dot labelled G, and from G draw one long vertical arrow pointing downwards labelled W. Add the note ‘all the small weights are parallel and act vertically downwards’.
All these parallel forces act in the same direction, vertically downwards towards the centre of the earth. They can therefore be replaced by a single force \(W\) of magnitude equal to the sum of all these forces,
\[W = w\_1 + w\_2 + w\_3 + \\dots\]
where \(W\) is the total weight of the body and \(w\_1\), \(w\_2\), \(w\_3\), … are the weights of its individual particles, all measured in newton (N).
Now the question arises: where should this single weight \(W\) be taken to act? The weight \(W\) is considered to act at a point G such that the algebraic sum of the moments due to the weights \(w\_1\), \(w\_2\), … of each particle about the point G is zero. The point G is called the centre of gravity of the body.
In other words, once the centre of gravity is known, the whole body can be treated as a single point particle of weight \(W\) placed at that point. This is exactly what we did without comment in Example 1.7, where the weight of a uniform metre rule was taken to act at its 50 cm mark.
Three points are worth noting about the centre of gravity.
- The position of the centre of gravity of a body of given mass depends on its shape, that is, on the distribution of the mass of its particles within it. It changes if the body is deformed. For example, the centre of gravity of a straight uniform wire is at the middle of its length; but if the same wire is bent into the form of a circle, its centre of gravity will then be at the centre of the circle.
- It is not necessary that the centre of gravity always be within the material of the body. For example, the centre of gravity of a ring, or of a hollow sphere, lies at its centre, where there is no material at all.
- By the concept of centre of gravity, a body of weight \(W\) can be considered as a point particle of weight \(W\) at its centre of gravity.
1.6.1 Centre Of Gravity Of Some Regular Uniform Objects
For bodies of regular shape made of uniform material, the position of the centre of gravity can be written down at once. These positions are collected in Table 1.3, and Fig. 1.6-2 shows the point G for several of them.
Table 1.3 — Position of the centre of gravity of some regular uniform objects
| Object | Position of centre of gravity |
|---|---|
| Rod | Mid-point of the rod |
| Circular disc | Geometric centre |
| Solid or hollow sphere | Geometric centre of the sphere |
| Solid or hollow cylinder | Mid-point on the axis of the cylinder |
| Solid cone | At a height \(h/4\) from the base, on its axis (if \(h\) = height of cone) |
| Hollow cone | At a height \(h/3\) from the base, on its axis (if \(h\) = height of cone) |
| Circular ring | Centre of the ring |
| Triangular lamina or scalene triangle | The point of intersection of the medians |
| Parallelogram, rectangular lamina, square or rhombus | The point of intersection of the diagonals |
Diagram
Fig. 1.6-2 — Centre Of Gravity Of Some Regular Objects Eight small outline sketches arranged in two rows, each with the centre of gravity marked by a dot labelled G. Row one: a circular ring labelled ‘circular ring’ with G at its centre; a circular disc labelled ‘circular disc’ with G at its centre; a triangular lamina labelled ‘triangular lamina’ with its three medians drawn and G at their point of intersection. Row two: a rectangle labelled ‘rectangle’ with both diagonals drawn and G at their intersection; a parallelogram labelled ‘parallelogram’ with both diagonals drawn and G at their intersection; a square labelled ‘square’ with both diagonals drawn and G at their intersection; a thin long rod labelled ‘rod’ with G at its mid-point; a cylinder drawn in outline labelled ‘cylinder’ with its axis shown as a dashed line and G at the mid-point of the axis.
1.6.2 Centre Of Gravity And The Balance Point
A solid body can be balanced by supporting it at its centre of gravity. For example, a uniform metre rule has its centre of gravity at the 50 cm mark. It can be balanced on a knife edge, or on a finger tip, by keeping the support exactly below the 50 cm mark, as shown in Fig. 1.6-3.
Diagram
Fig. 1.6-3 — A Metre Rule Supported On A Knife Edge At Its Mid-Point A horizontal metre rule with markings 10, 20, 30 … 90 shown along its length, resting on a knife edge drawn as a small triangle placed exactly below the 50 cm mark. Label the rule ‘metre rule’ and the support ‘knife edge’. Show the rule perfectly horizontal, with no weights hanging from it.
This is possible because the algebraic sum of the moments of the weights of all the particles of the rule about the knife edge is zero. Half the rule lies on either side of the support, and the anticlockwise moments of the particles on one side exactly balance the clockwise moments of those on the other.
Similarly, a square thin sheet, or lamina, can be balanced on the tip of a nail placed below the point where its diagonals cross, as shown in Fig. 1.6-4.
Diagram
Fig. 1.6-4 — A Square Sheet Balanced On The Tip Of A Nail A square lamina drawn in perspective, lying flat and horizontal, with both diagonals shown as dashed lines and their point of intersection marked G. Below G, draw a nail standing vertically with its tip touching the lamina at G; label it ‘nail’. Label the sheet ‘square lamina’.
There is a second, equally useful fact. When a body is freely suspended from a point, it comes to rest — that is, it stops oscillating — in such a position that its centre of gravity lies vertically below the point of suspension. This fact can be used to locate the position of the centre of gravity of an irregular lamina.
1.6.3 Determination Of The Centre Of Gravity Of An Irregular Lamina
The centre of gravity of an irregularly shaped lamina cannot be written down from a table. It is found experimentally, by the method of balance using a plumb line. A plumb line is simply a small heavy bob hung from a thread, which sets itself along the vertical.
Let A be an irregular lamina, as in Fig. 1.6-5, for which the position of the centre of gravity is to be determined. Make three fine holes at \(a\), \(b\) and \(c\), near the edge of the lamina.
Now suspend the given lamina, along with a plumb line, from the hole \(a\) on a pin or a nail clamped horizontally on a retort stand. Check that the lamina is free to oscillate on the nail about the point of suspension. When the lamina has come to rest, draw a straight line \(ad\) along the plumb line.
Diagram
Fig. 1.6-5 — Locating The Centre Of Gravity Of An Irregular Lamina An irregular flat lamina labelled A, with three small holes marked near its edge and labelled a (at the top), b (at the lower right) and c (at the lower left). From a, draw a straight line across the lamina to a point d on the opposite edge; from b, draw a straight line to a point e on the opposite edge; from c, draw a straight line to a point f on the opposite edge. Draw all three lines dashed, and mark their common point of intersection with a dot labelled G. Show a pin at hole a with a thread and a small bob hanging vertically alongside, labelled ‘plumb line’.
Repeat the procedure by suspending the lamina through the hole \(b\), and then through the hole \(c\), for which we get the straight lines \(be\) and \(cf\) respectively. It is noticed that the lines \(ad\), \(be\) and \(cf\) intersect each other at a common point G, which is the position of the centre of gravity of the lamina.
Two lines would in principle be enough to fix the point. The third is drawn as a check: if all three pass through one point, the work has been done carefully.
Finally, one condition worth remembering. For the stable equilibrium of a body, its centre of gravity must be as low as possible. It must be above the base and near the geometric centre of the body.
1.7 Uniform Circular Motion
When a particle moves with a constant speed in a circular path, its motion is said to be uniform circular motion. In such a motion, the particle travels equal distances along the circular path in equal intervals of time, so the speed of the particle is uniform.
But the direction of motion of the particle changes at each point of the circular path. A particle going round a circle is never headed the same way for two instants together. This continuous change in the direction of motion implies that the velocity of the particle is non-uniform, or variable, and therefore that the motion is accelerated.
1.7.1 Direction Of Velocity At Any Instant
Fig. 1.7-1 shows a particle moving in a circular path in a horizontal plane with uniform speed \(v\) in an anticlockwise direction. The particle travels each quarter of the circle AB, BC, CD and DA in the same interval of time \(t = T/4\), where \(T\) is the time taken by the particle to complete one round of the circular path.
Thus the speed of the particle is constant, or uniform, but the direction of motion of the particle is different at different points of the circular path. At any point, the direction of motion is along the tangent drawn at that point of the circular path.
Diagram
Fig. 1.7-1 — Direction Of Velocity In Uniform Circular Motion A circle with four points marked on it: A at the extreme right, B at the top, C at the extreme left and D at the bottom. Draw a curved arrow along the circle showing anticlockwise motion. At each of the four points draw a straight arrow labelled v, tangential to the circle: at A pointing upwards, at B pointing to the left, at C pointing downwards, at D pointing to the right. Alongside the circle draw a small compass cross with the four arms labelled north (up), west (left), south (down) and east (right).
Follow the particle round once. At the point A, the direction of motion of the particle is towards north. After completing a quarter of the circle, at the point B, the direction of motion is towards west. After completing half the circle, at the point C, the direction of motion is towards south. After completing three-quarters of the circle, when the particle is at the point D, its direction of motion is towards east.
Thus the velocity of the particle in circular motion is variable. In other words, circular motion is an accelerated motion even though the speed of the particle is uniform.
1.7.2 Uniform Circular Motion Compared With Uniform Linear Motion
Both motions are called uniform, but only one of them is unaccelerated, and the difference matters.
In uniform linear motion, the speed and the velocity are both constant, and the acceleration is zero. That is, uniform linear motion is an unaccelerated motion. In uniform circular motion the velocity is variable, although the speed is uniform, so it is an accelerated motion. Table 1.4 sets the two side by side.
An accelerated motion, as you know, requires a force. Which force keeps a body moving in a circle, and in which direction it acts, is taken up in the next section.
1.8 Centripetal And Centrifugal Force
1.8.1 Centripetal Force
We have read that a force is needed to change the direction of motion of a particle, or to change the velocity of a particle, that is, to produce acceleration. A particle moving in a circular path continuously changes its direction of motion at each point of its path. This change in direction of motion is not possible without a force.
Thus, motion in a circular path is possible only under the influence of a force, which is termed the centripetal force. The word centripetal means centre seeking, and the name is exact: at each point of the circular path this force is directed towards the centre of the circle, as shown in Fig. 1.8-1.
Diagram
Fig. 1.8-1 — Direction Of Force In Uniform Circular Motion A circle with a dot at its centre, and four points marked on it: A at the extreme right, B at the top, C at the extreme left and D at the bottom. At eight evenly spaced points around the circle, including A, B, C and D, draw straight arrows each labelled F, every arrow pointing from the circle inwards along the radius towards the centre. Add a curved arrow along the circle showing the body going round anticlockwise.
Notice what changes and what does not. The direction of this force, and therefore of the acceleration it produces, changes at each point of the circular path, since it always points to the centre. Its magnitude, however, remains the same throughout. So the acceleration in uniform circular motion is variable, or non-uniform. Hence, for a body moving in a circular path, a force is needed which acts as the centripetal force.
Where The Centripetal Force Comes From
Centripetal force is not a new kind of force with a supply of its own. In every case it is some ordinary force — a pull, a tension, an attraction — doing this particular job.
- In an atom, an electron moves around the nucleus in a circular path, for which the required centripetal force is obtained from the electrostatic force of attraction on the negatively charged electron by the positively charged nucleus.
- A planet moves around the sun in an elliptical path, for which the gravitational force of attraction on the planet by the sun provides the necessary centripetal force.
- The moon moves around the earth in an elliptical path, for which the gravitational force of attraction on the moon by the earth provides the required centripetal force.
- When a stone tied at the end of a string is whirled in a circular path, holding the other end in the hand, the tension in the string provides the centripetal force. In the absence of this tension, the stone will not turn to move in a circular path.
In all the above examples, the body moves in a circular or near circular path with a uniform speed under the influence of a centripetal force, and it is in a state of dynamic equilibrium.
1.8.2 Centrifugal Force
A centrifugal force is a force assumed, by an observer moving with the body, to be acting on the body in a direction away from the centre of the circular path.
Thus centrifugal force is in a direction opposite to the direction of centripetal force, and its magnitude is the same as that of the centripetal force. But the centrifugal force is not the force of reaction of the centripetal force, because action and reaction do not act on the same body, whereas both these forces are taken to act on the same body.
It is not a real force. It is a fictitious force assumed by an observer moving in a circular path along with the body. To understand this force, consider the following experiment.
An Experiment On A Merry-Go-Round
Fig. 1.8-2 shows a ball tied at one end of a string, the other end of which is tied at the centre of a merry-go-round. Initially, when the platform of the merry-go-round is stationary, the ball is seen stationary and the string is loose. As the platform starts rotating, the ball rolls towards the edge of the merry-go-round, and the string becomes tight due to the tension \(T\) developed in it.
Now let us consider the state of the ball as seen by two observers: one standing outside the merry-go-round on the ground at M, and one standing on the platform of the merry-go-round at A.
Diagram
Fig. 1.8-2 — A Ball Tied To A String On A Rotating Merry-Go-Round A large circle representing the platform of a merry-go-round, with a dot at its centre. A man drawn outside the circle on the left is labelled M and ‘man on the ground’. Inside the circle, mark points A, A′, A″ and A‴ at increasing angles round the centre, each marked as the changing position of the observer standing on the platform. From the centre, draw the string outwards to a ball marked P, with the string labelled ‘string under tension T’. Mark further ball positions P′, P″ and P‴ on a dotted circle, each lying directly outward from A′, A″ and A‴ respectively, and label this dotted circle ‘path of the tied ball as seen by the man at M’. Add a curved arrow showing the platform rotating.
The person standing on the ground at M, outside the merry-go-round, observes that the ball is moving in a circular path, shown by the dotted line. The person standing on the merry-go-round at A observes that the ball is stationary, placed just in front of him at P. As the merry-go-round rotates, the position of the person on the platform changes from A to A′, A″, A‴, and the ball reaches the positions P′, P″, P‴ respectively — as if it remains at rest always just in front of him.
The two persons see the same motion differently, and the difference is explained as follows.
For the person at M, the ball moves in a circular path because the tension \(T\) in the string provides the centripetal force needed for the circular motion.
The person at A observes the ball as stationary. He needs to consider the following two forces to be acting on the ball in order to understand his observation:
- the tension \(T\) of the string towards the centre of the merry-go-round, and
- the centrifugal force away from the centre.
Since the ball, according to his observation, always remains stationary in front of him, the above two forces must be equal and opposite, so that the net force on the ball is zero. Thus, a person on the rotating platform can explain his observation only when he considers the centrifugal force along with the force of tension in the string.
When The String Breaks
Suppose the string breaks when the ball is in the position P, as shown in Fig. 1.8-3. The force of tension \(T\) in the string ceases to act.
Now the person at M, standing on the ground, will observe that the ball is moving in a straight line along the path PT, the tangent drawn at the point P on the circular path. The person at A, standing on the merry-go-round, will observe the ball at positions P′, P″, P‴ when he himself is at positions A′, A″, A‴ respectively. That is, he will observe that the ball always remains in front of him, moving radially away from him. He attributes this motion to a centrifugal force acting on the ball away from the centre.
Diagram
Fig. 1.8-3 — The Ball After The String Breaks A large circle representing the merry-go-round platform, with a dot at its centre and a curved arrow showing rotation. A man outside on the left is labelled M and ‘man on the ground’. Inside the circle mark the observer positions A, A′, A″ and A‴ at increasing angles round the centre. Mark the ball at P on the rim, and further ball positions P′, P″ and P‴ lying on a straight dotted line drawn from P tangentially to the circle, ending at a point labelled T; label this straight line ‘actual path of the ball after the string is broken, as seen by the man at M’. Draw thin dotted lines joining A′ to P′, A″ to P″ and A‴ to P‴ to show that the ball stays in front of the moving observer.
Thus the person in a rotating frame, here the merry-go-round, has to assume the presence of the centrifugal force.
The conclusion is this. The centrifugal force is not a real force; it is a fictitious force. The only real force involved here is the force of tension in the string acting towards the centre, that is, the centripetal force. A force which really does not exist, but is considered in order to describe or understand a certain motion, is called a fictitious force, or a virtual force.
The two forces are compared in Table 1.5.
1.9 Summary
A force applied on a rigid body can only produce motion in it, while on a non-rigid body it produces a change in size or shape as well as motion.
Force is the rate of change of linear momentum of a body. For a body of constant mass, \(F = ma\), where \(F\) is the force in newton (N), \(m\) the mass in kilogram (kg) and \(a\) the acceleration in metre per second squared (m s⁻²).
Force is a vector quantity. Its SI unit is the newton (N); in gravitational units, \(1\\ \\text{kgf} = 9.8\\ \\text{N}\), taking \(g = 9.8\\ \\text{m s}^{-2}\).
A rigid body free to move undergoes linear or translational motion along the direction of the applied force. A body pivoted at a point undergoes rotational motion about the axis through the pivot.
The turning effect of a force depends on two factors: the magnitude of the force, and the perpendicular distance of the line of action of the force from the axis of rotation.
The moment of a force, or torque, is the product of these two factors: \[\\text{moment of force} = F \\times \\text{OP}\] where \(F\) is the force in newton (N) and OP is the perpendicular distance of its line of action from the axis of rotation, in metre (m).
The SI unit of moment of force is newton metre (N m); the C.G.S. unit is dyne centimetre. Also \(1\\ \\text{N m} = 10^{7}\\ \\text{dyne cm}\), \(1\\ \\text{kgf} \\times \\text{m} = 9.8\\ \\text{N m}\), and \(1\\ \\text{gf} \\times \\text{cm} = 980\\ \\text{dyne cm}\). The unit of torque is never written as joule.
For a given force, the moment is maximum when the perpendicular distance of its line of action from the axis of rotation is maximum. A force acting through the axis has zero moment about it.
An anticlockwise moment is taken positive and a clockwise moment negative. The direction of rotation can be changed either by changing the point of application of the force, or by changing the direction of the force.
Two equal and opposite parallel forces, not acting along the same line, form a couple. A couple is always needed to produce rotation.
The moment of a couple is \[\\text{moment of couple} = F \\times d\] where \(F\) is the magnitude of either force in newton (N) and \(d\) is the perpendicular distance between the two forces, called the couple arm, in metre (m). Its SI unit is newton metre (N m).
A body is in equilibrium when the forces acting on it produce no change in its state of rest, or of linear or rotational motion.
Equilibrium is of two kinds: static equilibrium, in which the body remains at rest, and dynamic equilibrium, in which the body remains in the same state of motion.
Two conditions must be satisfied for equilibrium: the resultant of all the forces on the body must be zero, and the algebraic sum of the moments of all the forces about the point of rotation must be zero.
By the principle of moments, for a body in equilibrium, \[\\text{sum of the anticlockwise moments} = \\text{sum of the clockwise moments}\] all moments being taken about the same axis of rotation. A beam balance works on this principle.
The principle is verified with a metre rule suspended horizontally by a thread, carrying two spring balances with slotted weights on either side; when the rule is horizontal, \(W\_1 l\_1 = W\_2 l\_2\), where \(W\_1\) and \(W\_2\) are the two weights and \(l\_1\) and \(l\_2\) their distances from the point of suspension.
The centre of gravity of a body is the point about which the algebraic sum of the moments of the weights of all its particles is zero. The entire weight \(W\) of the body can be considered to act at this point, howsoever the body is placed.
The position of the centre of gravity depends on the shape of the body, that is, on the distribution of its mass, and it changes if the body is deformed. It need not lie within the material of the body.
For uniform bodies of regular shape the centre of gravity lies at a known point: the mid-point of a rod, the geometric centre of a disc, sphere or cylinder, the centre of a ring, the intersection of the medians of a triangular lamina, the intersection of the diagonals of a parallelogram, and at a height \(h/4\) from the base of a solid cone or \(h/3\) from the base of a hollow cone, where \(h\) is the height.
A body can be balanced by supporting it at its centre of gravity. When a body is freely suspended, it comes to rest with its centre of gravity vertically below the point of suspension.
The centre of gravity of an irregular lamina is found by suspending it freely from three different holes near its edge with a plumb line, drawing the vertical line each time, and taking the common point of intersection of the three lines.
For stable equilibrium, the centre of gravity of a body must be as low as possible, above its base and near its geometric centre.
When a particle moves with a constant speed in a circular path, its motion is uniform circular motion. At any point, its direction of motion is along the tangent drawn at that point.
In uniform circular motion the speed is constant but the velocity is variable, so the motion is accelerated. Uniform linear motion, in which both speed and velocity are constant, is unaccelerated.
The centripetal force is the force acting on a body moving in a circular path, directed towards the centre of the path. Its magnitude stays the same while its direction changes at each point, so the acceleration is variable.
The centripetal force is always supplied by a real agent — tension in a string, gravitational attraction, or electrostatic attraction.
The centrifugal force is a force assumed by an observer moving along with the body to act on it away from the centre. It is equal and opposite to the centripetal force, but it is a fictitious force and is not the reaction of the centripetal force.
Flowchart
FC 1.1 — What A Force Does To A Rigid Body Force applied on a rigid body → Is the body free to move? → if yes: translational motion along the direction of the force → if no (the body is pivoted): the force has a moment about the axis → Is the algebraic sum of all moments zero? → if yes: no rotation, the body is in equilibrium → if no: rotational motion, anticlockwise or clockwise according to the sense of the net moment
1.10 Key Terms
- Anticlockwise moment — a moment of force whose effect is to turn the body anticlockwise; taken positive. SI unit: newton metre (N m).
- Axis of rotation — the fixed line about which a pivoted body turns.
- Centre of gravity — the point of a body about which the algebraic sum of the moments of the weights of all its particles is zero, and at which its entire weight may be taken to act.
- Centrifugal force — a fictitious force assumed by an observer moving with a body to act on it away from the centre of its circular path. SI unit: newton (N).
- Centripetal force — the real force acting on a body moving in a circular path, directed towards the centre of that path. SI unit: newton (N).
- Clockwise moment — a moment of force whose effect is to turn the body clockwise; taken negative. SI unit: newton metre (N m).
- Couple — two equal and opposite parallel forces that do not act along the same line, and which together produce rotation.
- Couple arm — the perpendicular distance between the two forces forming a couple. SI unit: metre (m).
- Dynamic equilibrium — the state of a body that remains in the same state of motion under the action of several forces.
- Equilibrium — the state of a body on which the forces acting produce no change in its rest, or in its linear or rotational motion.
- Force — a push or a pull that changes the state of rest or of motion of a body; equal to the rate of change of its linear momentum. SI unit: newton (N).
- Kilogram-force — a gravitational unit of force, equal to the weight of a mass of one kilogram; 1 kgf = 9.8 N.
- Line of action — the straight line drawn through the point at which a force acts, along the direction of the force.
- Linear motion — another name for translational motion.
- Moment of force — the product of a force and the perpendicular distance of its line of action from the axis of rotation. SI unit: newton metre (N m).
- Newton — the SI unit of force, symbol N.
- Pivot — the fixed point at which a body is held, so that it can turn but not move away.
- Plumb line — a small weight hung from a thread, used to mark the vertical direction.
- Principle of moments — for a body in equilibrium, the sum of the anticlockwise moments about an axis equals the sum of the clockwise moments about the same axis.
- Rigid body — a body whose shape and size do not change when a force acts on it.
- Rotational motion — the motion of a body in which it turns about a fixed axis, its particles moving in circles about that axis.
- Static equilibrium — the state of a body that remains at rest under the action of several forces.
- Torque — another name for the moment of a force. SI unit: newton metre (N m).
- Translational motion — the motion of a body along a straight path in the direction of the applied force, every particle moving through the same distance in the same direction.
- Uniform circular motion — the motion of a particle moving along a circular path with a constant speed, in which the direction of motion, and hence the velocity, changes continuously.
- Weight — the force with which the earth attracts a body or a particle towards its centre. SI unit: newton (N).
1.11 Points To Ponder
A force applied at the pivot has no turning effect at all, yet a force applied a hair’s breadth away from the pivot does have one. Where, then, does the turning effect begin, and what does this tell you about the way the moment grows with distance?
If rotation is always produced by a couple, why does a single push on a door seem quite enough to swing it open? What is supplying the second force, and why does it never appear in the calculation?
A body may satisfy the first condition for equilibrium and still fail the second. Can you imagine an arrangement of forces that satisfies the second condition but fails the first, and describe what the body would then do?
The whole weight of a body is taken to act at a single point, its centre of gravity, even though gravity really pulls on every particle separately. What makes this replacement legitimate, and can you think of a situation in which treating a body as a point at G would give a misleading picture?
A ring has its centre of gravity at a place where there is no material at all. What would happen if you tried to balance a ring by supporting it exactly at its centre of gravity, and why?
An observer on the ground and an observer riding on a turntable describe the same moving ball quite differently, and both accounts work. Does this mean physics gives two different answers to the same question, or does it mean something else about how motion is described?
1.12 Objective Questions
- (MCQ) The moment of a force about an axis depends on: (1 mark)
- the magnitude of the force only
- the perpendicular distance of the line of action from the axis only
- both the magnitude of the force and the perpendicular distance of its line of action from the axis
- neither of these
- (MCQ) A force acts on a body along a line that passes through the axis of rotation. The moment of this force about that axis is: (1 mark)
- maximum
- zero
- equal to the force
- equal to the distance from the axis
- (MCQ) Two equal and opposite parallel forces acting on a body along different lines will: (1 mark)
- move the body along a straight line
- keep the body completely at rest
- turn the body without moving it from its place
- have no effect whatever on the body
- (MCQ) The centre of gravity of a uniform triangular lamina lies at: (1 mark)
- the mid-point of its longest side
- the point of intersection of its medians
- the point of intersection of its diagonals
- one of its vertices
- (MCQ) In uniform circular motion, the quantity that keeps changing is: (1 mark)
- the speed
- the distance covered in each second
- the direction of motion
- the magnitude of the centripetal force
- (MCQ) A stone tied to a string is whirled in a horizontal circle. If the string suddenly snaps, the stone moves: (1 mark)
- along the radius, away from the centre
- along the radius, towards the centre
- along the tangent at the point where the string snapped
- straight downwards at once
- (Assertion–Reason) Assertion (A): A door is much harder to push open near its hinges than at its outer edge. (1 mark) Reason (R): The moment of a force is the product of the force and the perpendicular distance of its line of action from the axis of rotation.
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is false but R is true
- A is true but R is false
- (Assertion–Reason) Assertion (A): The centrifugal force is the force of reaction of the centripetal force. (1 mark) Reason (R): The centrifugal force is equal in magnitude and opposite in direction to the centripetal force.
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is false but R is true
- A is true but R is false
- (Match the following) Match each body in Column A with the position of its centre of gravity in Column B. (4 marks)
| Column A | Column B |
|---|---|
| a. Uniform rod | i. Point of intersection of the diagonals |
| b. Solid cone of height \(h\) | ii. Centre of the ring, where there is no material |
| c. Circular ring | iii. Mid-point of its length |
| d. Rectangular lamina | iv. At a height \(h/4\) from the base, on its axis |
(Fill in the blank) An anticlockwise moment is taken as …………… and a clockwise moment is taken as …………… . (1 mark)
(Fill in the blank) The perpendicular distance between the two forces forming a couple is called the …………… . (1 mark)
(Fill in the blank) A body freely suspended from a point comes to rest with its …………… lying vertically below the point of suspension. (1 mark)
(True or false) The centre of gravity of a body always lies within the material of the body. (1 mark)
(True or false) A body moving with a constant speed along a circular path has zero acceleration. (1 mark)
(Name the following) Name the kind of equilibrium of an aeroplane flying at a constant height with a constant velocity. (1 mark)
(Name the following) Name the force that keeps the moon moving in its path around the earth. (1 mark)
(Give one word or phrase for) Give one word or phrase for: the product of a force and the perpendicular distance of its line of action from the axis of rotation. (1 mark)
(State the SI unit of) State the SI unit of: (i) moment of a force, (ii) moment of a couple, (iii) weight. (1 mark)
(Odd one out) Pick the odd one out and give a reason: tension in a string, gravitational attraction, centrifugal force, electrostatic attraction. (1 mark)
1.13 Check Your Concepts
Define the moment of a force and state its SI unit. (2 marks)
State the two factors on which the turning effect of a force on a body depends. (2 marks)
Give a reason: the handle of a hand flour grinder is fixed near the rim of its upper stone, and not near the centre. (2 marks)
Define a couple and state the expression for the moment of a couple, naming each quantity in it. (2 marks)
State whether the moment of a force is a scalar or a vector quantity, and state the sign given to a clockwise moment. (2 marks)
Differentiate between translational motion and rotational motion, stating the basis of difference. (3 marks)
Differentiate between static equilibrium and dynamic equilibrium, stating the basis of difference, and give one example of each. (3 marks)
State the principle of moments, with the condition under which it holds. Name one device that works on this principle. (3 marks)
Why does the position of the centre of gravity of a body change when the body is deformed? Explain with the example of a straight uniform wire bent into a circle. (3 marks)
A uniform circular motion is said to be an accelerated motion, although the speed of the body does not change. Explain why. (3 marks)
Differentiate between centripetal force and centrifugal force, stating the basis of difference, and state which of the two is a real force. (4 marks)
What would happen to a pivoted body if the resultant of the forces acting on it were zero but the algebraic sum of their moments about the pivot were not zero? Explain your answer with an example. (4 marks)
Describe an experiment to verify the principle of moments, using a metre rule, a fixed support, two spring balances and some slotted weights. State clearly what is measured and what result verifies the principle. (5 marks)
Explain, with the help of the observations made by a person standing on the ground and a person standing on a rotating platform, why the centrifugal force is called a fictitious force. (5 marks)
1.14 Diagrams, Graphs And Constructions
Fig. 1.2-1 shows a body pivoted at O with a force \(F\) acting along the line AP. Redraw the figure, mark the perpendicular distance OP clearly, and state which two lengths in your figure must never be confused with each other. (3 marks)
Draw a diagram of a bar pivoted at its middle point O, with two equal and opposite parallel forces acting at its two ends. Label the couple arm, and mark with a curved arrow the sense in which the bar turns. (3 marks)
Draw a labelled diagram of the apparatus used to verify the principle of moments, as set up in Fig. 1.5-1. Label the support, the thread, the two spring balances, the slotted weights and the two distances measured from the point of suspension. (4 marks)
Copy the outline of a triangular lamina and construct the position of its centre of gravity, showing the lines you draw. State the geometrical name of the point you obtain. (3 marks)
Draw a diagram of an irregular lamina hanging from a pin near its edge, together with a plumb line. Mark the point of suspension by the letter S, draw the line you would rule along the plumb line, and state how many such lines are needed to fix the centre of gravity. (4 marks)
Fig. 1.7-1 shows a particle moving anticlockwise round a circle at a constant speed. Redraw the circle, mark four points on it, and draw an arrow at each point to show the direction of the velocity there. State what all four arrows have in common with respect to the circle. (4 marks)
Draw a circle to represent the path of a body in uniform circular motion, and mark on it the direction of the centripetal force at four different points. State how the direction and the magnitude of this force change as the body goes round. (3 marks)
1.15 Practice With Numericals
[Note: For a uniform rod or rule, its weight acts at its mid-point.]
A force of \(25\\ \\text{N}\) is applied normally at the end of a spanner handle of length \(0.20\\ \\text{m}\). Calculate the moment of the force about the nut. (2 marks)
A bolt turns only when a moment of \(18\\ \\text{N m}\) is applied to it. Calculate the least force that must be applied normally at the end of a spanner of handle length \(0.45\\ \\text{m}\). (2 marks)
Two forces, each of magnitude \(12\\ \\text{N}\), act in opposite directions at the two ends of a rod of length \(0.35\\ \\text{m}\), forming a couple. Calculate the moment of the couple. (2 marks)
A uniform metre rule is pivoted at its 50 cm mark. A weight of \(40\\ \\text{gf}\) hangs at the 30 cm mark. At which mark must a weight of \(25\\ \\text{gf}\) be hung to keep the rule horizontal? (3 marks)
A child of mass \(30\\ \\text{kg}\) sits on a see-saw at a distance of \(1.8\\ \\text{m}\) from the pivot. At what distance from the pivot, and on which side, must a person of mass \(54\\ \\text{kg}\) sit to balance the see-saw? (3 marks)
A uniform metre rule of mass \(50\\ \\text{g}\) is pivoted on a knife edge at its 30 cm mark. A mass \(m\) is hung at the 10 cm mark and the rule stays horizontal. Find the value of \(m\). (3 marks)
Express (i) a moment of \(2.5\\ \\text{kgf} \\times \\text{m}\) in N m, and (ii) a moment of \(3\\ \\text{N m}\) in dyne cm. (3 marks)
A uniform metre rule of weight \(100\\ \\text{gf}\) is pivoted at its 40 cm mark. A weight of \(60\\ \\text{gf}\) hangs at the 20 cm mark. Calculate: (i) the anticlockwise moment about the pivot, (ii) the clockwise moment about the pivot, (iii) whether the rule is in equilibrium, and (iv) the mark at which a weight of \(40\\ \\text{gf}\) must be hung to balance the rule. (5 marks)
A uniform bar AB of length \(2.0\\ \\text{m}\) and negligible weight rests horizontally on two supports, one at each end. A load of \(60\\ \\text{N}\) hangs from a point \(0.5\\ \\text{m}\) from the end A. Calculate the upward reaction at each support, and check that the two reactions together balance the load. (5 marks)