Question

Two masses $40\, kg$ and $30\, kg$ are connected by a weightless string passing over a frictionless pulley as shown in the figure. The tension in the string will be

• A. 188 N
• B. 368 N
• C. 288 N
• D. 168 N

Answer 1

Answer: (D)

$m_{1} g \sin 30^{\circ}-T=m_{1} a$
$T-m_{2} g \sin 30^{\circ}=m_{2} a$
Adding $(i) $ and $(ii)$, we get
$a=\frac{m_{1} g \sin 30^{\circ}-m_{2} g \sin 30^{\circ}}{m_{1}+m_{2}}$
$=\frac{(40-30) g / 2}{(40+30)}=\frac{10}{70} \times \frac{9.8}{2}=0.7 m s ^{-2}$
From eqn. $(ii)$, we get
$T =m_{2} g \sin 30^{\circ}+m_{2} a $
$=m_{2} g \sin 30^{\circ}+\frac{m_{2}}{m_{1}+m_{2}}\left(m_{1} g \sin 30^{\circ}-m_{2} g \sin 30^{\circ}\right) $
$=\frac{2 m_{1} m_{2} g \sin 30^{\circ}}{m_{1}+m_{2}}=\frac{2 \times 40 \times 30 \times 9.8 \times(1 / 2)}{40+30} $
$=\frac{1200}{70} \times 9.8=168 \,N$