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Friction problem -- 2 blocks and a rope on an inclined plane

  1. Feb 10, 2017 #1
    1. The problem statement, all variables and given/known data
    Capture.PNG
    So, these 2 blocks are connected by a rope. Aluminuim block is 2kg and copper is 6kg. They are kept in motionless state via some kind of mechanism on steel surface. θ is 30 degrees. I was given dynamic and static friction coefficient (Aluminium μs=0,61 i μd=0,47, Copper μs=0,53 i μd=0,36.
    Question is will these blocks move after mechanism that keeps them motionless turns off and I have to calculate force of tension in the rope. Friction of that little pulley is 0.

    Solution for rope tension is 10N.
    2. Relevant equations
    Ff=N*μ
    G=m*g

    3. The attempt at a solution

    I found G1 and G2:
    G1=19.62N
    G2=58.86N

    I also found G2x and G2y:
    G2x=sinθ*G2=29.43N
    G2y=cosθ*G2=50.97N

    After that I calculated force of static and dynamic friction:
    Fsf1=G1s=11.97N
    Fdf1=G1d=9.22N

    Fsf2=G2ys=27N
    Fdf2=G2yd=18.35N

    Unfortunately, this is as far as I got because I don't understand and don't know what to do next since I never had any problems with static friction.
    I hope someone can help me out on this one :)
     
    Last edited by a moderator: Feb 10, 2017
  2. jcsd
  3. Feb 10, 2017 #2
    I think I may figured it out but i'm not 100% sure. So, if T=G2x-Fdf2 then T=11.08N. But it's not 10N like it says it should be. Also, should I use static friction or dynamic friction for calculating tension ?
     
  4. Feb 10, 2017 #3

    haruspex

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    Static and kinetic friction are different. The static coefficient only sets an upper bound on the frictional force:
    |Fd|=μdN
    |Fs|≤μsN
    This distinction is important.
    You should use static if the surfaces do not slide over each other and dynamic if they do.
    So the first step is to determine whether the system can be static. Suppose it is static and work out the forces on that basis.
     
  5. Feb 10, 2017 #4
    Sorry, can you give me more info :). I have no idea were to start really.
     
  6. Feb 10, 2017 #5

    haruspex

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    What is the max frictional force on the block on the ramp?
    If it does not slip, what is the minimum tension?
    Can the frictional force on the other block supply that tension?
     
  7. Feb 10, 2017 #6
    Well max friction should be Fsf2=G2ys=27N right ?
     
  8. Feb 10, 2017 #7

    haruspex

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    Yes, so what is the minimum tension to prevent that block slipping?

    By the way, it is rarely a good idea to plug in numbers so soon. Far better to keep everything symbolic as long as possible. There are many advantages, not least for others trying to follow your working.
    In the present problem, for example, you might find that you did not keep enough significant digits to be able to resolve whether it is static, and you'll have to go all through it again.
     
  9. Feb 10, 2017 #8
    T=G2x-Fsf2=29.43-27=2.43N
     
  10. Feb 10, 2017 #9

    haruspex

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    Ok. If that is the tension, will the left hand block slip?
     
  11. Feb 10, 2017 #10
    I dont think so...
    Fsf1=G1s=11.97N
    Friction is higher than tension force pulling left block so it shouldn't slip right ?
     
  12. Feb 10, 2017 #11

    haruspex

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    Not exactly. As I pointed out, the static friction equation only gives you an upper bound for the frictional force. But you are right there is 11.97N available. So that's the first part answered.
    The actual tension is problematic. We have only found the minimum tension. At maximum tension, either the left hand block is about to slip to the right or the right hand block is about to slip up the slope! The maximum tension for equilibrium is the lower of the two.
    In practice, the actual tension depends on how the set-up was done. E.g. to get max tension we could start with a warm environment and cool it down. As the wire tries to shrink the tension increases until the left block slips right or the right block slips left. Or we could get this higher tension just by holding the wire a bit stretched while we place the blocks on the slope.
    In short, there is no way to answer the question of tension as posted. The best you can do is to calculate a range.
     
  13. Feb 10, 2017 #12
    Why does it say 10N for answer then?
     
  14. Feb 10, 2017 #13

    haruspex

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    I have no idea.
     
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