Question

The block $A$ in figure weighs $100N$. 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
$F_S={\mu }_{S}R={\mu }_S{W}_A=0.25\times 100=25N$
For equilibrium of block $A$,
$T_1=F_S=25N\dots (i)$
and for equilibrium of block $B$
$T_1=T_2\cos 45^{\circ }$ and $W_B=T_2\sin 45^{\circ }\dots$..(ii)
From Eq (ii)
$W_B=T_2\sin 45^{\circ }=\left(\frac{T_1}{\cos 45^{\circ }}\right)\sin 45^{\circ }$
$=T_1\tan 45^o=T_1$
$\therefore W_B=T_1=25N$