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Frictional force on an inclined plane

  1. Feb 11, 2016 #1
    1. The problem statement, all variables and given/known data
    A suitcase of mass 16 kg is placed on a ramp inclined at 15 ° to the horizontal. The coefficient of friction between the suitcase and the ramp is 0.4. Determine whether the suitcase rests in equilibrium on the ramp, and state the magnitude of the frictional force acting on the suitcase.


    2. Relevant equations
    Net F =ma

    3. The attempt at a solution
    The answer is the suitcase is not in equilibrium. But I'm not sure if my working is right to show it is not in equilibrium. This is what I did:

    Resolve forces parallel to the ramp:
    16gsin 15° - μR= ma
    16gsin15° - μ (mgcos15°) =16a
    a= -1.28 ms^-2
    So since a≠0 , it is not in equilibrium.
    But this kinda don't make sense.. How can frictional force be so large that it could pull the suitcase up? Is my normal reaction force, R=mgcos15° , correct?
     
  2. jcsd
  3. Feb 11, 2016 #2

    BvU

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    Note that the force you calculate with ##\mu N## is the maximum friction force. On a horizontal plane the friction is 0, ##\mu N## is not zero, but the case doesn't move !
     
  4. Feb 11, 2016 #3

    cnh1995

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    Consider the suitcase is placed on a horizontal plane. Since gravity can't set it into motion, it will stay at rest. Now, if the plane is tilted slowly, component of gravity along the incline will pull the case. At the same time, friction will oppose the sliding motion, keeping the case at rest. However, if we keep on tilting the plane, at a particular angle, the case will start sliding. Can you write the force equation at the instant "just before" the case starts sliding?
     
    Last edited: Feb 11, 2016
  5. Feb 12, 2016 #4
    Why is friction 0 on a horizontal plane? I'm sorry, but I don't understand.. how do we know that the case doesn't move when μN is not zero?
     
  6. Feb 12, 2016 #5
    ΣF=ma
    mgsin 15° - mgsin 15° = m(0)
    Frictional force is equal to the weight component parallel to the ramp, when it's about to slide.
    Oh so from there, we can determine the normal reaction force by mgsin 15°= (0.4)(R) , right?
    But why isnt R = mgcos 15°? (By resolving forces ⊥ to the plane) ?
     
  7. Feb 12, 2016 #6

    cnh1995

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    You forgot μ:wink:! You get the relation between the angle of incline and μ from that equation.
     
  8. Feb 12, 2016 #7

    BvU

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    When you push to move the case, you'll discover that you have to push harder than ##\mu N = \mu mg## before it starts moving. So apparently the maximum friction force pushes back to the tume of about 64 N.
    But if you stop pushing the case doesn't start accelerating towards you, so apparently the friction force is then zero ! Even though ##\mu N## is still 64 N.

    We're having two intertwined threads now; too complicated. I leave you to cnh, he's good !
     
  9. Feb 12, 2016 #8
    Hmm but I thought mgsin 15° is μR?
     
  10. Feb 12, 2016 #9
    Ok ok thanks for you help !
     
  11. Feb 12, 2016 #10

    cnh1995

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    Just before the case starts sliding, this is true. So, how would you modify the above equation and get some relation between angle of incline and μ?
     
  12. Feb 12, 2016 #11
    Frictional force, μR =mgsin θ ?
     
  13. Feb 12, 2016 #12

    cnh1995

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    Yes. But what is R in terms of mg and θ?
     
  14. Feb 12, 2016 #13
    R=mg sinθ / μ ?
     
  15. Feb 12, 2016 #14

    cnh1995

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  16. Feb 12, 2016 #15
    Oh, R= mg cos θ
    Hmm.. Then is my working above, at post #1 correct?
     
  17. Feb 12, 2016 #16

    cnh1995

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    Correct.
    Your acceleration was negative. That's not possible. BvU has explained it in an earlier post.You can now substitute the value of R in the above equation and get a relation between μ and θ.
     
  18. Feb 12, 2016 #17

    cnh1995

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    The whole point in doing all this is to understand how the angle of incline and friction co-efficient are related. From the relation, you can see if the case is at rest or is sliding at a particular angle of incline. You can also calculate its acceleration then.
     
  19. Feb 12, 2016 #18
    mgsin θ - μ (mgcos θ) = ma
    But μ(mgcos θ) > mgsinθ
    Does this means it is in equilibrium?
     
  20. Feb 12, 2016 #19

    cnh1995

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    Yes. As BvU said, μmgcosθ is the "maximum" frictional force exerted by the incline at that particular angle θ. Where exactly are you confused?
     
  21. Feb 12, 2016 #20

    cnh1995

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    Assume the inclination is θ, mgsinθ=10N and μmgcosθ=15N. This means, the incline will exert "maximum" 15N frictional force. In this case, frictional force will be only 10N since it is sufficient to keep the case at rest. If mgsinθ=14N, friction will also be 14N, which is required to keep the case at rest. If mgsinθ=15N, now maximum frictional force of 15N will be exerted, still keeping the case at rest. But if mgsinθ=16N, the friction will be 15N only, since it is the maximum possible value, decided by μ. If mgsinθ is anything above 15N, friction will always be 15N i.e. maximum friction. The incline will do its best to resist the motion by applying "maximum" friction. But maximum friction will come into picture only when mgsinθ≥ μmgcosθ. If mgsinθ<μmgcosθ, frictional force will also be less than μmgcosθ i.e.equal to mgsinθ. Hope this helps!
     
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