Question

A 43-N crate is suspended from the left end of a plank. The plank weighs 21 N, but it is not uniform, so its center of gravity does not lie halfway between the two ends. The system is balanced by a support that is 0.30 m from the left end of the plank. How far to the right of the support is the plank

Answer 1

Answer:

$d_{2}=0.585m$

Explanation:

The torque is the force by the distance so to determinate that both torque are the same magnitude so

$T_{1}-T_{2}=0$

$T_{1}=T_{2}$

$F_{1}*d_{1}=F_{2}*d_{2}$

$43N*0.30m=21N*d_{2}$

Solve to d2

$d_{2}=\frac{41N*0.30m}{21N}$

$d_{2}=0.585m$