Question

problem 5:
a person pulls open a door. the force on the door is 18 n. the distance from the hinge to the handle where the force is applied is 76 cm. the angle between the force and the displacement vector (from hinge to handle) is 35 degrees. find the torque acting on the door.

Explanation:

Step1: Convert the distance to meters

The formula for torque is \(\tau = rF\sin\theta\). Given \(r = 76\space cm=0.76\space m\), \(F = 18\space N\), \(\theta = 35^{\circ}\)

Step2: Calculate the torque

Substitute the values into the formula \(\tau=(0.76\space m)\times(18\space N)\times\sin(35^{\circ})\)
First, find \(\sin(35^{\circ})\approx0.5736\)
Then \(\tau = 0.76\times18\times0.5736\)
\(\tau=0.76\times10.3248\)
\(\tau = 7.846848\space N\cdot m\approx7.8\space N\cdot m\)

Answer:

The torque acting on the door is approximately \(7.8\space N\cdot m\)