In figure, the tension in the diagonal string is 60N. What is the weight of the suspended block ?

 At P

$F_2=T\cos 45°=\frac{T}{\sqrt{2}}$        ………….(i)

$w=T\cos 45°=\frac{T}{\sqrt{2}}$               ………..(ii)

At $Q. F_1=\frac{T}{\sqrt{2}}$               ………..(iii)

From these three equations, we can see that

$$\begin{gathered} F_1=F_2=w=\frac{T}{\sqrt{2}} \\ =\frac{60}{\sqrt{2}}=30\sqrt{2} \mathrm{N} \end{gathered}$$