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Problem related to pulley system and velocity

  1. Sep 18, 2013 #1
    1. The problem statement, all variables and given/known data

    Diagram.jpg

    In the given figure if the masses A and B move down with a velocity 'v' at any moment then what will be the velocity with which mass W rises.

    A purely logical answer would be appreciated and please don't use calculus based approach, as I don't know calculus. Also please see my attempt at the solution and correct me.

    2. Relevant equations

    Trigonometric ratios


    3. The attempt at a solution

    If the masses are moving downwards with a velocity 'v' then the string is also moving with the same velocity that is 'v' so the velocity with which the mass W should rise should be equal to
    'v cos θ' but the correct answer is 'v/cos θ'. Please do tell me why my approach is wrong and what is the correct method
     
  2. jcsd
  3. Sep 18, 2013 #2
    your answer would be right only if V was a force which its obviously isnt
    also , if you had the force then the upward force would equal 2F * cos theta . where f is the force at each block and since they look the same then F1=F2
    if you had acceleration or time you could apply on this equation F= MA but since you dont , i dont really know , maybe you should law of conservation of momentum but its just a wild guess
     
  4. Sep 18, 2013 #3
    I think you are correct and the answer of v/cos theta is wrong.
     
  5. Sep 18, 2013 #4

    haruspex

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    Quite so, but that means for W to stay connected to each string it must also have a velocity component v towards each pulley. If its speed is u then u cos θ = v.
     
  6. Sep 18, 2013 #5

    ehild

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    Let be the pieces of string between the pulleys and the middle weight of length L initially. A and B go downward by vΔt in a short time Δt. Then the pieces of the string between the pulley and the middle block get shorter, the new lengths are L'=L-vΔt. The weight raises by Δh.

    Apply the Law of Cosine for the blue triangle : L'2=L2+(Δh)2-2LΔh cos(θ), and substitute L'=L-vΔt. Expand, simplify. vΔt and Δh are small, you can ignore their square.Isolate Δh. The speed of the middle block is Δh/Δt.

    ehild
     

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  7. Sep 19, 2013 #6
    Could you please elaborate I think I'm getting it.
     
  8. Sep 19, 2013 #7
    Correct me if I am wrong but I think that the conservation of momentum applies here. The momentum of A and B is must be equal and opposite to the momentum of W.
     
  9. Sep 19, 2013 #8

    HallsofIvy

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    No, you don't need "conservation of momentum" or any reference to a force. You are told that "the masses are moving downwards with a velocity 'v'" so it is simply a matter of geometry- how does the contraction of the lengths of the sides of the triangle affect the height of the object.
     
  10. Sep 19, 2013 #9
    lets see
    if the string falls for a distance X in time T
    then the Velocity of the strings is V=X/T
    and since the string moves X downwards , the strings holding the weight also moves X upwards
    thus , raising the weight with X Cos theta .
    So Velocity of the weight = X Cos theta / T
    and velocity of weight that is falling down = X/T
    now the velocity of Weight is equal to X cos theta / t right ?
    that is Velocity of Falling body multiplied by Cos theta
    so velocity of the body moving upwards is V * cos theta * 2 since T is now half because two bodies are lifting it
     
  11. Sep 19, 2013 #10

    ehild

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    It does not move upward. It gets shorter by X. As the weight W does not change its horizontal position, the angle of the string will change with respect to the vertical. Try to follow the method I suggested, with Law of Cosines.

    ehild
     
  12. Sep 19, 2013 #11
    In echild's figure in response 5, the three sides of the right triangle to the right are h(t)=Lcosθ, Lsinθ, and L. After a time interval Δt, the lengths of the three sides of the triangle become h(t+Δt), Lsinθ, and L-vΔt. These three sides of the right triangle have to satisfy the Pythagorean theorem, so:

    [tex]h(t+Δt)=\sqrt{(L-vΔt)^2-(L\sinθ)^2}=\sqrt{L^2\cos^2θ-2vΔtL+v^2(Δt)^2}=\sqrt{h(t)^2-2v(h(t)/cosθ)Δt+v^2(Δt)^2}[/tex]
    The average upward velocity of the center weight during the time interval Δt is given by:

    [tex]w_{ave}=\frac{h(t)-h(t+Δt)}{Δt}=\frac{h(t)-\sqrt{h(t)^2-2v(h(t)/cosθ)Δt+v^2(Δt)^2}}{Δt}[/tex]

    If we multiply numerator and denominator of this equation by [itex]h(t)+\sqrt{h(t)^2-2v(h(t)/cosθ)Δt+v^2(Δt)^2}[/itex], we obtain:

    [tex]w_{ave}=\frac{2v(h(t)/cosθ)+v^2(Δt)}{h(t)+\sqrt{h(t)^2-2v(h(t)/cosθ)Δt+v^2(Δt)^2}}[/tex]
    If we take the limit of this equation as the time interval Δt becomes small, we obtain the instantaneous velocity w:
    [tex]w=\frac{v}{cosθ}[/tex]
     
  13. Sep 19, 2013 #12

    ehild

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    Nice explanation! I applied the Law of Cosines for the blue triangle in my drawing, but it is the same in principle.

    L'2=L2+(Δh)2-2L(Δh)cos(θ)

    L'=L-(vΔt), plugging in, expanding, simplifying, ignoring squares of vΔt and Δh, you get the vertical velocity of the weight as u= Δh/Δt.


    ehild
     
    Last edited: Sep 19, 2013
  14. Sep 19, 2013 #13

    haruspex

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    Chet, ehild, you're making this much too complicated. See my post #4.
     
  15. Sep 19, 2013 #14

    haruspex

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    The section of string on the weight's side of the pulley does not move upwards at speed v; it moves towards the pulley at speed v. The end connected to the weight must also be moving towards the pulley at speed v (or the string would change length), as well as moving tangentially, so as to produce a net vertical motion. I.e. the component of its vertical motion that is towards the pulley must be v. Therefore its vertical speed is v sec theta.
     
  16. Sep 19, 2013 #15
    Yes. I like #4 much better. Sorry I missed it the first time through.

    Chet
     
  17. Sep 19, 2013 #16
    i said that it covers Distance X * cos theta * / Time
    so V= V * cos theta
    will you explain where did you get the sec ?
    i thought the component must be Cos since The change in Vertical position is equal to CosTheta the Change in the position of string towards the pully
     
  18. Sep 20, 2013 #17

    ehild

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    The length of the string can not change. If the vertical piece gets longer by X, the other one, connected to the weight W is shortened by X. So the end of the string has to move upward, its direction will change. See attachment.

    ehild
     

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  19. Sep 20, 2013 #18

    haruspex

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    In time dt it must be v dt closer to (each) pulley. Its net movement must be vertical. If it moves upwards distance dx and as a result is v dt closer to a pulley then dx cos theta = v dt. Thus dx/dt = v / cos theta = v sec theta.
     
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