Question

Block B weighs 710 N. The coefficient of static friction between the block and horizontal surface is 0.25. Calculate the maximum weight of block A for which the system will be in equilibrium.

Answer 1

The free body diagram of the system is shown as,

According to free body diagram,

`W_A=T\sin 45`

The frictional force can be given as,

`f=\mu W_B`

According to free body diagram,

`f=T\cos 45`

Plug in the known values,

`\begin{gathered} \mu W_B=T\cos 45 \\ T=(\mu W_B)/(\cos 45) \end{gathered}`

Substitute the known value in above condition,

`W_A=((\mu W_B)/(\cos45))\sin 45`

Substituting values,

`\begin{gathered} W_A=(\frac{(0.25)(710\text{ N)}}{(0.707)})(0.707) \\ =177.5\text{ N} \end{gathered}`

Therefore, the maximum weight of block A for which system will be in equilibrium is 177.5 N.