Question

Two masses $40kg$ and $30kg$ 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
$a=\frac{m_{1}g\sin 30^{\circ }-m_{2}g\sin 30^{\circ }}{m_1+m_2}$
$=\frac{(40-30)g/2}{(40+30)}=\frac{10}{70}\times \frac{9.8}{2}=0.7m{s}^{-2}$
From eqn. $(ii)$, we get
$T=m_{2}g\sin 30^{\circ }+m_{2}a$
$=m_{2}g\sin 30^{\circ }+\frac{m_2}{m_1+m_2}\left(m_{1}g\sin 30^{\circ }-m_{2}g\sin 30^{\circ }\right)$
$=\frac{2m_1{m}_{2}g\sin 30^{\circ }}{m_1+m_2}=\frac{2\times 40\times 30\times 9.8\times (1/2)}{40+30}$
$=\frac{1200}{70}\times 9.8=168N$