Two blocks A and B of equal masses are attached to a string passes over a smooth pulley fixed to a wedge as shown find the magnitude of acceleration of the center of mass of the two blocks when they are released from rest . neglect friction ?

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1 Answer
Jul 27, 2018

a = 3.6 m/s^2

Explanation:

I am going to add 3 words to the question:
"magnitude of the acceleration of the center of mass".

Assume that mass A is on the left and mass B is on the right. Define right to be positive direction (for force, and acceleration).

The component of each block's weight parallel with the respective incline:

W_A = -m*g*sin30^@

W_B = m*g*sin60^@

The weights are translated to upward pull on the string. Therefore, the net force acting on the each mass along the respective incline:
Mass A

F_"netA" = -W_A + W_B = -m*g*sin30^@ + m*g*sin60^@
F_"netA" = m*g*(-sin30^@ + sin60^@)
F_"netA" = m*g*(-0.5 + 0.866) = 0.366*m*g

Mass B

F_"netB" = -W_A + W_B = -m*g*sin30^@ + m*g*sin60^@
F_"netB" = m*g*(-sin30^@ + sin60^@)
F_"netB" = m*g*(-0.5 + 0.866) = 0.366*m*g

Acceleration of A:

F_"netA" = m*a_A
a_A = F_"netA"/m = (0.366*cancel(m)*g)/cancel(m)

Notice that if you go thru the same calculation for B, you get the same result. Therefore the center of mass will accelerate at

a = 0.366*9.8 m/s^2 = 3.6 m/s^2 (2 sig figs)

I hope this helps,
Steve