Homework Help: 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.