# Why does Newton's first law create two different answers to this question?

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?

• 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. – LegionMammal978 Sep 6 '19 at 20:28
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.! 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.