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A pulley is attatched to one arm of a balance and a string passed around it carries two masses m1 and m2. The pulley is provided with a clamp due to which m1 and m2 do not move.On removing the clamp, m1 and m2 starts moving. Show that the change in counter mass to be made to restore the balance is (m1-m2) 2 / (m1+m2)

4152-1425_3899_baj1.bmp


A pulley is attatched to one arm of a balance and a string passed around it carries two masses m1 and m2. The pulley is provided with a clamp due to which m1 and m2 do not move.On removing the clamp, m1 and m2 starts moving. Show that the change in counter mass to be made to restore the balance is  (m1-m2)2 / (m1+m2)

Grade:12

1 Answers

pratham ashish
9 Points
14 years ago
hi, case 1: when system is no in motion mass on LHS = mass on RHS = M1 +M2 case 2: when system is in motion tension(T) in the string connecting two masses M1 & M2 on solving the eqns: 1) M1-T=M1*a (a= accelaration of M1& M2) 2) T-M2 =M2*a on dividing these eqns v get: T= 2 (M1*M2)/M1+M2 now tension in string connecting the pulley & balance wud b: (tension on both the side of the pulley) 2T=4*(M1*M2)/M1+M2 mass on for balance RHS = 2T=4*(M1*M2)/M1+M2 now change in mass on RHS = (M1+M2)- 4*(M1*M2)/M1+M2 ={(M1-M2)^2}/ M1+M2 GOOD LUCK & ENJOY THE PHYSICS

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