Equilibrium of a Rigid Body

Chapter 10

First condition for equilibriumNet force equals zero

0

0

x

y

F

F

Second condition for equilibrium

Net torque equals zero

0

about any axis

Where is the axis of rotation?Wherever you wantBe consistentPick the easiest one – usually at the

center of gravity

Center of gravityWeight acts on the entire objectWe can calculate the torque from weight

as if it acted only on the center of gravity

Same as the center of massSee book for proof

Center of Gravity

1 1 2 2

1 2

...

...

wx wxw x w xX

w w w W

1 1 2 2

1 2

...

...

wy wyw y w yY

w w w W

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

1. 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

1. 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 tryYou 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°

30°