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 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 force that the wire exerts on the sign?

Question

Quizplus · Accepted Answer

To find the force that the wire exerts on the sign, we need to consider the forces acting on the sign. There are two forces acting on the sign - the weight (due to gravity) and the tension force by the wire.
We can first find the weight of the sign:
Weight = mass x gravitational acceleration
Weight = 20 x 9.81 = 196.2 N
Since the sign is uniform, the center of gravity is at the center of the sign, which is at 1.5 m from either end.
To find the tension force by the wire, we can take moments about the bolt:
Clockwise moment due to weight = Weight x distance from bolt to center of gravity = 196.2 x 1.5 = 294.3 Nm
Counter-clockwise moment due to tension force = tension force x distance from wire to center of gravity = T x 1.5
At equilibrium, the moments must be equal, so:
T x 1.5 = 294.3
T = 196.2 N
Therefore, the force that the wire exerts on the sign is 196.2 N, which is closest to option B) 230 N.

A Store's Sign Has a Mass of 20 Kg and Is