Question
A light string passes over a frictionless pulley. To one of its ends a mass of 8 kg is attached. To its other end two masses of 7 kg each are attached. The acceleration of the system will be
10.2 g
5.10 g
20.36 g
0.27 g
Solution
D.
0.27 g
free body diagram,
For body A, having mass m1 forces are given by
m1a = T1 - m1 g ....(i)
For body B
m2a = m2 g + T2 - T1 .....(ii)
For body C
m3 a = m3g - T2 ......(iii)
On solving Eqs. (i), (ii) and (iii), we get
a =
As m1 = 8 kg, m2 = m3 = 7 kg
a =
a =
a = 0.27 g
