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Force Paradigm Question

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  1. Jul 25, 2015 #1
    1. Statement of the problem statement, including all variables
    I am working the following example in my book:

    A light string of length b is attached at point A, passes a pulley B located a distance 2d away, and finally attached to a mass m1. Another pulley with mass m2 attached passes over the string, pulling it down between A and B. Calculate the distance x1 when the system is in equilibrium. The pulleys are massless.

    The book works the problem in terms of the energy paradigm, however I am working the problem in terms of the force paradigm (I need the practice). I have had a bit of trouble with seeing what forces are acting where, but think I have finally nailed it and am looking for a second opinion to see if I am right for the right physical reasons. Please excuse my use of the attached pictures As I am currently without my laptop and therefore am without an adequate and convenient means of displaying LateX.
    2. Relevant equations
    See attached equation file.

    3. The attempt at a solution
    The basic idea is that I use the equality derived from the Sine equation listed to give me an quality that I may use to solve for x1 in terms of the masses, b, and d. I get the original Sine equality from the fact that the Tension becomes split among the two triangles when the pulley with the second mass is added in the manner indicated in the attached diagram (System2). Thus the upward felt force on the second mass from one of these split Tension vectors is equal to half the weight of the second mass due to the fact that the Tension is split evenly and that neither mass experiences any acceleration. Does this seem valid? When I do this I get that the equilibrium position of x1 is b minus the term d times the ratio of 4m1 to the square root of 4(m1)2-(m2)2.
     

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    Last edited: Jul 25, 2015
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  3. Jul 25, 2015 #2

    SammyS

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    Your tension equation is incorrect.

    If there is no friction in the pulley movement, then T = Ta = Tb .
     
    Last edited: Jul 26, 2015
  4. Jul 25, 2015 #3
    So in another words my Ta should be pointed in the same direction as Tb? The vector is only split evenly but no changes to the direction are made?
     
  5. Jul 26, 2015 #4

    SammyS

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    No.

    The magnitudes of all 3 tensions are the same. They do not point in the same directions. They point along the direction of the string.
     
  6. Jul 26, 2015 #5
    Forgive me , but I seem to have a slight problem with visualizing this. If they point along the direction of the string then wouldn't the vertical components of Ta and Tb cancel as at the point where the second mass hangs one of those vectors is directed down and to the right and the other is directed up and to the right along the string ?
     
  7. Jul 26, 2015 #6

    jbriggs444

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    Tension is not a unidirectional force like that. It is a condition within the string. You can think of it as acting in both directions all along the length of the string. No matter what point you pick on the string, the tension applied on that point from the portion of the string to the right pulls rightward along the axis of the string and the tension applied to that point from the portion of the string to the left pulls leftward along the axis of the string.

    With that in mind, what three external forces act on the pully with mass m2? In what direction do they act?
     
  8. Jul 26, 2015 #7
    The first force is the weight of m2 acting downward. The second is a resistive pair of forces. The first of the pair acts up and to the left and is the pull on the pulley from the first mass m1. The second of the pair acts up and to the right and is the upward pull on the pulley with the second mass from the point at which the string begins (think of the problem in which a rope is tied to a car and a pole and you pull on it at the center. The pole and the car resist your pull on the rope).
     
  9. Jul 26, 2015 #8
    The force pair adds up to result in a vertical upward resultant force that counterbalances the weight of the second mass at equilibrium.
     
  10. Jul 26, 2015 #9

    TSny

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    FallenLeibnitz,
    There appear to be two equations in your solution that are incorrect, if I am interpreting your notation correctly. I have boxed these two errors in the attachment below.

    ##\vec{T} \neq \vec{T}_a +\vec{T}_b##. If I understand your notation, ##\vec{T}## is the upward tension force supporting m1 (as shown in your diagram).
    However, ##\vec{T}_a +\vec{T}_b## is the net upward tension force supporting ##m_2##. Since ##m_1## and ##m_2## are not necessarily the same, it is not correct to write ##\vec{T} = \vec{T}_a +\vec{T}_b##

    For the same reason, it is not correct to write ##0 = T_y-m_2g##. It should be ##0 = T_{a,y}+T_{b,y}-m_2g##. [EDIT: From your diagram, I now see that ##T_y## here is probably not meant to denote the ##y## component of the force ##\vec{T}## acting on ##m_1##. Or is it? But I think you can see how your notation is confusing.]

    However, I believe everything else you wrote is correct :smile:. In particular, I think your final equation is correct. So, it should give the correct answer for ##x_1##.
     

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    Last edited: Jul 26, 2015
  11. Jul 26, 2015 #10
    Yes,that wa
    Yes I do apologize. ##\vec{T}=\vec{Ta-Tb}## should be ##\vec{Ty}=\vec{Ta-Tb}##. My bad. I will update the equation list when I get back to my high tech white board . The above mentioned physics is still correct though right?
     
  12. Jul 26, 2015 #11

    SammyS

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    Judging by your figure, that should be ##\ \vec{T_y}=\vec{T_a}+\vec{T_b}\ ## .

    system2-jpg.86403.jpg

    Solving for ##\ \sin(\phi) \ ## first, then finding x1 seems to me to be a cleaner way to solve this problem.
     
  13. Jul 26, 2015 #12

    TSny

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    I don't see why you would now want to subtract ##\vec{T}_a## and ##\vec{T}_b##. I assume you are still defining these two vectors as shown in your diagram where you have ##\vec{T}_a## pointing up to the right and ##\vec{T}_b## pointing up to theleft.

    My confusion is with how you are using the symbols ##T## and ##\vec{T}##. In your original diagram, you have denoted the tension force on ##m_1## as ##T \hat{j}##. So, ##T## is the magnitude of the tension throughout the string. I'm not sure how you are using the symbol ##\vec{T}##. Does it mean the same as ##T \hat{j}##, or does it have some other meaning?

    [EDIT: I think everything in your original post is OK if you are defining ##\vec{T}## to be the sum of ##\vec{T}_a## and ##\vec{T}_b##. I got confused because I was interpreting the symbol ##\vec{T}## to be the same as the force ##T \hat{j}## acting on ##m_1##.]
     
    Last edited: Jul 26, 2015
  14. Jul 26, 2015 #13
    OK, so I took a look at my equations and there is something incorrect. The sum should be ##\vec{T_y}=\vec{T_a}+\vec{T_b}##. Furthermore the magnitudes of ##\vec{T_a}## and ##\vec{T_b}## are the same as the Tension (I believe a poster mentioned this above, but I didn't see what was meant till now). The sum of the two vectors add to give the upward pull that resists the weight of the second mass. I will add that ##T \hat{j}## and is the effect of the Tension on the first mass, however my use of ##\vec{T}## is incorrect. I will make the corrections now.
     
  15. Jul 26, 2015 #14
    Here is the corrected equations list. I also want to thank the posters who have replied so far. I had been stuck on this for a really long time.
     

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