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Condition to not let the block descend

  1. Aug 27, 2016 #1
    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##
    Picture related to the question.jpg

    2. Relevant equations
    F.B.D of ##m_1##
    alturist.PNG
    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##
    dabura.PNG
    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##
    vegeta.PNG
    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. jcsd
  3. Aug 27, 2016 #2

    gneill

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    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 '$'.
     
  4. Aug 27, 2016 #3
    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##.
     
  5. Aug 27, 2016 #4

    gneill

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    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:

    upload_2016-8-27_13-54-51.png

    What can you say about any forces that might exist between m1 and m2 if the system is being accelerated to the right?
     
  6. Aug 27, 2016 #5
    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.
     
  7. Aug 27, 2016 #6

    gneill

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    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.
     
  8. Aug 27, 2016 #7
    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##
    gadha.PNG
    2)F.B.D of ##m_2##
    pagal.PNG

    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.
     
  9. Aug 28, 2016 #8

    gneill

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    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?
     
  10. Aug 28, 2016 #9
    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:-
    bamboo.PNG
    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, $$F=T+(m_1+m_2)a_{m_1} \implies F=(m_1+m_2+m_3)\frac{m_2}{m_3}g$$
     
  11. Aug 28, 2016 #10

    gneill

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    Bravo! Looks good.
     
  12. Aug 28, 2016 #11
    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##.
     
  13. Aug 28, 2016 #12

    SammyS

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    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.
     
  14. Aug 28, 2016 #13
    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.
     
  15. Aug 28, 2016 #14

    SammyS

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    Yes. Of course that's right.

    I missed the change in conditions.
     
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