Question

In the figure given, the system is in equilibrium. What is the maximum value that $W$ can have if the friction force on the $40N$ block cannot exceed $12.0N$ ?

• A. 3.45 N
• B. 6.92 N
• C. 10.35 N
• D. 12.32 N

Answer 1

Answer: (B)

For equilibrium at point $p$,

$T_1=T\cos 30^{\circ }$ and $T\sin 30^{\circ }=W$
$T_1=\mu R=12N$
$\therefore 12=T\cos 30^{\circ }$
$\Rightarrow T=\frac{12\times 2}{\sqrt{3}}=\frac{24}{\sqrt{3}}$
$\therefore T_1\sin 30^{\circ }=W$
$W=\frac{24}{\sqrt{3}}\times \frac{1}{2}=6.92$