# Condition to not let the block descend

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1. Aug 27, 2016

### decentfellow

1. The problem statement, all variables and given/known data
In the system shown in Fig. $2E.5 (a)$. block $m_2$ is being prevented from descending by pulling $m_1$ to the right with force $F$. Assuming all the surfaces to be frictionless, Find $F$

2. Relevant equations
F.B.D of $m_1$

From the F.B.D of $m_1$ we get the following equations:-
$$F=m_{1}a_{m_1}\tag{1}$$
$$N_1+m_1g-N_2=0 \implies N_1+m_1g=N_2\tag{2}$$

F.B.D of $m_2$

From the F.B.D of $m_2$, we get the following equation:-
$$m_2g-T=m_2a\tag{3}$$

F.B.D of $m_3$

From the F.B.D of $m_3$ , we get the following set of equations:-
$$m_3g-N_1=0 \implies m_3g=N_1\tag{4}$$
$$T=m_3a\tag{5}$$
3. The attempt at a solution

From eq. $(3)$ and $(5)$, we get
$$a=\frac{m_2}{m_2+m_3}g$$

Now, for $m_2$ to not descend, the condition required is $$\vec{a}_{m_3/m_2}=0\implies \vec{a}_{m_3}-\vec{a}_{m_2}=0\implies \vec{a}_{m_3}=\vec{a}_{m_2}$$
From $(1)$, we have
$$\vec{a}_{m_1}=\frac{\vec{F}}{m_1}$$
$$\therefore F=\frac{m_1m_2}{m_2+m_3}g$$

The book gives the answer as $$F=(m_1+m_2+m_3)\frac{m_2}{m_3}g$$.

Last edited: Aug 27, 2016
2. Aug 27, 2016

### Staff: Mentor

Hi decentfellow, Welcome to PF!

If all surfaces are frictionless as stated in the problem, how can $m_2$ pull $m_1$ in a horizontal direction? What supports the pulley?

By the way, so insert inline LaTeX, use '$' as a delimiter rather than ''. 3. Aug 27, 2016 ### decentfellow Okay, now that I see, the book had not attached a support for the pulley to the mass$m_1$, so do consider that. I think that attaching a support to the mass$m_1$,resolves the issue as to how does the$m_3$gets pulled due to the weight of mass$m_2$. Sorry for poorly stating my question. Also in my solution consider that the$N_2$in the F.B.D of$m_1$at the top of the mass is$N_1$. 4. Aug 27, 2016 ### gneill ### Staff: Mentor I think I see where some confusion might arise from the way the problem is stated. It's not m2 that pulls on the block m1. Consider that some external agency is providing a force F on block m1 and is accelerating the system to the right: What can you say about any forces that might exist between m1 and m2 if the system is being accelerated to the right? 5. Aug 27, 2016 ### decentfellow If what you say is that the mass$m_2$does not pull$m_3$, then there is no need of any external force to be put on the block$m_1$, so as to not let$m_2$slip. I think the weight of$m_2$does produce some tension in the string, which connects it to$m_3$, hence producing an acceleration in$m_3$along the$x-$direction. 6. Aug 27, 2016 ### gneill ### Staff: Mentor No,$m_2$definitely exerts a force on$m_3$via the tension in the string, but the reason that this does not cause block$m_2$to descend is that something else counters that tension force. Suppose for a moment that block$m_1$was fixed in place. Then I think you'd agree that$m_2$would descend and pull$m_3$to the right. What the question proposes is that block$m_1$is free to move and that some external force F is applied to it in such a way that the result is block$m_2$does not descend. The idea is to find that force F which accomplishes this "static" situation for$m_2$and$m_3$, and the whole system is accelerated by F without$m_2$descending. 7. Aug 27, 2016 ### decentfellow From the figure that you provided, I get that the masses$m_1$and$m_2$are in contact hence due to a force acting on$m_1$, there will be a normal reaction acting between them, so the F.B.D of$m_1$and$m_2$should be changed as follows:- 1)F.B.D of$m_1$2)F.B.D of$m_2$From the above FBDs, we get the following set of equations:- N_1+N_2=m_1g\tag{1} F-N_3=m_1a_{m_1}\tag{2} m_2g-T=m_2a\tag{3} N_3=m_2a_{m_1}\tag{4} From the above set of equations, we get a_{m_1}=\dfrac{F}{m_1+m_2} Also, as we have already found$a_{m_3}=\dfrac{m_2}{m_3}g$. Now for no relative motion between$m_3$and$m_1$,$a_{m_1m_2}=0$So, a_{m_1}=a_{m_3}\implies \dfrac{F}{m_1+m_2}=\dfrac{m_2}{m_3}g\implies F=(m_1+m_2)\dfrac{m_2}{m_3}g Still, the book's answer is different from mine so I think I am still doing something wrong, if that is so please point it out. 8. Aug 28, 2016 ### gneill ### Staff: Mentor If the pulley is attached to$m_1$then the horizontal and vertical forces applied to it via the tension in the ropes will be transferred to$m_1$(we assume a massless pulley here). So$m_3$should have a contribution to the overall acceleration of the system via the tension in the string. Looking at it another way (perhaps a bit more intuitively), as none of the masses are allowed to move with respect to each other under the specified conditions, the entire system of blocks moves as a whole and so all of them are being accelerated by force F. So what's the total mass being accelerated by F? 9. Aug 28, 2016 ### decentfellow Finally understood it, thanks for all the time, I will write the final solution below. So from the first part of your suggestion the modified F.B.D of$m_1$will be:- So, we get F-T-N_3=m_1a_{m_1} Now, as N_3=m_2a_{m_1} \therefore F=T+(m_1+m_2)a_{m_1} As, on applying the force all the masses are going to be accelerated with te same acceleration and$m_2$would not be descending so we have T=m_3a_{m_1} And,T=m_2g \therefore a_{m_1}=\frac{m_2}{m_3}g So, F=T+(m_1+m_2)a_{m_1} \implies F=(m_1+m_2+m_3)\frac{m_2}{m_3}g 10. Aug 28, 2016 ### gneill ### Staff: Mentor Bravo! Looks good. 11. Aug 28, 2016 ### decentfellow One, thing I noticed was that when the mass$m_1$doesn't move, i.e. there is no force acting on it, then the tension in the string is the weight of the reduced mass of the system which includes$m_3g, m_2g$and$T$. I will be showing my work below. From equations$(3)$and$(5)$in the first post, we get T=m_3a And, m_2g-T=m_2a So, we get T=\dfrac{m_2m_3}{m_2+m_3}g As we know, the reduced mass of$m_2$and$m_3$is given by$\mu=\dfrac{m_2m_3}{m_2+m_3}$So,$T=\mu g$Why is it that the tension comes out to be the reduced mass of$m_2$and$m_3$. 12. Aug 28, 2016 ### SammyS Staff Emeritus No. T only gives the vertical component of force on m2. Therefore,$\ m_2g-T=0\ .$The vertical component of the acceleration of m2 is zero. 13. Aug 28, 2016 ### decentfellow Yes, that would have been the case if i had been looking for the solution and would have considered that$m_2$should not descend, but as i stated in my last post "One, thing I noticed was that when the mass$m_1$doesn't move, i.e. there is no force acting on it", which clearly says that I am not referring to the condition that the question wants, but another case where the block$m_1## is stationary.

14. Aug 28, 2016

### SammyS

Staff Emeritus
Yes. Of course that's right.

I missed the change in conditions.