Equilibrium of a Rigid Body Chapter 10

First condition for equilibrium Net force equals zero

Second condition for equilibrium Net torque equals zero

Where is the axis of rotation? Wherever you want Be consistent Pick the easiest one – usually at the center of gravity

Center of gravity Weight acts on the entire object We can calculate the torque from weight as if it acted only on the center of gravity Same as the center of mass See book for proof

Center of Gravity

Example: Example (exercise 1): A ball with radius r1 = 0.060 m and mass m1 = 1.00 kg is attached to a light (massless) rod 0.400 m in length to a second ball with radius r2 = 0.080 m and mass m2 = 3.00 kg. Where is the center of gravity of this system? 0.4 m m2 m1

Example Example: The leaning Tower of Pisa is 55 m high and 7.0 m in diameter. The top of the tower is displaced 4.5 m from the vertical. Treat the tower as a uniform, circular cylinder. What additional displacement, measured at the top, would bring the tower to the verge of toppling?

Example Example: An automobile with a mass of 1360 kg has 3.05 m between the front and rear axles. Its center of gravity is located 1.78 m behind the front axle. With the automobile on level ground, determine the magnitude of the force from the ground on each front wheel (assuming equal forces on front wheels) and on each rear wheel (assuming equal forces on rear wheels).

You try You try: One end of a uniform beam that weighs 222 N is attached to a wall with a hinge. The other end is supported by a wire. Find the tension in the wire. 30°