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:

(source: draw.to)

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:

(source: draw.to)

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:

(source: draw.to)

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 Answer 1


I think your question contains false premises.

  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.

  2. You don't need to know the direction of the acceleration to solve this problem. If you get a negative sign, then acceleration is in the negative direction. That being said, you do need to know that an upward acceleration of one weight corresponds to a downward acceleration of the other; but this is embedded in your inextensible equation.

A different way to solve the same problem is to use Lagrangian mechanics or any of the other formulations of classical mechanics. But I suspect you'll run into the inextensible statement, and associated consequences, there too.

  • $\begingroup$ Perfect answer, the penny finally dropped now. Thanks a bunch! $\endgroup$
    – larrydavid
    Jan 21, 2015 at 20:19

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