Question

You are trying to tilt a very tall refrigerator (20 m high, 1.0 m deep, 1.4 m wide, and 150 kg) so that your friend can put a
Part A Determine the force that you need to exert on the front of the refrigerator at the start of its tipping. You push hori
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Answer #1

Given is:-

Horizontal length to move L = 1.5 m

The Free body diagram of the given problem is as shown below:

Fridge Axis of rotation

The axis of rotation is along one edge of the refrigerator.

Thus the expression of torque will be

T = Fj(0) – F, C5) + Fapp(y)

by equating it to zero, because the refrigerator does not move

thus

Fapp = เเล้

or

Fapp = (mgd 24

by plugging all the values we get

Fapp = ((150kg) (9.8m/s))(1m) 2(1.5m)

which gives us

Fapp = 490N

This is the required amount of force.

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