Question

The handle of a nutcracker is 16cm long and a nut is placed 2cm from its hinge. If a force of 4kgf is applied at the end of the handle to crack it, what weight if simply, placed on the nut will crack it.

Answer 1

1. Formula used:

Mechanical advantage = $\frac{Effort}{Load}$ = $\frac{Loadarm}{Effortarm}$

2. Solution:

Given that,

length of the nutcracker handle = 16cm

distance of nut from hinge = 2cm

effort = 4Kgf

let the load be L

then, L x load arm = E x effort arm

L x 2 = 4 x 16

L = 32Kgf

3. Conclusion:

Hence, the value of load = 32Kgf