In table 7.2, F represents the fulcrum of the lever, E the force or effort applied to ‘do the job’ and L the load, which is the output force produced that
overcomes the resistance when the effort is applied.
The benefit of some levers lies in the fact that quite often the force to be used is rather less that the value of the load being overcome. For example, we can use a claw-hammer (like that shown in figure 7.15 (a)) whose handle is always very much longer than the distance between the butt end (or heel) of the hammer and the ‘V’ of the claw. This gives an advantage to the user of the hammer and so we say that the hammer offers to the user a mechanical
advantage (M.A. for short), mechanical because the forces involved are all
mechanical and the benefit constitutes an advantage to the user. To understand the significance and meaning of the advantage, consider the line diagram of figure 7.16 in which a claw-hammer is being used to draw a nail out of a piece of wood. The user applies an effort E at right angles to the handle of length a. This effort gives rise to an output force (equal to the load L) which will be used to overcome the resistance of the wood. Assuming that the frictional resistance of the wood is R and that it is equal to the load, L, we can use the lever law and write that
effort × arm of effort = load × arm of load or E × a = L × b
See figure 7.16.
This equation gives the mechanical advantage, by definition, load/effort, as M.A. = LE = ab
(which is much larger than 1, since the handle is always so long – the longer the handle the better!).
You will be able to conclude from this that the theory of the crow-bar as a machine is very similar to that of the hammer. In the case of the crow-bar, with the much longer handle, it is possible to obtain a much better M.A. and so the effort needed to overcome a given load is made smaller.
An alternative way of looking at the matter above, is to observe that the ratio E/L = b/a and so
E = (b/a) × L
which shows that the effort used to draw the nail is much smaller than the frictional resistance of the wood to the nail, (b/a) being a small ratio.
A further mathematical point is to observe that since E = (b/a) × L Then E = (b × L)/a
For a hammer with a fixed value of b and a given job to be done (L constant) the effort needed varies inversely as the length of the handle. So if the hammer handle is too short to do the job, then the solution could be to lengthen it in some way (by using a length of steel pipe?) since, by doubling the handle length, we should need only half the effort. Of course, this is only true if the machine is ideal or 100% efficient. The question of efficiency is discussed in chapter 9.
You will note that from the information given in the table 7.2 above, the M.A. of a lever used as a machine could be either greater than 1 or less than 1. See figure 7.15 for examples of levers whose mechanical advantages are shown in table 7.3.
ITQ8
The handle of a claw-hammer is 30 cm and the distance from the butt (heel) to the ‘V’ of the claw is 3 cm. What is the effort needed to draw the nail from a piece of hardwood whose frictional resistance to the nail is 150 N?
MATHEMATICS: inverse variation E L handle of hammer claw nail a b Figure 7.16
Table 7.3 Classes of lever.
Class of lever 1 2 3
Values of M.A. greater than 1 greater than 1 less than 1 Examples crow-bar, claw-
hammer nut-cracker, wheelbarrow long-handled broom, garden fork
Stability
Sometimes if we accidentally knock against a vase of flowers it does not return to rest but topples over. Why? We will find that moments and the position of the centre of mass play an important part.
Consider an object resting on a horizontal surface. Figure 7.17 illustrates the effect of tilting an object whose base is getting increasingly smaller and smaller, but whose centre of mass is always the same height above the base. As the base reduces in size, the line of action of the weight of the object moves closer to the edge about which the body is tilted.
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Figure 7.17 Toppling of tall objects.
As long as the line of action of the weight remains within the limits of the base, there is a moment exerted by the weight that tends to turn the body clockwise. This clockwise moment will have the effect of putting the body back to rest. There will be no toppling.
At positions (a) and (b) the body is able to return to rest, and so it is said to be stable. At (c), where the line of action just passes through the turning point, the body is on the point of toppling. At (d), where the line of action of the weight passes outside the turning point, the moment becomes anticlockwise and the body topples. When it is at position (d), it is unstable, because, although it is at rest while being held, it will not return to rest when it is released. Therefore, ‘stable’ means ‘able to return to rest after being slightly displaced from rest through a small angle’.
Factors that increase stability are the width of the base and the height of the centre of mass above the base. Figure 7.18 (a) shows that an object with the
stable ❯
same base size is stable when the centre of mass is low but not when the centre of mass is above a certain height. If the centre of mass is low and the object is tilted, the moment of the weight is clockwise and the object returns to its original position. If the centre of mass is high, then even a small tilt results in the turning moment of the weight being anticlockwise and the object topples. Figure 7.18 (b) shows why an object with a wide base is stable, even though the C.M. is high and why one with a narrow base is not.
(a)
(b)
W
W W W
W W
Figure 7.18 The factors that affect toppling are the position of the centre of mass and the size of
the base. A curved arrow shows the direction in which turning takes place. In (a) as the C.M. gets higher the body topples earlier; in (b) as the base gets wider the body topples later.
ITQ9
Use the diagrams of figures 7.19 (a)–(e) to explain why the rod in the position labelled (a) is much more stable than the same rod in position labelled (d)
W C (a) W W (c) α α (d) C W (e) C bench top W (b) Figure 7.19
Chapter summary
• Coplanar forces are forces whose lines of action are all in the same plane. • Coplanar forces may be parallel, anti-parallel or non-parallel.
• Anti-parallel forces are forces with parallel lines of action, but two opposite directions. • Coplanar parallel forces always have a resultant.
• Coplanar anti-parallel forces may or may not have a resultant.
– If they have a resultant, the body on which the forces act will move in the direction of the resultant.
– If they do not have a resultant, the centre of mass of the body on which they act will remain at rest.
• Rotation is caused by the turning effect of a force, called a moment.
• The moment is the product of the force and the perpendicular distance from the turning point to the line of action of the force. This perpendicular distance is called the
Answers to ITQs
ITQ2 (a) 2.5 N cm (b) 0.025 Nm
ITQ3 0.14 N. If the newton-meter is not light it will read the tension as well as its own weight, thus making the reading of the tension inaccurate.
ITQ4 R = (0.136 N × 40)15 = 0.36 N
ITQ5 R = (0.5 N × 40)55 = 0.36 N
ITQ6 6 N
ITQ7 14 cm from the 500 g mass
ITQ8 15 N
Examination-style questions
1 (i) State whether each of the set of forces shown is parallel, anti-parallel or non-parallel.
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K
L
M
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(ii) State whether equilibrium is possible or not possible in each of the set of forces shown. Give a reason for each answer.
2 (i) State the magnitude of the resultant of the forces shown.
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(ii) Show the direction of this resultant.
(iii) Show the magnitude and direction of the force that will produce equilibrium when put • If a body remains at rest under anti-parallel forces, then the resultant of the forces
acting in one direction balances the resultant of those acting in the opposite direction. • In equilibrium, the total clockwise moments of forces about any point in the plane of
the forces will balance the total anticlockwise moments.
• The stability of a body is a measure of the ability of the body to maintain its equilibrium when it is displaced to one side through a small angle.
• The stability of a body depends on:
– the height of its centre of mass above the base; – the width of its base.
• For good stability:
– the centre of mass should be as low as possible, and – the base should be as wide as possible.
3 Parallel forces of 4 N and 3 N act on a sheet of cardboard separated by a distance of 14 cm.
(i) What is the magnitude of the resultant of these forces?
(ii) Where between the two given forces must a balancing force be applied to make the resultant force on the sheet of cardboard zero?
(iii) If the resultant force did become zero what would happen to the sheet? 4 (i) State Hooke’s law.
(ii) Give an example to illustrate the meaning of Hooke’s law.
(iii) A certain spring requires an initial force of 5 N to separate the coils before the spring begins to stretch. The spring then stretches 1 cm for every 12 N of load applied. If the initial length of the spring is 40 cm, how long will the spring be when the load applied to it is 48 N?
5
5 JT
defi ne the terms distance, displacement, speed, velocity and acceleration state Aristotle’s argument in support of his hypothesis that the speed of anobject varies directly as the force acting on the object
state Newton’s three laws of motion use Newton’s laws of motion to solve problems on motion in a straight line understand the difference between distance–time and displacement–time graphs understand the difference between speed–time and velocity–time graphs use displacement–time and velocity–time graphs to solve problems defi ne linear momentum as the product of mass and linear velocity describe situations which demonstrate the laws of motion apply the laws of motion to solve problems use Newton’s laws of motion to explain the working of common dynamicalsystems
straight line motion
uniform motion uniformly accelerated motion
constant velocity changing velocity
no resultant force present
constant momentum Newton’s laws of motion changing momentum
First Law Second Law Third Law
constant resultant force present