91Ó°ÊÓ

Weight of the block B,$\mathrm{W}=711\mathrm{N}$

Coefficient of static friction between block and table,${\mathrm{μ}}_{\mathrm{s}}=0.25$

Angle,$\mathrm{θ}={30}^0$

To find the weight of the block A, use Newton's 2nd law for the stationary system. According to Newton's 2nd law of motion, a force applied to an object at rest causes it to accelerate in the direction of the force.

Formula:

$F_{net}=∑ma$

where, F is the force, m is mass and a is an acceleration.

Free Body Diagram of the system:

By using Newton’s 2nd law of motion along the vertical direction to the block B,

$$\begin{gathered} N-W=0 \\ N=W \end{gathered}$$

Thus, the frictional force,

$$\begin{gathered} f_s={\mathrm{μ}}_{s}N \\ ={\mathrm{μ}}_{s}W \end{gathered}$$

Now, applying Newton’s 2nd along horizontal direction to the block B,

Using the Newton’s 2nd law along vertical direction to the block A,

Thus, the maximum weight of block A for which the system will be stationary is $102\mathrm{N}$.