Question

The length of an iron bar is 1.9m. the distance of the hand twisting the bar from the support is 1.6m. The distance of the force applied on the rock to be twisted is 0.3m. How much force is applied on the rock if the tensional force is 450N.

Answer 1

$\vec{N}=450N$

$l_1 =1.6m$

$l_2 = 0.3m$

$F_1*l_1 = F_2*l_2$

$1.6F_1= 0.3F_2$

$F_1+F_2 = |\vec N|$

$F_1 = 450-F_2$

$1.6*(450-F_2)= 0.3F_2$

$F_2\approx378.95 N$

$\text{Answer: }378.95\ N$