Equilibrium of a Rigid Body

Lab# 13 Equilibrium of a Rigid Body

Figure 1: Equipment for the Equilibrium of a Rigid Body experiment showing a meter stick, which acts asour rigid body, with the knife edges (which also are the mass positioners), knife edge support, masses and mass hangers.

Introduction:The objective of this experiment is to investigate and understand the conditions for the equilibrium of a rigid body. In particular, we wish to understand that the equilibrium of an extended object depends not only on the magnitudes and directions of the forces being applied, but also where on the object these forces act (see Figures Figure 1 and Figure 2).

Equipment: One meter stick, 3 knife edges/mass positioners, 1 knife edge support stand, set of masses, 2 mass hangers, 1 body of unknown mass.

Theory:An object in equilibrium has no net influences to cause it to move, either in translation (linear motion) or rotation. In other words, the object experiences no net force acting on it and no net torque. Thus, the two conditions for a rigid body to be in equilibrium are:

Fi 0,ithe resultant of all the forces acting on the body must equal zero. And,

i 0,ithe resultant torque acting on a body (computed about any axis) must equal zero.

(1)

(2)

As discussed in your textbook, a torque may be identified with the ability of a force to produce rotation. That is, an object may be fixed in position, but free to rotate about some axis. A merry- go-round, for example, is fixed in position so it can not move across a playground, but is free to rotate about an axis through its center. A force acting on the merry-go-round will cause it to rotate, so long as the line of action of the force does not pass through the merry-go-rounds axis of rotation. It seems than that torque depend on three factors:1. The magnitude of the force.

2. The direction of the force.3. The point of application of the force on a rigid body relative to the axis of rotation.

These factors may be combined in the equation:(rsin)F,

(3)

where is the magnitude of the torque produced by the force F, F is the force, r is the vector from the axis of rotation to the point of application of the force and is the angle between the force F and the vector r (see Figure 2).

Figure 2: Applied force F, its line of action, perpendicular component of F and the perpendicular distancefrom axis to line of action of F.

Thus the torque would take on its maximum value when = 90 so that sin = 1. This is related to the fact that a force whose line of action passes through the axis of rotation produces no rotation, so only the component of the force perpendicular to the vector r (which passes through the axis of rotation) produces a torque about that axis. In equation (3) the combination Fsin() pulls out the component of F perpendicular to the vector r. Notice, one may also think in terms of the perpendicular distance between the axis of rotation and the line of action of the force. That distance is rsin(), as shown in Figure 2.

It is convenient to regard a torque which would produce rotation in a counter-clockwise direction as (+) and a torque which produces rotation in a clockwise direction as (-). Suppose a bar, a meter stick, for example, is supported at some point along its length, and suitable weights suspended at other points along the meter stick. There will be some combinations of weights and positions for which the stick will be in equilibrium in a horizontal position. The weight of the bar acts at its center of gravity and produces an effect similar to any other weight.

(Hint: Fill in the open circles in front of the procedure steps to help you perform the experiment)Procedure: