Two boxes, $A$ and $B$, are connected to each end of a light vertical rope. A constant force of $70.0N$ is applied to box $A$. Starting from rest, box $B$ descends $11.6m$ in $4.40s$ The tension force is $40.0N$, find the mass of box $A$ and $B$.

I could find the mass of box $B$ easily to be $4.65kg$ by drawing a force diagram. Indeed, the result is that $mg - 40N = ma$, where $a$ is found by solving $11.6m = \frac{1}{2}a(4.40s)^2$.

However, I'm not sure how to find the mass of $A$. I'm trying to find a viable force statement to set up so I can just algebra my way through this, but I can't. I think part of the reason that I can't is that I don't really understand the problem.

So far we have only dealt with ropes that don't change in length, so it confused me when it said that box $B$ descended. Does that imply that the rope is elastic and stretching to allow it to drop, or, if the rope is rigid, does that mean that box $A$ is also moving at that same acceleration? If not, is box $A$ at rest? Accelerating upwards? I don't see how I can make a statement about forces on box $A$ in either the $x$ or $y$ direction when I don't know its net force in the $y$ direction and when there are no forces being applied to it in the $x$.

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    $\begingroup$ The question is not clear enough, I don't even understand it all, I can imagine some cases, but the question is horribly formulated.. This shouldn't be an answer, but I'm new and I can't post a comment, yet. P. D. : It really bothers me that "Qmechanic" only edits the questions and he doesn't even helps to answer them!! $\endgroup$
    – CIJ
    Commented Sep 22, 2014 at 3:03
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    $\begingroup$ Hi @Carlos: I have currently answered around 2% of all questions. I didn't answer this question because I think the question is off-topic. But let's see what the community thinks. $\endgroup$
    – Qmechanic
    Commented Sep 22, 2014 at 9:18

1 Answer 1


I have looked up your question on other sites, and found that the diagram your question is based off is probably the following:

enter image description here

So, I'll answer the question based on this diagram.

Your method for finding the mass of B is spot on. As a general rule, if you aren't told about any of the material properties for the rope in these sort of questions, you can assume that the rope is inextensible. So, therefore, the acceleration of both boxes are the same.

To find the mass of A, apply Newton's 2nd Law on the free body diagram of A:

$$40 + m_Ag - 70 = m_Aa$$

Rearrange, and you get $3.48kg$!

Alternatively, apply 2nd Law on the global diagram(note that internal forces like tension are not included in global diagrams):

$$m_Ag + m_Bg - 70 = (m_A + m_B)a$$

$$m_A + m_B = 8.13kg$$

And you should get the same result!


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