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Homework Help: Help with finding the coefficient of kinetic friction!

  1. Dec 11, 2012 #1
    ScreenShot2012-12-11at104406AM.png

    Ok so I tried making myself a free-body diagram but I reached a dead end..

    Step 1) fkinetic = μkn
    fkinetic = μk(251)
    fkinetic / 251 = μk

    Step 2)
    Fx:
    T1

    Fy
    n = (m)(g) = 251 N
    120.5 N

    Step 3)

    ƩFx = m(-a) since it's moving the left or ƩFx = 0 (since it's moving at CONSTANT velocity)
    ƩFy = ma

    Step 4)

    ƩFy: 251 + 120.5 = (25.58 + 12.28)a

    *** I got the masses by solving 120.5 = m(9.81) and 251 = m(9.81) ***

    ƩFy: 371.5 = (37.86)a
    ƩFy: 371.5/37.86 = a
    ƩFy: 9.81 = a


    Step 5)

    T1 = m(-a)
    T1 = (25.58)(-9.81) = 250.9 = 251 :(

    Help?
     
  2. jcsd
  3. Dec 11, 2012 #2
    One thing I would advise you to do is not to substitute the numbers in until the end - it's better to work with symbols as long as possible.

    Using your free-body diagrams, write out equations for the horizontal and vertical forces acting on W1 and the vertical forces acting on W2.
    Then see how you can combine them to get a simple expression for μ
     
  4. Dec 11, 2012 #3
    W1

    Forces in x direction
    T1
    fk

    Forces in y direction
    Fn
    _________________________

    W2

    Forces in x direction
    none

    Forces in y direction
    T2
    _________________________

    ƩFx: T1 + fk = ?
    T1 + 251μk = ?​
    ƩFy: Fn + T2 = ?
    (m1g) + (-m2a) = ?​
    (25.58*9.81) + (-12.28*9.81) = ?​
    251 -120.5 = ?​
    130.5 = ?​


    What next?
     
    Last edited: Dec 11, 2012
  5. Dec 11, 2012 #4
    ƩF = ma. In this case, you're told that a=0, and this applies to both masses.
    Also, you need to remember the direction of the forces.
    So,
    ƩFx = T1 - fk = 0
    → T1 = fk

    Similar problem with signs for the forces on W2
     
  6. Dec 11, 2012 #5
    Ohh ok.
     
  7. Dec 11, 2012 #6
    Ohh ok.


    ƩFx: T1 + (-fk) = ma
    T1 - 251μk = m(0)​
    T1 - 251μk = 0​
    T1 = 251μk

    Wait but I still have to find T1...How do I do that?

    ƩFy: Fn + T2 = ?
    (m1g) + (-m2a) = ?​
    (25.58*9.81) + (-12.28*9.81) = ?​
    251 - 120.5 = ?​
    130.5 = ?​
     
  8. Dec 11, 2012 #7
    Your first equation is ok.

    Apply the same reasoning to the second equation:
    ƩFy = T2 - W2 = 0
    → T2 = W2

    Now, T1 = T2 since the pulley just changes the direction of the tension force and not its value.

    So, you can combine the 2 equations to eliminate the tension force - the answer should be clear after doing that.
     
  9. Dec 11, 2012 #8
    T1 = T2
    T1 = 120.5

    So for

    T1 = 251μk
    120.5 = 251μk
    0.48 = μk

    Thanks!

    One quick question though...
    Whenever I see a problem like this - two blocks and a pulley...

    T1 always equals T2?
     
  10. Dec 11, 2012 #9
    As long as the rope and pulley are 'ideal', ie don't have any mass, then the 2 tensions will be the same.
     
  11. Dec 11, 2012 #10
    Ok thanks a bunch!!
     
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