Question

A woman opens a 1.15 m wide door by pushing on it with a force of 55.5 N directed perpendicular to its surface.

HINT

(a)

What magnitude torque (in N · m) does she apply about an axis through the hinges if the force is applied at the center of the door?

N · m

(b)

What magnitude torque (in N · m) does she apply at the edge farthest from the hinges?

N · m

Answer 1

An axis passes through the hinges is the axis of rotation.

Torque = T = rF (since, r and F are perpendicular),

r = distance of point of application of force F = 55.5 N from axis of rotation.

For center of the door, r = ( 1.15 / 2 ) m,

and for the farthest edge from the hinges of the door, r = 1.15 m.

(a) Torque at the centre of the door = T = ( 1.15 / 2 ) * 55.5 N.m = 31.9 N.m

(b) Torque at the farthest edge from the hinges of the door = T = ( 1.15 * 55.5 ) N.m = 63.8 N.m