Question

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

Answer 1

Answer:

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

Answer 2

Answer:

85 N

Explanation:

Force exerted on handle (f)  = 17 N

Distance from the pivot ( d ) = 5 L

force exerted on nut ( Fnut) = ?

we assume all forces are at a right angle to the lever arms

before the nut cracks the torque around the system would be zero

torque about pivot points is represented as:

d$f_{exerted} sin 90$ = L$f_{nut} sin90$  equation 1

remember d = 5 L

equation 1 becomes

5 L $f_{exerted} sin90$ = L$f_{nut} sin90$ equation 2

dividing both sides by ( L sin90)  equation 2 becomes

5 $f_{exerted}$ = $f_{nut}$

5(17) = Fnut = 85 N