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I have been having some difficulty with a recent question. Take the following pulley system:

The three blocks have masses $M$, $m_1$, and $m_2$. All are subjected to a gravitational force $g$. The pulley and string are of negligible mass, and all surfaces are frictionless. The problem is (to paraphrase):

What magnitude of force $F$ is necessary for $m_1$ and $m_2$ to be motionless relative to $M$?

When I first solved this, I just considered Newton's first law for each of the three masses and added the additional conditions $a_{xm_1}=a_{xM}$, $a_{xm_2}=a_{xM}$, $a_{ym_1}=0$. After eliminating most of the variables, I ended up with $F=g\frac{m_1}{m_2}(M+m_1)$.

However, the textbook gives the answer $F=g\frac{m_1}{m_2}(M+m_1+m_2)$. In attempting to find the discrepancy, I solved the problem again somewhat differently: Let $a=a_{xM}=a_{xm_1}=a_{xm_2}$. Since $m_1$ does not accelerate vertically, $T=m_1g$. Since $m_2a_{xm_2}=T=m_1g$, we have that $a=g\frac{m_1}{m_2}$. Finally, since $F$ is pushing on $M$, which in turn is pushing on $m_1$ via its normal-force interaction, we have $F=(M+m_1)a=g\frac{m_1}{m_2}(M+m_1)$, the same answer I previously came to.

I asked my teacher about this problem to determine where my error occurred, and his reasoning was as follows: As with the earlier reasoning, $a=g\frac{m_1}{m_2}$. Since $M$, $m_1$, and $m_2$ are motionless relative to each other, we can treat the three as a single system and ignore internal forces:

Now, we simply have $F=M_sa=g\frac{m_1}{m_2}(M+m_1+m_2)$.

Both lines of reasoning are compelling, so my overall question is this: Which of these two answers is correct, and how is the other answer incorrect?

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The first approach is indeed valid and should have given you the correct answer. I just used that approach and got it correct, so you must have made a mistake.

Without seeing your work I can only guess what the mistake would be. However, the most likely guess is that you forgot to include the force from the pulley on the free body diagram for M.

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  • $\begingroup$ It seems that I indeed forgot to account for the force on $M$ from the pulley. Adding that to the equations produces the correct result. I've accepted this answer, even though both are equally valid, since it is the earlier one. $\endgroup$ – LegionMammal978 Sep 6 at 20:28
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You have made a mistake in the equation $$F=(M+m_1)a$$ You have missed the fact that the tension is also works on M.enter image description here! The sum of the two tensions add up and produce a force on the pulley. However, as the pulley is(should be) massless, the block M experiences a opposite force, which has a horizontal component of $$T√2*cos45° = T$$ So the equation should be $$F-T=(M+m_1)a$$ which then gives same answer as your teacher.

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