A trap door, of length and width 2.23 m, is held open at an angle of 56.5° with

Question

A trap door, of length and width 2.23 m, is held open at an angle of 56.5°
with respect to the floor. A rope is attached to the raised edge of the door and fastened to the wall behind the door in such a position that the rope pulls perpendicularly to the trap door. If the mass of the trap door is 15.6 kg,what is the torque exerted on the trap door by the rope?

A trap door, of length and width 2.23 m, is held open at an angle of 56.5°
with respect to the floor. A rope is attached to the raised edge of the door and fastened to the wall behind the door in such a position that the rope pulls perpendicularly to the trap door. If the mass of the trap door is 15.6 kg,what is the torque exerted on the trap door by the rope?

Explanation / Answer

Weight of door = 15.6 * 9.8 = 152.88 N
The weight of the door is located at the center of the door = 2.23 ÷ 2 = 1.115m from the edge of the door at an angle of 56.5°
So the perpendicular distance from the edge of the door = 1.115* cos 56.5°

Torque caused by weight of door = 152.88 * 1.115* cos 56.5°
= 94.08 N-m

Since the rope is perpendicular to the door, the torque caused by the rope = Tension in rope * 2.23 m

Torque caused by rope = torque caused by weight of door
T * 2.23 =152.88 * 1.115* cos 56.5°

Tension = 42.19 N

The answer to your question, “If the mass of the trap door is 15.6 kg, what is the torque exerted on the trap door by the rope?”, is simple.

Torque caused by rope must equal the torque caused by weight of door, because those are the only 2 torques.

Torque caused by weight of door = 152.88 * 1.115* cos 56.5°= 94.08 N-m
Torque exerted on the trap door by the rope = 94.08 N-m

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