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Pulley + ramp problem

  1. Mar 24, 2014 #1
    Problem:
    Block A is suspended from a vertical string. The string passes over a pulley and is attached to block B. The mass of block B is 1.8 kg. The mass of block C is 0.75 kg. If the coefficient of kinetic friction is 0.100 between block B and the ramp and the coefficient of static friction between block B and C is 0.900, determine the maximum mass of block A, such that block C does not slip off.

    nm15hg.jpg


    2. Relevant equations
    f = ma


    3. The attempt at a solution

    I drew a FBD for block C and was able to calculate its force of friction which is: 5.61 N (μmgcosθ)

    I also found out that the force of gravity for Block B is (M)(G)(COSθ). But I don't know what the force of Block C on B would be. If I could find that, I will be able to find the Force of Normal for Block B and solve for the force of friction for block B as well.

    After that, I am not too sure what I must do next. I am thinking of solving for acceleration but not too sure if that will help.

    Any help or any push in the right direction would be very helpful.

    Thanks
     
    Last edited: Mar 24, 2014
  2. jcsd
  3. Mar 24, 2014 #2

    BvU

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    Hello Ashi, and welcome to PF.
    Your ##\mu\, mg\,\cos\theta## is the maximum force block B can exercise on block C without block C slipping, right ?

    Can you calculate the friction force block B exercises on block C in the case that block A is so light that block C and B do not move ?
     
  4. Mar 24, 2014 #3
    Thank you BvU!

    Okay I think I got it but I am not too sure:

    I made a mistake on the Force of friction between block B and C. It should have been (mgμcosθ)/m where m is the mass of block c.

    So force of friction must be greater than acceleration for block c to slip off.

    Therefore: a = fnet/total mass = force of friction of block c

    force of friction of block c = 7.48 N

    total mass = M + 2.55

    fnet = Mg - (2.55)(g)(0.1)(cos32) - (2.55)(g)(sin32)

    acceleration = [Mg - (2.55)(g)(0.1)(cos32) - (2.55)(g)(sin32)] / [M + 255]


    [Mg - (2.55)(g)(0.1)(cos32) - (2.55)(g)(sin32)] / [M + 255] = 7.48

    M = 14.0KG

    I believe that is the right answer but I am not 100% sure. Can someone please check my work and let me know if they believe my answer is correct.
     
    Last edited: Mar 24, 2014
  5. Mar 25, 2014 #4

    BvU

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    No: ##\mu\, g\,\cos\theta## has the dimension m/s^2, an acceleration. So it cannot be a force.

    You mean smaller ?

    An acceleration can not be a force.

    Where does the 7.48 come from ? Before, you had 5.616 as the maximum force of friction, right ? (I did ask rather explicitly).
    However, with ##\mu\, g\,\cos\theta = ## 7.49 m/s2 (not a force, but an acceleration) you do have a value for the maximum acceleration of block C (and it is NOT this 7.49 m/s2! (*)). And thereby a value for the maximum acceleration of all three of A, B and C.

    An earlier small mistake I need to point out:
    No: just mg.

    So back to the drawing board: free body diagrams for C and for B !

    (*) Can you calculate the friction force block B exercises on block C in the case that block A is so light that block C and B do not move ? (Didn' t I ask for that already?)
     
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