Ajanitor opens a 1.10 m wide door by pushing on it with a force of 47.5 n directed perpendicular to its surface. hint (a) what magnitude torque (in n · m) does he apply about an axis through the hinges if the force is applied at the center of the door? (b) what magnitude torque (in n · m) does he apply at the edge farthest from the hinges?

The magnitude of torque applied about the axis of the door when the force is applied at the center is 26.125 N.m

The magnitude of the torque applied at the edge farthest from the center  is 52.25 N.m.

The given parameters;

The torque applied by the janitor is the product of applied force and perpendicular distance.

τ = F.r

The torque applied about the axis of the door when the force is applied at the center.

The magnitude of the torque applied at the edge farthest from the center ;

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(a) 26.1 Nm

The magnitude of the torque exerted by a force acting perpendicularly to a surface is given by:

where

F is the magnitude of the force

r is the distance from the pivot

In this situation,

F = 47.5 N is the force applied

is the distance from the hinges (the force is applied at the center of the door)

So, the magnitude of the torque is

(b) 52.3 Nm

In this case, the force is applied at the edge of the door farthest from the hinges. This means that the distance from the hinges is

r = 1.10 m

So, the magnitude of the torque is

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explanation: