A person exerts a force of 50 N on the end of a door 71 cm wide. (a) What is the magnitude of the torque if the force is exerted perpendicular to the door? (b) What is the magnitude of the torque if the force is exerted at a 45 degree angle to the face of the door?

Question

Karli Downs

Physics

6 August, 01:53

A person exerts a force of 50 N on the end of a door 71 cm wide. (a) What is the magnitude of the torque if the force is exerted perpendicular to the door? (b) What is the magnitude of the torque if the force is exerted at a 45 degree angle to the face of the door?

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Answers (1)

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Kiley · Answer 1

a) 35.5

b) 25.1

Explanation:

Torque=distance * (perpendicular component of the force to the distance)

a) force=50 N distance=0,71m (we need the distance in meters to have the same units with the Newton)

T=50*0.75

b) only one component of the force makes torque, so we multiply all the 50N with sin (45) to find the perpendicular component to the distance

T=50*sin (45) * 0.71