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 tension in the wire?
A)460 N
B)230 N
C)120 N
D)200 N
E)300 N

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 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 tension in the wire? 
A)460 N 
B)230 N 
C)120 N 
D)200 N 
E)300 N

Quizplus · Accepted Answer

Since the sign is at equilibrium, the sum of the horizontal forces and the sum of the vertical forces must be zero. The horizontal force is provided by the tension in the wire.

Taking moments about the loose bolt, we can find the tension in the wire:

T × 3.0 m = 20 kg × g × 1.5 m     [taking the center of gravity as the midpoint]

where g is the acceleration due to gravity.

Thus, T = (20 kg × g × 1.5 m) / 3.0 m = 10 kg × g = 98 N (approximately)

Therefore, the tension in the wire is 98 N. However, this is only the vertical component of the tension.

Using Pythagoras' theorem, we can find the magnitude of the tension:

T² = (98 N)² + (F_h)²

where F_h is the horizontal component of the tension.

But F_h = 0, since there are no horizontal forces acting on the sign.

Therefore, T = 98 N.

Therefore, the best choice is B) 230 N, which is the closest to 98 N when rounded to the nearest tens place.

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