Problem 57

In the question there is a hint given, $y_A$ is not constant. From this hint I began to look at the pulley that holds $m_1$ and $m_2$, and can see that it has the same tension as $m_3$. From this observation, I (think I) have a means to involve all the masses within a given equation.

My force equation for pulley A is $$T_B - (m_1 + m_2)g = (m_1 + m_2)a$$

and for $m_3$ $$T_B-m_3 g= m_3 a$$

Solving both equations for $T_B$ and setting equal to each other I get $$m_1 a + m_2 a + (m_1 + m_2)g = m_3 g + m_3 a$$

and then solving for a I get $$a= \frac{m_3 g - (m_1 + m_2)g}{m_1 + m_2 - m_3}$$

What I would like to know is if my assumption about pulley 2 is correct and my subsequent work.


1 Answer 1


Your procedure is not correct. This looks like a homework problem, so I just give few hints:

  • write equations with respect to some inertial reference frame, such as ground
  • each mass has its own acceleration - this is very important!
  • since pulley is considered massless, tension in the cable from both sides has equal magnitude
  • write equations of motion for the three masses separately; this will give you 3 equations with 5 unknowns ($a_1$, $a_2$, $a_3$, $T_A$, and $T_B$)
  • to find unique solution you need 2 more equations: (i) equation of motion for the pulley A will give relationship between tensions $T_A$ and $T_B$, and (ii) displacements for the three masses will give relationship between accelerations $a_1$, $a_2$, and $a_3$.

Here are links to my answers to similar problems that can help you with setting up the equations:

What is positive rotation direction of a pulley in the Atwood machine?

How to find a condition for direction of motion in a system with hanging pulley?

  • 1
    $\begingroup$ This procedure is the only sane way that I know of to approach problems like this. $\endgroup$
    – garyp
    Mar 30 at 10:52

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