Q. Two weights $w_{1}$ and $w_{2}$ are suspended from the ends of a light string over a smooth fixed pulley. If the pulley is pulled up with acceleration $g$, the tension in the string will be
A) $\frac{4w_{1}w_{2}}{w_{1} + w_{2}}$
B) $\frac{2w_{1}w_{2}}{w_{1} + w_{2}}$
C) $\frac{w_{1}{- w}_{2}}{w_{1} + w_{2}}$
D) $\frac{w_{1}w_{2}}{2(w_{1} + w_{2})}$
Explanation
(a) For solving the problem, we assume that observer is situated in the frame of pulley (non-inertial reference frame)
$m_{1}g = w_{1}$
$m_{2}g = w_{2}$
From force diagram,
[image:5_image291.emf]
$T - m_{2}a_{0} - w_{2} = m_{2}a$
or $T - m_{2}g - w_{2} = m_{2}a\ \ \ \ (\because a_{0} = g)$
or $T - 2w_{2} = m_{2}a$ …(i)
From force diagram,
$m_{1}a_{0} + w_{1} - T = m_{1}a
$or $m_{1}a_{0} + w_{1} - T = m_{1}a$
or $2w_{1} - T = m_{1}a\ \ \ \ $…(ii)$\ \ (\because a_{0} = g)$
From Eqs. (i) and (ii),
$T = \frac{4w_{1}w_{2}}{w_{1} + w_{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.