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Coefficients of kinetic and static friction in a pulley

  1. Jul 25, 2016 #1
    1. The problem statement, all variables and given/known data
    A block of mass ## M_1 ## is on an inclined plane (the plane is inclined at 30 degrees), and is attached with a light cord over a light, frictionless pulley to another mass ## M_2 ##. ## M_1 ## is fixed at 5.00kg, and ## M_2 ## can be varied.

    When ## M_2 = 3.20 kg ##, ## M_1 ## just begins to slide up the slope, acclerating at ## 0.20 m/s^2 ##.

    Determine the coefficients of static and kinetic friction between ## M_1 ## and the plane.

    2. Relevant equations
    ## F_{fr_static} = \mu (static) F_N ##
    ## F_{fr_kinetic} = \mu (kinetic) F_N ##
    F = ma

    3. The attempt at a solution
    After resolving the components of mg into x, and y: ## F_N = y = 5.00gCos30 = 42.435244 N ##
    The component of mg in the x direction: mgsin30 = 24.5N

    To determine the coefficient of static friction, there must be no net horizontal force. I am not sure of this, but I am wondering if I just ignore the pulley and the mass at the end to calculate the static friction coefficient? But then that assumes that it will sit on the plane without sliding down, and I am not sure if I should make that assumption.

    If I did make that assumption, and ignored the pulley for the moment, then:
    ## F_{fr} - 5.00gsin30 = 0 --> \mu F_N = 24.5 --> \mu (static) = 24.5/42.43 = 0.577 ##

    To determine the coefficient of kinetic friction:
    ## net F_x = 3.20 * 0.20 ##
    ## F_{tension} - F_{fr_kinetic} - F_{mg-xcomponent} = 3.20 * 0.2 ##
    ## F_{tension} = 3.20g ##
    ## F_{fr_kinetic} = (F_{tension} - F_{mg-xcomponent})/ (3.20 * 0.2) ##
    ## F_{fr_kinetic} = (3.20g - 24.5)/ (3.20 * 0.2) = 10.71875 ##
    ## \mu (kinetic) = F_{fr_kinetic} / F_N = 10.71875/ 42.435244 = 0.253 ##

    I'm not sure in particular about the coefficient of static friction.
     
  2. jcsd
  3. Jul 25, 2016 #2

    andrevdh

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    Homework Helper

    I don't think you can assume the tension is 3.20 x g.
    The tension in the rope reduces when the system accelerates.
    Analyze the motion of M2 to determine the tension in the rope.
     
  4. Jul 25, 2016 #3
    ## mg - F_{tension} = 0.20 * 3.2 --> F_{tension} = 3.2g - 0.64 = 30.72 ##

    Is this right? Also, was I right in my assumptions with the static friction coefficient?
     
  5. Jul 25, 2016 #4
    Correct .

    No . You cannot ignore the mass at the end . You also need to take tension into account .

    Have you copied the question correctly ? I am finding the data inconsistent . I might be wrong but it would be nice if you could check the question .
     
  6. Jul 25, 2016 #5
    Vibhor, the question is correct
     
  7. Jul 25, 2016 #6
    Ok . Let us first deal with static friction case .

    What are the forces acting on M1 along the plane ? What would be the direction of static friction ? Can you resolve the forces along the inclined plane and use Newton's II Law ? Please use symbols . Do not put any numbers .
     
  8. Jul 25, 2016 #7
    Regarding taking the tension into account, given the mass is variable, and that it begins to move at 3.20kg, then ## F_{tension} < 3.2g ##. That seems an imprecise way of expressing it.

    Just saw your follow-up post Vibhor.

    ## F_{tension} + F_{fr-static} - F_{mg-x} = 0 ##
    ## m_2 g + \mu m_1 g cos30 - m_1 g sin 30 = 0 ##
    ## \mu = (m_1 g sin30 - m_2g) / m_1 g cos30 ##
    ## \mu = tan30 - (m_2/m_1cos30) ##

    But I don't know what to do about m_2 as it is varied, and all I think we can say about the max mass before it moves is that it is less than 3.2. But how much less? 3.1, 3.1999?
     
  9. Jul 25, 2016 #8
    Forget about M2 being variable . Think that when M2 = 3.2 Kg M1 just starts moving up the slope . In other words , When M2 =3.2 Kg , M1 and M2 are as good as not moving . You can treat them as in equilibrium . But once M1 and M2 start moving , things change ( we will deal with it later in kinetic friction case). They will no more be in equilibrium .

    So for the case of static friction , consider M1 and M2 to be in equilibrium and apply Newton's II law .

    Just write the equation in terms of symbols T , M1 , M2 , g , θ , μs .
     
  10. Jul 25, 2016 #9
    Like I did? (albeit with 30 instead of theta and m2 g instead of T)
     
  11. Jul 25, 2016 #10
    What would be the direction of friction ?
     
  12. Jul 25, 2016 #11
    I thought it would be in the same direction as ## F_T ##, given it is opposing the x component of mg
     
  13. Jul 25, 2016 #12
    Static friction acts so as to oppose the (potential) relative motion between the surfaces in contact . If there were no friction between M1 and the plane , in which direction would M1 be moving , up the slope or down the slope ?
     
  14. Jul 25, 2016 #13
    That would depend on the mass of M2 wouldn't it? M2 at 3.2, and it moves up the slope (so in this case friction would go in the direction of mg_x). I am not sure if it would in any instance slide down the slope.
     
  15. Jul 25, 2016 #14
    What is mg_x ? I do not understand the notation .
     
  16. Jul 25, 2016 #15
    It is the x component of m1 * g
     
  17. Jul 25, 2016 #16
    And what is the x direction :smile: ?
     
  18. Jul 25, 2016 #17
    the horizontal of the incline. The direction of the x component of mg is down the incline.
     
  19. Jul 25, 2016 #18
    Ok . So now do you see what is the error in post#7 ?
     
  20. Jul 25, 2016 #19
    Sure, it should be: ## F_T - F_{fr-static} - F_{x-mg} = 0 ##

    ## m_2 g - \mu m_1 g cos30 - m_1 g sin 30 = 0 ##
    ## \mu = (m_2 g - m_1 gsin30)/ m_1 g cos30 ##
    ## \mu = (m_2/m_1cos30) - tan30 ##
    ## \mu = 3.2/(5cos30) - tan30 = 0.162 ##

    Is the rest okay?
     
  21. Jul 25, 2016 #20
    Correct .

    "rest" what ?
     
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