Question

(a) a woman opens a 1.05 m wide door by pushing on it with a force of 45.5 n applied at the center of the door, at an angle perpendicular to the doors surface. what magnitude of torque (in n·m) is applied about an axis through the hinges? n·m (b) a second person opens the same door, using the same force, again directed perpendicular to the surface, but now the force is applied at the edge of the door. what magnitude of torque (in n·m) is applied about the axis through the hinges now? n·m

Explanation:

Step1: Recall the torque formula

The formula for torque is \(\tau = rF\sin\theta\). Since the force is perpendicular to the door's surface, \(\sin\theta = 1\).

Step2: Calculate the torque for part (a)

The width of the door \(d = 1.05\space m\). The distance from the hinges to the center of the door \(r=\frac{d}{2}=\frac{1.05}{2}= 0.525\space m\), and \(F = 45.5\space N\).
Using \(\tau = rF\), we substitute \(r = 0.525\space m\) and \(F=45.5\space N\)
\(\tau=(0.525)(45.5)\)
\(\tau = 23.9875\space N\cdot m\)

Step3: Calculate the torque for part (b)

The distance from the hinges to the edge of the door \(r = d=1.05\space m\), and \(F = 45.5\space N\)
Using \(\tau = rF\), we substitute \(r = 1.05\space m\) and \(F = 45.5\space N\)
\(\tau=(1.05)(45.5)\)
\(\tau=47.775\space N\cdot m\)

Answer:

(a) \(24.0\space N\cdot m\)
(b) \(47.8\space N\cdot m\)