Question

In figure, the bar is uniform and weighing $500\text{N}$. How large must $W\text{in N}$ be if $T_1$ and $T_2$ are to be equal?

Answer 1

Taking torque about the attachment point for $W$ , we get
$-T_1\left(0.4L\right)+T_2\left(0.3L\right)+500\left(0.2L\right)=0$
where $T_1=T_2=T$
$T\left(0.1\right)=100\Rightarrow T=1000N$
By balancing forces
$\sum F_y=0\Rightarrow 2T-W-500=0$
$\Rightarrow W=1500N$