Solving for acceleration of two blocks with two pulleys

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SUMMARY

The discussion focuses on solving the acceleration of two blocks connected by two pulleys using Newton's second law. The equations derived include T = Ma1, mg - 2T = ma2, and a1 = 2a2, leading to the final expression for acceleration a1 as a1 = mg / (4M + m). The consensus among participants confirms the correctness of the calculations, with the final acceleration for block B expressed as aB = mg / (m + 4M).

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Homework Statement
If block A has mass M and block b has mass m, find the acceleration of block B assuming that the acceleration of block A equals two times the acceleration of block B.
Relevant Equations
Equation 1: T = Ma1

Equation 2: mg - 2T = ma2

Equation 3: a1 = 2a2
241179
Equation 1: T = Ma1

Equation 2: mg - 2T = ma2

Equation 3: a1 = 2a2

Since a1 = 2a for equation one I get T = 2Ma1

mg - 4Ma1 = ma1

mg = 4Ma1 + ma1

mg = 4a1(M+m)

a1= mg / 4M+m

Not sure if this is correct. Can someone please help to make sure I'm doing this right?
Thanx
 
Physics news on Phys.org
Looks correct, I got same answer,
##a_{\rm B}=\frac{mg}{m+4M} ##
 

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