The free body diagram of the system is shown as,
According to free body diagram,
[tex]W_A=T\sin 45[/tex]
The frictional force can be given as,
[tex]f=\mu W_B[/tex]
According to free body diagram,
[tex]f=T\cos 45[/tex]
Plug in the known values,
[tex]\begin{gathered} \mu W_B=T\cos 45 \\ T=\frac{\mu W_B}{\cos 45} \end{gathered}[/tex]
Substitute the known value in above condition,
[tex]W_A=(\frac{\mu W_B}{\cos45})\sin 45[/tex]
Substituting values,
[tex]\begin{gathered} W_A=(\frac{(0.25)(710\text{ N)}}{(0.707)})(0.707) \\ =177.5\text{ N} \end{gathered}[/tex]
Therefore, the maximum weight of block A for which system will be in equilibrium is 177.5 N.
