Question

The block $A$ in figure weighs $100 \,N$. The coefficient of static friction between the block and table is $0.25$. The maximum weight of block $B$ for which the system is in equilibrium is

• A. 25 N
• B. 30 N
• C. 35 N
• D. 40 N

Answer 1

Answer: (A)

The frictional force $F_{S}$ on block A, when it tends to move is
$F_{S}=\mu_{S} R=\mu_{S} W_{A}=0.25 \times 100=25 \,N$
For equilibrium of block $A$,
$T_{1}=F_{S}=25 N \ldots(i) $
and for equilibrium of block $B$
$T_{1}=T_{2} \cos 45^{\circ}$ and $W_{B}=T_{2} \sin 45^{\circ} \ldots$..(ii)
From Eq (ii)
$W_{B}=T_{2} \sin 45^{\circ}=\left(\frac{T_{1}}{\cos 45^{\circ}}\right) \sin 45^{\circ}$
$=T_{1} \tan 45^{o}=T_{1}$
$\therefore W_{B}=T_{1}=25\, N$