How Does a Pulley System Affect Tension and Acceleration?

[tsw99]

Homework Statement

This is from the Serway Book Prob 5.63a
A cart of mass M with a (light) pulley at the corner, which connects with two blocks, one on the top of cart (m_1), one at the right side (m_2), and in contact with the cart. assume frictionless surfaces, wheels and pulley.
Also assume m_2 can move vertically only.

Initially the system is held at rest. At the instant after the system is released, find the tension in the string.
(Note: The pulley accelerates along with the cart)

The Attempt at a Solution

I draw free body diagrams for M, m_1 and m_2, and get (take right as positive)
$$-N=MA$$
$$T=m_{1}(a-A)$$
$$m_{2}g-T=m_{2}a$$
where N is the action/reaction force between M and m_2

Then I was stuck. how to eliminate N? also it seems weird to me, becuase the only horizontal force act on m_2 is the reaction force from the cart (N), how can it just move vertically only when the cart is accelerating to left (as it should be? the only force on the cart is -N)

Any "tips" or "hints" is appreciable. Thanks

 

[AvroArrow]

A:The trick here is to note that the pulley must remain in contact with both $m_1$ and $m_2$ (assuming no slipping). This means that the two masses must accelerate at the same rate, $\ddot y$. This implies that the tension in the string must be the same irrespective of whether it is connected to $m_1$ or $m_2$. Let's denote the tension in the string as $T$. Then:$$ T = m_1 (a-A) $$$$ T = m_2 a $$Equating the two, we get:$$ m_1 (a-A) = m_2 a $$Now, since the pulley and cart have the same acceleration $A$, we can substitute this into the equation above:$$ m_1 (a-A) = m_2 (a-A) $$$$ m_1 = m_2 $$This implies that the tension $T$ is the same in the string regardless of which mass it is connected to. To find the tension, simply solve for $T$ in either of the equations you had.