First slide
Connected bodies
Question

In the system shown in figure mB = 4kg, and mA = 2 kg. The pulleys are massless and friction is absent everyrwhere. The acceleration of block A  is

Moderate
Solution

For movable pulley

aB = O+aA2

aA = 2aB--------(i)

Free body diagrams of A and B:

 

Equation of motion of block A and B

For A : T-mAg2 = mAaA = mA(2aB)-----(ii)

For B : mBg-2T = mBaB-------------(iii)

From (ii) and (iii) [2(ii) + (iii)]

mBg-mAg = [4mA+mB]aB

aB = [mB-mA]g[4mA+mB] = [4-2]×10[4×2+4]

           = 2012 = 53m/s2

Hence  aA = 2aB = 103m/s2

 

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