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Using Torque and Tension to find the rate of decreasing tension

  1. Apr 7, 2013 #1
    1. The problem statement, all variables and given/known data

    a 5 kg block with a 10cm length sits at the top of a frictionless, 6.0m 12kg uniform ramp. This ramp is held up by a rope of tension T at a 35* angle to the ground. The block is released and allowed to accelerate. What is the rate at which the tension in the supporting rope is decreasing when the block has traveled for 0.5s

    Crappy Sketch of the senario

    2. Relevant equations
    Related Rates (I believe)
    Fparallel = FgSinθ
    Fperpendicular = FgCosθ
    Torque = FdSinθ
    Kinematics formulas

    3. The attempt at a solution

    To find the rate at which tension is decreasing, youd need to find tension at both points. To find tension at point time= 0.5s, you have to calculate how far along the block has moved.

    Point 1) Top of the rope (time=0)

    Illustration for the top of the ramp

    Since the 5kg mass is 10cm long, the center of mass is in the middle (at 5cm)

    The toques clockwise and counterclockwise (FdSinθ) are:

    Tcw = Tccw
    T * 6 * Sin 90 = (49 * (6-0.05) * Sin 35) + (117.6 * 3 *Sin35)
    T = 61.6 N


    Point 2) Down the rope (time = 0.5)

    Fnet = Fparallel (as there is no friction force)
    = MgSinθ
    a = (9.8)Sin35
    a = 5.621 m/s^2

    Vo = 0
    Vf = X
    t = 0.5
    d = ?
    a = 5.621

    d=VoT+1/2at^2
    d= (0) + 1/2(5.621)(.5)^2
    d = 0.703 m

    Illustration for the Torques

    Since the center of mass is in the center of the 5kg block, the point that will rotate is (0.10/2) + (0.703), correct?

    Following that logic, the Torques Clockwise and CounterClockwise (FdSinθ) are:

    Tcw Tccw
    T*6*Sin90 = (49*(6-.753)*Sin35) + (117.6 * 3 * Sin35)

    T = 58.3 N



    Now heres where Im stuck. I could calculate the Tension when the block is at the top, and use the formulate rate = ΔTorque/ΔTime.

    Rate = (58.3 - 61.6) / 0.5s
    rate = -6.6 N / s

    This would give me a logical answer and I think it would work fine. However, I have a feeling i'm supposed to use related rates to solve this. Any help would be amazing.
     
  2. jcsd
  3. Apr 7, 2013 #2
    You could use your method if the tension decreased in a linear fashion, but of course it doesn't. I tried it without making the approximation and got a more negative value which you'd expect.

    What you need is the tension, T, in terms of time, t. Then you can simply differentiate to find the answer.

    Firstly: why does the tension vary; i.e. what variable could you define that is related to the variation of tension? Hint: it is a variable which itself varies with t.
     
  4. Apr 7, 2013 #3
    The tension varies because the object is accelerating down the ramp. If I want to find the rate at which tension is decaying (dTen/dt) id have to find an equation to express tension using values (t=?)

    So ill try that now:

    T= (FdSinθ + FdSinθ) / Fd Sinθ

    However im not sure what to do here, because obviously those Tetas and F and d are different, but I can't sub in the values can I?

    Is the variable you are talking about acceleration? as it directly influences the tension? Or is it simply distance as it is going Further and further down the ramp?
     
  5. Apr 7, 2013 #4

    haruspex

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    Just use the method you used originally, but instead of putting in t = 0.5s, leave t as an unknown. This will give you an equation for T as a function of t.
    Btw, the LHS in this line is wrong:
    T*6*Sin90 = (49*(6-.753)*Sin35) + (117.6 * 3 * Sin35)
     
  6. Apr 7, 2013 #5
    T * 6 * Sin90 = F(6-(1/2at^2)) * Sin35 + (117.6 * 3 * Sin35)

    Like So? Also, whats incorrect on the left hand side? The angle is 90 and the length is 6m.
     
  7. Apr 7, 2013 #6

    haruspex

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    (6-0.05-(1/2at^2))?
    Can you differentiate that now?
    My mistake - didn't notice you had a sketch attached and I took the cable to be vertical.
     
  8. Apr 7, 2013 #7
    I can differentiate that, but am I not missing the rest of the equation?

    T = (6-0.05 - (1/2at^2))

    dTension/dt = (2t * da/dt)

    Is that correct?
     
  9. Apr 7, 2013 #8

    haruspex

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    No no. I was just pointing out that you had dropped the -.05. The rest of the equation was ok.
     
  10. Apr 8, 2013 #9
    Ahh, alrighty, makes sense :).

    T * 6 * Sin90 = 49(6- 0.05 - (1/2at^2)) * Sin35 + (117.6 * 3 * Sin35)

    I have no idea how to differentiate that entire equation :S. How would I go about doing that?
     
  11. Apr 8, 2013 #10

    haruspex

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    The constant terms disappear, so you only have to worry about these parts: T * 6 * Sin90 = 49(- (1/2at^2)) * Sin35.
     
  12. Apr 8, 2013 #11
    Alright, my best effort here:

    Sin90*6*T = 49(6-(1/2at^2) *Sin35
    6T= 294-(24.5*Sin35 * (at^2))
    6 dten/dt = 0 - 14.053 * da/dt * 2t

    Yea, thats about as far as I got.
     
    Last edited: Apr 9, 2013
  13. Apr 8, 2013 #12

    haruspex

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    That isn't how the product rule works. If u and v are both functions of t then d(uv)/dt = u dv/dt + v du/dt (not du/dt dv/dt). So if a were varying you would have (d/dt)(at2/2) = (t2/2) da/dt + at. But in this instance a is constant (= g sin(35), as you worked out), so it's simply at. Looks like you lost the factor 1/2 also.
    You should have 6dT/dt= -24.5*Sin35 * at
    So plug in the values you have for a and t.
     
  14. Apr 9, 2013 #13
    The 1/2 is still there, but I multiplied Sin35 into the 24.5 already, yielding 14.053. Alright, that makes sense, but I didn't have to use the product rule here, just expanded and used the -24.5*sin35 as a constant. Does that work?

    Also, I added an extra 1/2, not lost one right? I multiplied the 49 in, but never removed the 1/2 when I rewrote the equation. Fixed it in the post before.

    Looking over the equation and re doing it, this is what I end up with.

    6 * dT/dt = -49 * Sin35 * at

    6 * dT/dt = -28.105 * (9.8 * Sin35) * 0.5

    dT/dt = 13.17 N/s

    Thanks
     
    Last edited: Apr 9, 2013
  15. Apr 9, 2013 #14

    haruspex

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    You turned (d/dt)(24.5*Sin35 * (1/2at^2)) into 14.053 * da/dt * 2t.
    OK, I see that the LHS was wrong, it should have been 49, not 24.5.
    The (49*Sin35)/2 turns into 14.053 ok, so that means you turned (d/dt) (at^2) into da/dt * 2t. This implies you differentiated both a and t2, but left the results as a product instead of being in two separate terms. That's a violation of the product rule. (Since a is constant, da/dt = 0, so you should have ended up with 0 as the answer.)
    dT/dt. Looks right.
     
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