$We are now going to concentrate on the motion of rigid bodies rather than on the motion of general systems of particles.
We shall recapitulate what effect the external forces have on a rigid body. (Henceforth we shall omit the adjective ‘external’ because unless stated otherwise, we shall deal with only external forces and torques.) The forces change the translational state of the motion of the rigid body, i.e. they change its total linear momentum in accordance with Eq. (7.17). But this is not the only effect the forces have. The total torque on the body may not vanish. Such a torque changes the rotational state of motion of the rigid body, i.e. it changes the total angular momentum of the body in accordance with Eq. (7.28 b).
A rigid body is said to be in mechanical equilibrium, if both its linear momentum and angular momentum are not changing with time, or equivalently, the body has neither linear acceleration nor angular acceleration. This means
(1) the total force, i.e. the vector sum of the forces, on the rigid body is zero;
$\mathbf{F_1}+\mathbf{F_1}+...+\mathbf{F_n}=\Sigma_{i=1}^{n}\mathbf{F_i}=0$
If the total force on the body is zero, then the total linear momentum of the body does not change with time. Eq. (7.30a) gives the condition for the translational equilibrium of the body.
(2) The total torque, i.e. the vector sum of the torques on the rigid body is zero,
τ$_1$ + τ$_2$ + ... + τ$_n=\Sigma_{i=1}^{n}$ τ$_i$ = 0
If the total torque on the rigid body is zero, the total angular momentum of the body does not change with time. Eq. (7.30 b) gives the condition for the rotational equilibrium of the body.
One may raise a question, whether the rotational equilibrium condition [Eq. 7.30(b)] remains valid, if the origin with respect to which the torques are taken is shifted. One can show that if the translational equilibrium condition [Eq. 7.30(a)] holds for a rigid body, then such a shift of origin does not matter, i.e. the rotational equilibrium condition is independent of the location of the origin about which the torques are taken. Example 7.7 gives a proof of this result in a special case of a couple, i.e. two forces acting on a rigid body in translational equilibrium. The generalisation of this result to n forces is left as an exercise.
Eq. (7.30a) and Eq. (7.30b), both, are vector equations. They are equivalent to three scalar equations each. Eq. (7.30a) corresponds to
$\Sigma_{i = 1}^{n}F_{ix}=0$; $\Sigma_{i = 1}^{n}F_{iy}=0$ and $\Sigma_{i = 1}^{n}F_{iz}=0$
where $F_{ix}$, $F_{iy}$ and $F_{iz}$ are respectively the x, y and z components of the forces $\mathbf{F_i}$. Similarly, Eq. (7.30b) is equivalent to three scalar equations
$\Sigma_{i = 1}^{n}\tau_{ix}=0$; $\Sigma_{i = 1}^{n}\tau_{iy}=0$ and $\Sigma_{i = 1}^{n}\tau_{iz}=0$
where $\tau_{ix}$, $\tau_{iy}$ and $\tau_{iz}$ are respectively the x, y and z components of the torque τ$_i$.
Eq. (7.31a) and (7.31b) give six independent conditions to be satisfied for mechanical equilibrium of a rigid body. In a number of problems all the forces acting on the body are coplanar. Then we need only three conditions to be satisfied for mechanical equilibrium. Two of these conditions correspond to translational equilibrium; the sum of the components of the forces along any two perpendicular axes in the plane must be zero. The third condition corresponds to rotational equilibrium. The sum of the components of the torques along any axis perpendicular to the plane of the forces must be zero.
The conditions of equilibrium of a rigid body may be compared with those for a particle, which we considered in earlier chapters. Since consideration of rotational motion does not apply to a particle, only the conditions for translational equilibrium (Eq. 7.30 a) apply to a particle. Thus, for equilibrium of a particle the vector sum of all the forces on it must be zero. Since all these forces act on the single particle, they must be concurrent. Equilibrium under concurrent forces was discussed in the earlier chapters.
A body may be in partial equilibrium, i.e., it may be in translational equilibrium and not in rotational equilibrium, or it may be in rotational equilibrium and not in translational equilibrium.
Consider a light (i.e. of negligible mass) rod (AB) as shown in Fig. 7.20(a). At the two ends (A and B) of which two parallel forces, both equal in magnitude and acting along same direction are applied perpendicular to the rod.
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| Fig.1 |
Let C be the midpoint of AB, CA = CB = a. the moment of the forces at A and B will both be equal in magnitude (aF ), but opposite in sense as shown. The net moment on the rod will be zero. The system will be in rotational equilibrium, but it will not be in translational equilibrium;
$\Sigma \mathbf{F} \neq 0$
The force at B in Fig. 1 is reversed in Fig. 2. Thus, we have the same rod with two forces of equal magnitude but acting in opposite diretions applied perpendicular to the rod, one at end A and the other at end B. Here the moments of both the forces are equal, but they are not opposite; they act in the same sense and cause anticlockwise rotation of the rod. The total force on the body is zero; so the body is in translational equilibrium; but it is not in rotational equilibrium. Although the rod is not fixed in any way, it undergoes pure rotation (i.e. rotation without translation).
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| Fig.2 |
A pair of forces of equal magnitude but acting in opposite directions with different lines of action is known as a couple or torque. A couple produces rotation without translation.
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| Fig.3: Our fingers apply a couple to turn the lid. |
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| Fig.4 |
Fig.4: The Earth’s magnetic field exerts equal and opposite forces on the poles of a compass needle. These two forces form a couple.
Example 1
Show that moment of a couple does not depend on the point about which you take the moments.
Answer
Consider a couple as shown in Fig. 5 acting on a rigid body. The forces F and -F act respectively at points B and A. These points have position vectors r$_1$ and r$_2$ with respect to origin O. Let us take the moments of the forces about the origin.
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| Fig.5 |
The moment of the couple = sum of the moments of the two forces making the couple
= r$_1$ × (–F) + r$_2$ × F
= r$_2$ × F – r$_1$ × F
= (r$_2$ – r$_1$) × F
But r$_1$ + AB = r$_2$, and hence AB = r$_2$ – r$_1$. The moment of the couple, therefore, is AB × F.
Clearly this is independent of the origin, the point about which we took the moments of the forces.
Principle of moments
An ideal lever is essentially a light (i.e. of negligible mass) rod pivoted at a point along its length. This point is called the fulcrum. A seesaw on the children’s playground is a typical example of a lever. Two forces $F_1$ and $F_2$, parallel to each other and usually perpendicular to the lever, as shown here, act on the lever at distances $d_1$ and $d_2$ respectively from the fulcrum as shown in Fig. 6.
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| Fig.6 |
The lever is a system in mechanical equilibrium. Let R be the reaction of the support at the fulcrum; R is directed opposite to the forces $F_1$ and $F_2$. For translational equilibrium,
R – $F_1$ – $F_2$ = 0
For considering rotational equilibrium we take the moments about the fulcrum; the sum of moments must be zero,
$d_1F_2$ – $d_2F_2$ = 0
Normally the anticlockwise (clockwise) moments are taken to be positive (negative). Note R acts at the fulcrum itself and has zero moment about the fulcrum.
In the case of the lever force $F_1$ is usually some weight to be lifted. It is called the load and its distance from the fulcrum $d_1$ is called the load arm. Force $F_2$ is the effort applied to lift the load; distance $d_2$ of the effort from the fulcrum is the effort arm.
Eq. (ii) can be written as
$d_1F_2$ = $d_2F_2$ (7.32a)
or load arm × load = effort arm× effort
The above equation expresses the principle of moments for a lever. Incidentally the ratio $F_1$/$F_1$ is called the Mechanical Advantage (M.A.);
$M.A=\frac{F_1}{F_2}=\frac{d_2}{d_1}$
If the effort arm $d_2$ is larger than the load arm, the mechanical advantage is greater than one. Mechanical advantage greater than one means that a small effort can be used to lift a large load. There are several examples of a lever around you besides the see-saw. The beam of a balance is a lever. Try to find more such examples and identify the fulcrum, the effort and effort arm, and the load and the load arm of the lever in each case.
You may easily show that the principle of moment holds even when the parallel forces $F_1$ and $F_2$ are not perpendicular, but act at some angle, to the lever.
Example 1
A metal bar 70 cm long and 4.00 kg in mass supported on two knife-edges placed 10 cm from each end. A 6.00 kg load is suspended at 30 cm from one end. Find the reactions at the knifeedges. (Assume the bar to be of uniform cross section and homogeneous.)
Answer
Fig.7
Figure 7 shows the rod AB, the positions of the knife edges $K_1$ and $K_2$, the centre of gravity of the rod at G and the suspended load at P.
Note the weight of the rod W acts at its centre of gravity G. The rod is uniform in cross section and homogeneous; hence G is at the centre of the rod; AB = 70 cm. AG = 35 cm, AP = 30 cm, PG = 5 cm, A$K_1$ = B$K_2$ = 10 cm and $K_1G$ = $K_2G$ = 25 cm. Also, W= weight of the rod = 4.00 kg and $W_1$ = suspended load = 6.00 kg; $R_1$ and $R_2$ are the normal reactions of the support at the knife edges.
For translational equilibrium of the rod,
$R_1$ + $R_1$ – $W_1$ – $W$ = 0 (i)
Note $K_1$ and W act vertically down and $R_1$ and $R_2$ act vertically up.
For considering rotational equilibrium, we take moments of the forces. A convenient point to take moments about is G. The moments of $R_2$ and $K_1$ are anticlockwise (+ve), whereas the moment of $R_1$ is clockwise (-ve).
For rotational equilibrium,
–$R_1$($K_1G$) + $W_1$(PG) + $R_2$($K_2G$) = 0 (ii)
It is given that W = 4.00g N and $W_1$ = 6.00g N, where g = acceleration due to gravity. We take g = 9.8 $m/s^2$.
With numerical values inserted, from (i)
$R_1$ + $R_2$ – 4.00g – 6.00g = 0
or $R_1$ + $R_2$ = 10.00g N (iii)
= 98.00 N From (ii),
– 0.25$R_1$ + 0.05$W_1$ + 0.25$R_2$ = 0
or $R_1$ – $R_2$ = 1.2g
N = 11.76 N (iv)
From (iii) and (iv),
$R_1$ = 54.88 N, $R_2$ = 43.12 N
Thus the reactions of the support are about 55 N at $K_1$ and 43 N at $K_2$.
Example 2
A 3m long ladder weighing 20 kg leans on a frictionless wall. Its feet rest on the floor 1 m from the wall as shown in Fig.8. Find the reaction forces of the wall and the floor.
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| Fig.8 |
The ladder AB is 3 m long, its foot A is at distance AC = 1 m from the wall. From Pythagoras theorem, BC = 2 2 m. The forces on the ladder are its weight W acting at its centre of gravity D, reaction forces $F_1$ and $F_2$ of the wall and the floor respectively. Force $F_1$ is perpendicular to the wall, since the wall is frictionless. Force $F_2$ is resolved into two components, the normal reaction N and the force of friction F. Note that F prevents the ladder from sliding away from the wall and is therefore directed toward the wall.
For translational equilibrium, taking the forces in the vertical direction,
N – W = 0 (i)
Taking the forces in the horizontal direction,
$F$ – $F_1$ = 0 (ii)
For rotational equilibrium, taking the moments of the forces about A,
$2\sqrt{2}F_1$ − $(1/2)W$= 0
Now W = 20 g = 20 × 9.8 N = 196.0 N
From (i) N = 196.0 N
From (iii) $F_1=W/4\sqrt{2}=196/4\sqrt{2}$ = 34.6 N
From (ii) $F=F_1$ = 34.6 N
$F_2=\sqrt{F^2+N^2}$
The force $F_2$ makes an angle α with the horizontal,
$tan \ \alpha = N/F=4\sqrt{2}$
$\alpha = tan^{-1}(4\sqrt{2}) \approx 80^0$
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