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Pulley on an incline

  1. Oct 15, 2009 #1
    1. The problem statement, all variables and given/known data

    Flat weights on a picture are connected by a light rope by a light pulley. How much at most (at least) does mass m2 have to be to balance the pulley, if mass m1 = 1 kg. With how much acceleration do the weights move, if the mass m2 of the weight is replaced with a twice heavier (lighter) weight? Ratio of friction between the weight with mass m1 and the surface is 0.3

    2. Relevant equations

    F1= m1g sinα μk
    F2= m2g
    F1 – F2= a(m1 + 2m2)

    3. The attempt at a solution

    m1= 1 kg
    μk= 0.3
    α= 30°
    m2=?
    a= ?

    For finding the m2:
    F1 = F2
    m1g sinα μk = m2g
    m2= 0.15 kg

    For finding the acceleration, if m2 is double the weight:
    F1 – F2= a(m1 + 2m2)
    g(m1 sinα μk – 2m2) = a(m1 + 2m2)
    a= 1.51 m/s²

    For finding the acceleration, if m2 is half the weight:
    F1 – F2= a(m1 + ½m2)
    g(m1 sinα μk – ½m2) = a(m1 + ½m2)
    a= 3.87 m/s²
     
  2. jcsd
  3. Oct 15, 2009 #2
    Can someone please tell me if I calculated correctly? Thank you!
     
  4. Oct 15, 2009 #3
    I don't understand where you're getting your angles from, unless one of the masses is also on a ramp? Can you describe the geometry of the setup a little better?
     
  5. Oct 15, 2009 #4
    I attached the picture of the pulley system. Hope it will make the problem clearer.
    Thanks for helping!
     

    Attached Files:

    • FI2.bmp
      FI2.bmp
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  6. Oct 15, 2009 #5

    rl.bhat

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    Your calculation of F1 is wrong.
    On m1 two forces are acting. What are they? Draw FBD.
     
  7. Oct 15, 2009 #6
    OK, I added the forces to the picture.

    F1 is decomposed into F1a and F1b and because F1a and -f1a cancel each other I assumed there is only one force left working on m1, which is F1b (= m1g sinα μk).

    Where did I go wrong?
     

    Attached Files:

    • FI3.bmp
      FI3.bmp
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  8. Oct 15, 2009 #7
    After looking at the picture and thinking really, really hard, I remembered the tension of the rope. Does this too has to be considered when calculating F1?
     
  9. Oct 15, 2009 #8

    rl.bhat

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    Forces acting on m1 are the component of weight and frictional force in the downward direction and tension in the upward direction.
    What is the frictional force?
     
  10. Oct 15, 2009 #9
    Fk= μkN
    N= mg
     
  11. Oct 15, 2009 #10
    So the sum of the forces working in y direction are 0, because they cancel each other out.
    But the sum of forces working on x direction taking tension into the consideration would be
    T- m1g sinα?
     
  12. Oct 15, 2009 #11
    I think I lost you.
    So, we have tension of the rope, which is pulling in the direction upwards. Does this mean that tension is connected with m2 and I can write T= m2g?

    If my upper conclusions are correct, does the formula for F1 now rewrites to:
    F1= T- m1g sinα= m2g -m1g sinα, and I can now from this formula calculate m2?

    I'm desperate!
     
    Last edited: Oct 15, 2009
  13. Oct 15, 2009 #12

    rl.bhat

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    N is m1g*cosθ

    Τ - m1*g*sinθ - μ*m1*g*cosθ = m1a ...(1)
    For m2
    m2*g - T = m2*a....(2)
    From eq 1 and 2 find a and equate them. Then solve for T.
    Then substitute T is either eq 1 or 2 to find a.
     
    Last edited: Oct 15, 2009
  14. Oct 16, 2009 #13
    Hello!!! for the second equation are you sure that the tention is negative? because i read that it always should be positive! and could you explain what to do to solve the acceleration and tension, that would help a lot! thank you in advance!!

    If the tension is positive in the second equation i think the m2 should be .24 kg, right!!!
     
  15. Oct 16, 2009 #14

    rl.bhat

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    Body accelerates in the direction of the net force.In the second case, if 2m2 is moving down, then 2m2*g must be greater than T. So net force = 2m2*g - T = 2m2*a.
     
  16. Oct 16, 2009 #15
    OK, let's start from the beginning.

    The first part of the problem is asking us to find weight of m2 that will make the system to be in equilibrium.

    As you see I made some approaches to solve this part of the problem, but I obviously didn't succeed. Can you please give me some hint?

    I think that having m2 known will make the rest of the problem much much easier to solve, because I think I understand your explanation how to merge the formulas and finding T and a.

    Thank you for your time and patience!
     
  17. Oct 16, 2009 #16

    rl.bhat

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    Τ - m1*g*sinθ - μ*m1*g*cosθ = m1a ...(1)
    For m2
    m2*g - T = m2*a....(2)
    In the above equations, put a = 0 and solve for m2.
     
  18. Oct 16, 2009 #17
    So by putting a=0 into the above equations I ended with equation:

    m1*g*sinθ + μ*m1*g*cosθ = m2*g

    and solved for m2= 0.76 kg.

    I really hope I'm correct!
     
    Last edited: Oct 16, 2009
  19. Oct 16, 2009 #18
    So, if I understood correctly, by adding the equations for m1 and m2 I should get the formula for acceleration which would be

    a= (m2g - m1*g*sinθ - μ*m1*g*cosθ) / (m1 + m2)

    and playnig with weight (once double the weight, once half the weight for m2) I should be able to get the correct results?
     
  20. Oct 16, 2009 #19
    I did the the calculations and get this results:

    for 2m2 (double the weight) the a= 2.96 m/s²
    for 1/2 m2 (half the weight) the a= - 2.7 m/s²

    Did I get it right?
     
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