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

Answer:

#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