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Two masses, A and B, are connected to a rope. A constant upward force 86.0N is applied to box A. Starting from rest, box B descends 12.1m in 4.70s . The tension in the rope connecting the two boxes is 32.0N.

What is the mass of B? What is the mass of A?

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My work and what I am struggling with:

I am trying to find the acceleration experienced by B, with that I will find the mass. I find the acceleration using: $$2\Delta_y/t^2=a$$ $$a=1.0955m/s^2$$

Then I have a problem for B do I use this equation: $$\sum F_y =32-M_b*g=M_ba$$ $$Or$$ $$\sum F_y =86-M_b*g=M_ba$$

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  • $\begingroup$ Hint: There are two forces acting on B; (1) the force due to the rope tension $F_T = ?$ and (2) the force due to gravity $F_G = -m_Bg$ $\endgroup$ – Alfred Centauri May 25 '14 at 22:46
  • $\begingroup$ @AlfredCentauri Hint: I already posted what $F_t$ equals to and my equation already demonstrates that I am aware of $F_G = -m_Bg$. So what was the point of your "hint"? $\endgroup$ – CharlieK May 26 '14 at 5:07
  • $\begingroup$ If the answer isn't obvious to you after my hint then nothing short of giving you the answer outright will do now will it? Isn't it clear that all you have to do is fill in the question mark $F_T = ?$ $\endgroup$ – Alfred Centauri May 26 '14 at 10:56
  • $\begingroup$ @AlfredCentauri "If the answer isn't obvious to you after my hint then nothing short of giving you the answer outright will do now will it?" My question to you is: How is your "hint" helpful if I already gave you that information? It is like me saying $2+2=4$ and you say: hint: add $2+2$. $\endgroup$ – CharlieK May 26 '14 at 21:33
  • $\begingroup$ On the contrary, if you already knew that $F_T$ is the force acting on B along with gravity, you would have known that the correct equation for B is the first one with 32N and not the one with 86N. Now, given that you asked which equation to use, it follows that you did not know that $F_T$ is the force to use in the equation. But this is precisely the information I provided in my hint which I quote in part: "There are two forces acting on B; (1) the force due to the rope tension...". Since the rope tension is given as 32N, that is the answer to your question. $\endgroup$ – Alfred Centauri May 26 '14 at 21:49
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The two possible equations you bring up only differ by the magnitude of the upper force. So you're asking: should you use $32\ \text{N}$ or $86\ \text{N}$ as the upward force on box B?

To help you arrive at your own answer: Ask yourself which object is actually exerting the force on box B. Is it whatever is exerting the force "F" on object A? Is it the rope connecting the two boxes? Is it box A? Something else?

Once you identify which object is responsible, your task of determining which force to use becomes easier.

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  • $\begingroup$ I'm thinking only the rope connection both objects. By the way is it 32-mg or mg-32? $\endgroup$ – CharlieK May 25 '14 at 22:04
  • $\begingroup$ Your choice of upward being the "positive direction" or "negative direction" determines if it's $f-mg$ or $mg-f$. But be consistent throughout your entire problem. $\endgroup$ – BMS May 25 '14 at 22:07
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Split the rope and see what force are acting on B

$$T-M_b g = M_b a$$

and on A

$$F-T-M_a g = M_a a$$

Since you know $T=20\,{\rm N}$, $F=86\,{\rm N}$, and you have calculated $a=-1.09955\,{\rm m/s^2}$ it is trivial to find $M_a=\frac{F-T}{a+g}$ and $M_b = \frac{T}{a+g}$.

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