Question

The moment of force required to open a door of width 50cm is 100Nm, what is the minimum force required to open it​

Answer 1

Answer:

200 N

Explanation:

Min force is applied on the door at its farthest end from the pivot point about which it rotates.

Since the width of the door is 50 cm min force will be applied at a perpendicular distance of 50 cm from the axis of rotation.

Torque = Force × Perpendicular distance

100 Nm = force ( min) × 50/ 100 m

100 = f( min) × 1/ 2

F min = 200 N

Therefore, the min force applied on the door to produce the required torque is 200 N.

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Hope this helps!