Question

The sign in the figure weighs 200 N. The boom is of uniform construction. If the force exerted by the hinge on the boom is 300 N and acts at an angle of 20º above the horizontal, the weight of the boom is

A. ~250 N

B. ~540 N

C. ~460 N

D. ~413 N

E.None of these is correct.

Answer 1

Torque = Force * distance * sin θ
The cable is attached to the end of the boom, so distance = L
Torque caused by tension in cable = T * L

The 200 N weight is attached to the end of the boom, so distance = L
Torque caused by tension in cable = 200 * L * sin 20° clockwise

The 300 N weight of the boom is at the center of the boom so distance = L/2
Torque caused by weight of boom = 300 * ½ * L * sin 20° clockwise

Torque counter clockwise = torque clockwise

T * L = 300 * ½ * L * sin 20° + 200 * L * sin 20°
L cancels out

T = 150 * sin 20° + 200 * sin 20° = 350 * sin 20°

T = 119.7 N

now

Horizontal component of the tension = T * cos 20°

= 119.7*cos20^o

T = 112.48 N

total weight is 200+119+112

= 431N

so nearly equal to 413 N

so option (D) is correct.