# Rotational equilibrium: Force and its moment (torque)

The figure shows a person whose weight is W = 644 N doing push-ups. Find the normal force exerted by the floor on (a)each hand and (b)each foot, assuming that the person holds this position.

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Push Up

The figure shows a person whose weight is W = 644 N doing push-ups. Find the normal force exerted by the floor on (a)each hand and (b)each foot, assuming that the person holds this position.

For translation equilibrium of the person the net force on it must be Zero.

Let the force on the hands is F1 and at the feet is F2.

The resultant force on the person is given by

Gives ------------------------------------ (1)

For rotational equilibrium of the person the net torque him must be equal to zero.

The magnitude of torque of a force is given by the product of the force and the perpendicular distance from the axis of rotation (arm length).

Taking torque about the center of mass of the person we get

Or

Or

Or

Substituting the value of F1 in equation (1) we get

Or F2 = 644/3.05 = 211 N

And hence

F1 = 2.05*F2 = 211*2.05 = 433 N

Thus the force on each hand will be F1/2 = 433/2 = 216.5 N

And the force on each foot will be F2/2 = 211/2 = 105.5 N

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