Question

(a) A person opens a 1.05 m wide door by pushing on it with a force of 51.5 N applied at the center of the door, at an angle perpendicular to the door's surface. What magnitude of torque (in N m) is applied about an axis through the hinges? N m (b) A girl 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

Answer 1

Torque (t) = r*F*sin(theta)

A)-

So, r= 1.05/2 = 0.525m

Torque = 0.525 * 51.5 * sin(90) = 27.0374 Nm = 27.04Nm

B) - if r=1.05m

Then . Torque = 1.05* 51.5* sin(90) = 54.075 Nm = 54.08Nm

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