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Question

# In a simple Atwood machine, two unequal masses m1 and m2 are connected by a string going over a clamped light smooth pulley. In a typical arrangement (Figure) m1=300 g and m2=600g. The system is released from rest. (a) Find the distance travelled by the first block in the first two seconds. (b) Find the tension in the string. (c) Find the force exerted by the clamp on the pulley.

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Solution

## Given that, m1 = 0.3 kg m2= 0.6 kg T−(m1g+m1a)=0 ...(i) ⇒T=m1g+m1a Again, T+m2a−m2g=0 ....(ii) ⇒T=m2g−m2a From equations (i) and (ii) , we have m1g+m1a+m2a−m2g=0 ⇒a(m1+m2)=g(m2−m1) ⇒a=gm2−m1m1+m2 = 9.8×0.6−0.30.6+0.3 (a) t = 2 sec, acceleration a = 3.266ms−2 Initial velocity , u = 0 So, distance travelled by the body S = ut+12at2 =0+12(3.266)22=6.5m From equation (i), T = m1(g+a) = 0.3 (3.8+3.26) = 3.9 N The force exerted by the clamp on the pulley is given by, F - 2T = 0 ⇒F=2T=2×3.9=7.8N

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