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Can't understand 2 pulley 3 masses problem

  1. Feb 18, 2016 #1
    1. The problem statement, all variables and given/known data
    System is in equilibrium. 1 mass known: 79.4 kg (or 778.6 N) and 2 other unknown. Other masses unaccounted for. Everything friction-less. The Tensions are unknown also.

    2. Relevant equations
    ∑F = 0
    ∑Fx = -T1*cos135 + -T3*cos37 = 0
    ∑Fy = (-T1*sin135 - m1g) + (-T3*sin37 - m3g) + T2 - 778.6 N = 0

    3. The attempt at a solution
    I don't have any idea how to comprehend this problem. Of course I am missing something but I don't know how to ''think'' the problem from the get go. Any pointers would be very appreciated, I have tried to figure it out for the last hours...The T1 and T3 arrows should be parallel to the rope on the drawing 1epe8x.jpg
     
  2. jcsd
  3. Feb 18, 2016 #2

    TSny

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    The key is to consider the forces acting on the red knot where the three strings meet. The knot is in static equilibrium. Below is a figure of the knot and the three strings. Use that as the beginning of a free-body diagram of the knot. You should draw the force vectors acting on the knot and label them. Be sure to put arrow heads on the force vectors to indicate the directions of the forces.
     

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  4. Feb 18, 2016 #3
    So if I understand correctly, you mean that I should consider that T1,T2 and T3 vectors are pulling away from the red knot instead of towards it? Wouldn't that make them go in the same direction as the m1g, m2g and m3g vectors? Or is it an error on my part to consider the masses of the objects as vectors instead of mere units of mass? Thank you very much for your help!
     
  5. Feb 18, 2016 #4

    TSny

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    Yes, the three tension forces acting on the knot all point away from the knot along the strings. There are no other forces acting on the knot.
     
  6. Feb 18, 2016 #5
    One thing I can't get my head around is if I must substract the masses with both ∑Fx and ∑Fy

    ∑Fx = -T1*cos135 - m1g + T3*cos37 + m3g = 0
    ∑Fy = (T1*sin135 + m1g) + (T3*sin37 + m3g) - T2 - 778.6 N = 0

    Are these valid to start the problem? I have a hard time understanding how to ''translate'' the force vectors of the free-body diagram into mathematically sound equations for both X and Y axis. Thanks for your help anyone!
     
  7. Feb 18, 2016 #6

    TSny

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    There are only three forces acting on the knot: T1, T2, and T3. So when you write out ∑Fx and ∑Fy, you should only use these three forces.

    When you write the x-component of T1, you should write it as T1cos135o and not as -T1cos135o. If you calculate cos135o, you will see that it is a negative number. So, T1cos135o is already negative. You shouldn't add an extra negative sign in front.

    However, if they had specified the angles as shown below, then the x-component of T1 would be written with a negative sign as
    -T1cos45o
     

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  8. Feb 18, 2016 #7
    Thank you very much! in fact, both angles had to be measured in real life but I thought I had to consider the angle in relation to the positive X axis (I don't know if this is clear), so I ended up with 135 degrees. However, I suppose it is more logical or easier to go with 45 degrees now that you told me this...! What got me bugged is how I am including the weights in the equations (m1g, m2g (778.6N) and m3g). Does it seem like I'm on the right track?
     
  9. Feb 18, 2016 #8

    TSny

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    Try to write out ∑Fx = 0 and ∑Fy = 0 for the knot. The only forces that should appear are the x or y components of the three tension forces.

    However, you can then proceed to relate the tension forces to the weights of the three objects at the other end of the ropes. To do this, construct a separate free body diagram for each of the three objects and apply the condition for equilibrium.

    You can assume that the tension at one end of a rope is equal to the tension at the other end of that rope.
     
  10. Feb 19, 2016 #9
    Hi again! Here is what I found so far, if correct I only have one other question!

    ∑F = T1 + T2 + T3 (with vector arrows)
    ∑Fx = -T1cos45 + T3cos37 = 0
    ∑Fy = T1sin45 - T2 + T3sin37 = 0

    1) -T1cos45 + T3cos37 = 0
    T1 = (T3cos37)/cos45 = 1,129T3

    2) T3(1,129*sin45 + sin37) - T2 = 0
    T3(0,7986 + sin37) - 778,6N = 0
    T3(1,400) = 778,6N
    T3 = 778,6N / 1,400 = 556,1 N

    3) T1 = 1,129 * 556,1N = 627,8 N

    If the above is OK, I just wonder if the masses for the objects are equal to double their equivalent in Newtons, so: m1 = 2*T1 (in N) ??

    Thanks a lot for your help TSny! :)
     
  11. Feb 19, 2016 #10

    TSny

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    Your results for the tensions look correct and your work is nicely written out.

    I'm not sure why you would think that. Suppose you want to find m1. Draw a free body diagram for the object with mass m1. How many forces act on this mass? What direction do they point? Is m1 in equilibrium? Do you know the value of any of the forces acting on m1?
     
  12. Feb 19, 2016 #11
    What I mean to say is, I thought that a pulley divided by a factor of 2 the force necessary to lift it. I'm not sure exactly where I have read that, I cannot find it in my notes...Maybe I am mistaken! I sure know now that T1 is 627,8N but my brain seems to tell me there is more to it than simply T1 = m1 (which needs to be converted into kg). Am I just crazy?

    Thanks for the encouragement by the way!
     
  13. Feb 19, 2016 #12
    Aaaah I think I get why I thought it divides the force by 2. I somehow subconsciously considered this system as a combined pulley system, with each pulley being fixed of course but the ''knot'' acting as a movable pulley, which is not the case!

    I now assume that it is clearly T1 = m1, since :
    '' There is a force (tension) on the rope that is equal to the weight of the object.
    This force or tension is the same all along the rope. In order for the weight
    and pulley (the system) to remain in equilibrium, the person holding the end
    of the rope must pull down with a force that is equal in magnitude to the
    tension in the rope. For this pulley system, the force is equal to the weight, as
    shown in the picture. The mechanical advantage of this system is 1'' I cannot post the picture but it shows a simple pulley with a weight attached. Found this online.
     
  14. Feb 19, 2016 #13

    TSny

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    Yes. However, keep in mind that mass is not the same as weight. So, you would not say that T1 = m1. Otherwise, I think your reasoning is good!
     
  15. Feb 19, 2016 #14
    Yes I was about to add that in fact, the force being already known, it would rather go like: T1 = m1*g → M1 = T1/g

    Which gives the mass!

    Thanks a lot again for your help!
     
  16. Feb 19, 2016 #15

    TSny

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    Yes, good work.
     
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