When you push down on the handle of a doll-like wooden nutcracker, its jaw pivots upward and cracks a nut. If the point at which you push down on the handle is 5 times as far from the pivot as the point at which the jaw pushes on the nut, how much force will the jaw exert on the nut if you exert a force of 17 N on the handle

The force exerted by the jaw on nut is 85 N

Explanation:

The wooden nutcracker forms an angle 90° with the jaw pivot to crash a nut

The net torque on this system = 0

we can say distance of force applied sin90° = length of force on nut sin90°

Note that distance of the applied force to the nut is 5 times = 5L

dFapp(sin90) = LFnuts(in90)

but sin90 = 1

and distance d = 5L

and exerted force = 17N

substituting for d, we will have

5LF = LF

divide both sides by L and we will have

5Fapp = Fnut

F on nut = 5*17

Force exerted by jaw on nut = 85 N

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