Question

a store's sign has a mass of 20 kg and is 3.0 m long. it is uniform, so its center of gravity is at the 13) center of the sign. it is supported horizontally by a small loose bolt attached to the wall at one end and by a wire at the other end, as shown in the figure. what is the magnitude of the net force that the bolt exerts on the sign?

Answer 1

The magnitude of the net force that the bolt exerts on the sign is equal to the weight of the sign, which is 196 N.

Step-by-step explanation:

The magnitude of the net force that the bolt exerts on the sign can be calculated using the principles of equilibrium. Since the sign is uniform and the center of gravity is at its center, the weight acts downward at that point. The force at the bolt must balance the weight of the sign. Assuming no friction, the magnitude of the net force that the bolt exerts on the sign is equal to the weight of the sign.

Weight of the sign = mass * gravity = 20 kg * 9.8 m/s^2 = 196 N