# Acceleration in a system equal for every body — why? A 1 kg cart can slide frictionlessly on the table. The black weights each weigh 1 kg. The pulleys are frictionless. The task is to determine the acceleration of the cart.

For the left-most weight we have the following free-body diagram:

Applying Newton's II law, $$\textbf{F}= m \textbf{a}$$, where in this case $$m$$ is 1 kg, we have $$(T_1 - g) \textbf{e}_y = a_1 \textbf{e}_y \ ,$$

where of course $$T_1 > 0$$ but $$a_1 \in \mathbb{R}$$.

For the cart we have:

Newton's II law in the $$x$$-direction (once again, $$m$$ is 1 kg) yields: $$(T_2 - T_1) \textbf{e}_x = a_2 \textbf{e}_x \ ,$$

where $$T_2 > 0$$ but $$a_2 \in \mathbb{R}$$.

For the right-most weight:

In this case $$m$$ is 2 kg and Newton's II law gives us: $$(T_2 - 2g)\textbf{e}_y = 2a_3 \textbf{e}_y \ ,$$

where, again, $$T_2 > 0$$ but $$a_3 \in \mathbb{R}$$.

Hence we the following three simultaenous equations: $$\begin{cases} T_1 - g = a_1 \\ T_2 - T_1 = a_2 \\ T_2 - 2g = 2a_3 \end{cases}$$

As of now, the system is indeterminate. We have to make the assumption that every single body involved in the system has the same acceleration magnitude i.e. $$|a_1| = |a_2| = |a_3|$$. Sure, that definitely seems to be the case purely by intuition and could most certainly be verified through experiment. But cannot the problem be solved without this a priori assumption that every body must experience the same acceleration? Cannot we derive it perhaps?

Even if I do make the assumption $$|a_1| = |a_2| = |a_3|$$, the system of equations below does not become any easier to solve. $$\begin{cases} T_1 - g = a_1 \\ T_2 - T_1 = a_2 \\ T_2 - 2g = 2a_3 \\ |a_1| = |a_2| = |a_3| \end{cases}$$

I know that the traditional way to solve the problem is to assert two things a priori:

(1) Every body in the system experiences, in magnitude, the same acceleration.

(2) The left-most weight will accelerate upward, the cart rightward and the right-most weight downward.

This produces the following system of equations: $$\begin{cases} T_1 - g = a \\ T_2 - T_1 = a \\ T_2 - 2g = -2a \end{cases}$$

where $$a > 0$$, and can readily be solved to yield $$a = 2.5$$ m/s^2.

However I wanted to approach this problem a little differently, without having to make "too many" a priori assumptions. But I am having trouble arriving at the same result through my more "thorough" method.

1. The "assumption" that the acceleration is the same for all bodies isn't actually an assumption. It is a statement that the string is inextensible: $$y_1=x_\text{cart}-\text{const}=y_2-\text{const}\implies \dot y_1 = \dot x_\text{cart} = \dot y_2\implies \ddot y_1 = \ddot x_\text{cart} = \ddot y_2,$$ where I've arbitrarily chosen upward, rightward, and downward to be positive for the left weight, cart, and right weight, respectively. If you don't include the inextensible string in your equations somehow, you'll have an undetermined system since any other string would produce different results.