Question

A person exerts a horizontal force of 82 N on the end of a door 82 cmwide.Part AWhat is the magnitude of the torque if the force is exerted perpendicular to the door? Part B What is the magnitude of the torque if the force is exerted at a 60 degrees angle to the face of the door

Answer 1

`\begin{gathered} A)67.24Nm \\ B)58.23Nm \end{gathered}`

Step-by-step explanation

Parameters given:

Force exerted on door, F = 82N

Width of door, r = 82 cm = 0.82 m

A) Torque is described as a force that tends to cause rotation and it is given mathematically as:

`\tau=r\cdot F\cdot\sin \theta`

where F = force, r = radius, θ = angle between the force and the arm of the lever.

From the question, the radius r is 82 cm (0.82 m) since the door pivots at its edge and because the force is acting perpendicularly, the angle θ is 90 degrees.

Therefore, the torque is:

`\begin{gathered} \tau=0.82\cdot82\cdot\sin 90 \\ \tau=67.24Nm \end{gathered}`

B) Now, the force is acting at an angle of 60 degrees to the face of the door.

This means that θ is 60 degrees.

Therefore, the torque is:

`\begin{gathered} \tau=0.82\cdot82\cdot\sin 60 \\ \tau=58.23Nm \end{gathered}`