EQUILIBRIUM OF RIGID BODIES
When a body is at rest without any motion on a table, we say that there is no force acting on the body. Actually it is wrong because, there is gravitational force acting on the body downward and also the normal force exerted by table on the body upward. These two forces cancel each other and thus there is no net force acting on the body. There is a lot of difference between the terms “no force” and “no net force” acting on a body. The same argument holds good for rotational conditions in terms of torque or moment of force.
A rigid body is said to be in mechanical equilibrium when both its linear momentum and angular momentum remain constant.
When the linear momentum remains constant, the net force acting on the body is zero.
In this condition, the body is said to be in translational equilibrium. This implies that the vector sum of different forces … acting in different directions on the body is zero.
If the forces … act in different directions on the body, we can resolve them into horizontal and vertical components and then take the resultant in the respective directions. In this case there will be horizontal as well as vertical equilibria possible.
Similarly, when the angular momentum remains constant, the net torque acting on the body is zero.
Under this condition, the body is said to be in rotational equilibrium. The vector sum of different torques 1 , 2 , 3 … producing different senses of rotation on the body is zero.
Thus, we can also conclude that a rigid body is in mechanical equilibrium when the net force and net torque acts on the body is zero.
As the forces and torques are vector quantities, the directions are to be taken with proper sign conventions.
Based on the above discussions, we come to a conclusion that different types of equilibrium are possible based on the different conditions. They are consolidated in Table 5.2.
Consider a thin uniform rod AB. Its center of mass is at its midpoint C. Let two forces which are equal in magnitude and opposite in direction be applied at the two ends A and B of the rod perpendicular to it. The two forces are separated by a distance of 2r as shown in Figure 5.13.
As the two equal forces are opposite in direction, they cancel each other and the net force acting on the rod is zero. Now the rod is in translational equilibrium. But, the rod is not in rotational equilibrium. Let us see how it is not in rotational equilibrium. The moment of the force applied at the end A taken with respect to the center point C, produces an anticlockwise rotation. Similarly, the moment of the force applied at the end B also produces an anticlockwise rotation. The moments of both the forces
cause the same sense of rotation in the rod. Thus, the rod undergoes a rotational motion or turning even though the rod is in translational equilibrium.
A pair of forces which are equal in magnitude but opposite in direction and separated by a perpendicular distance so that their lines of action do not coincide that causes a turning eﬀect is called a couple. We come across couple in many of our daily activities as shown in Figure 5.14.
Consider a light rod of negligible mass which is pivoted at a point along its length. Let two parallel forces F1 and F2 act at the two ends at distances d1 and d2 from the point of pivot and the normal reaction force N at the point of pivot as shown in Figure 5.15. If the rod has to remain stationary in horizontal position, it should be in translational and rotational equilibrium. Then, both the net force and net torque must be zero.
For net force to be zero, − F1 + N − F2 = 0
For net torque to be zero, d1F1 − d2 F2 = 0
The above equation represents the principle of moments. This forms the principle for beam balance used for weighing goods with the condition d1 = d2; F1 = F2. We can rewrite the equation 5.33 as,
If F1 is the load and F2 is our effort, we get advantage when, d1< d2. This implies that F1> F2. Hence, we could lift a large load with small effort. The ratio (d2/d1) is called mechanical advantage of the simple lever. The pivoted point is called fulcrum.
There are many simple machines that work on the above mentioned principle.
Each rigid body is made up of several point masses. Such point masses experience gravitational force towards the center of Earth. As the size of Earth is very large compared to any practical rigid body we come across in daily life, these forces appear to be acting parallelly downwards as shown in Figure 5.16.
The resultant of these parallel forces always acts through a point. This point is called center of gravity of the body (with respect to Earth). The center of gravity of a body is the point at which the entire weight of the body acts irrespective of the position and orientation of the body. The center of gravity and center of mass of a rigid body coincide when the gravitational field is uniform across the body. The concept of gravitational field is dealt in Unit 6.
We can also determine the center of gravity of a uniform lamina of even an irregular shape by pivoting it at various points by trial and error. The lamina remains horizontal when pivoted at the point where the net gravitational force acts, which is the center of gravity as shown in Figure 5.17. When a body is supported at the center of gravity, the sum of the torques acting on all the point masses of the rigid body becomes zero. Moreover the weight is compensated by the normal reaction force exerted by the pivot. The body is in static equilibrium and hence it remains horizontal.
There is also another way to determine the center of gravity of an irregular lamina. If we suspend the lamina from different points like P, Q, R as shown in Figure 5.18, the vertical lines PP', QQ', RR' all pass through the center of gravity.
Here, reaction force acting at the point of suspension and the gravitational force acting at the center of gravity cancel each other and the torques caused by them also cancel each other.
Let us consider a cyclist negotiating a circular level road (not banked) of radius r with a speed v. The cycle and the cyclist are considered as one system with mass m. The center gravity of the system is C and it goes in a circle of radius r with center at O. Let us choose the line OC as X-axis and the vertical line through O as Z-axis as shown in Figure 5.19.
The system as a frame is rotating about Z-axis. The system is at rest in this rotating frame. To solve problems in rotating frame of reference, we have to apply a centrifugal force (pseudo force) on the system which will be mv2/r
This force will act through the center of gravity. The forces acting on the system are, (i) gravitational force (mg), (ii) normal force (N), (iii) frictional force (f) and (iv) centrifugal force (mv2/r) . As the system is in equilibrium in the rotational frame of reference, the net external force and net external torque must be zero. Let us consider all torques about the point A in Figure 5.20.
For rotational equilibrium,
The torque due to the gravitational force about point A is ( mgAB) which causes a clockwise turn that is taken as negative. The torque due to the centripetal force is ( mv2/r BC) which causes an anticlockwise turn that is taken as positive.
negotiating a circular level road of radius r at velocity v, a cyclist has to
bend by an angle θ from vertical given by the above
expression to stay in equilibrium (i.e. to avoid a fall).
Solved Example Problems for Types of Equilibrium
Arun and Babu carry a wooden log of mass 28 kg and length 10 m which has almost uniform thickness. They hold it at 1 m and 2 m from the ends respectively. Who will bear more weight of the log? [g = 10 ms-2]
Let us consider the log is in mechanical equilibrium. Hence, the net force and net torque on the log must be zero. The gravitational force acts at the center of mass of the log downwards. It is cancelled by the normal reaction forces RA and RB applied upwards by Arun and Babu at points A and B respectively. These reaction forces are the weights borne by them.
The total weight, W = mg = 28 × 10 = 280 N, has to be borne by them together. The reaction forces are the weights borne by each of them separately. Let us show all the forces acting on the log by drawing a free body diagram of the log.
The net force acting on the log must be zero.
Here, the forces RA an RB are taken positive as they act upward. The gravitational force acting downward is taken negative.
The net torque acting on the log must be zero. For ease of calculation, we can take the torque caused by all the forces about the point A on the log. The forces are perpendicular to the distances. Hence,
Here, the reaction force RA cannot produce any torque as the reaction forces pass through the point of reference A. The torque of force mg produces a clockwise turn about the point A which is taken negative and torque of force RB causes anticlockwise turn about A which is taken positive.
As RB is greater than RA, it is concluded that Babu bears more weight than Arun. The one closer to center of mass of the log bears more weight.
Solved Example Problems for Bending of Cyclist in Curves
A cyclist while negotiating a circular path with speed 20 m s-1 is found to bend an angle by 30o with vertical. What is the radius of the circular path? (given, g = 10 m s-2)
Speed of the cyclist, v = 20 m s-1
Angle of bending with vertical, θ = 30o