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Finding the acceleration of 2 blocks

  1. May 5, 2016 #1
    1. The problem statement, all variables and given/known data
    A block of mass m1 = 2 kg and a block of mass m2 = 6 kg are connected by a massless string over a pulley in the shape of a solid disk having radius R = 0.250 m and mass M = 10 kg. The fixed, wedge-shaped ramp makes an angle of 30 degrees and the coefficient of kinetic friction is 0.360 for both blocks. Determine the acceleration of the two blocks. The only part of this problem that i don't understand is how to get the forces acting on the object:
    T1 = μm1g+m1a
    T2 = m2gsinθ−μm2gcosθ−m2a
    I know you have to divide it into the sin and cos components but I don't understand why the cosine was used when it is sliding down the ramp. I thought only the sine should be used and that the cos would cancel out with the normal (N). Thanks in advance for any clarification!


    2. Relevant equations


    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. May 5, 2016 #2
    Try to visualize it. Think about all the force vectors that are acting on either mass that run parallel to the surface their respective mass is on (I hope that's clear enough). I count 3. It should be easy from there.
     
  4. May 5, 2016 #3
    I don't understand the term: μm2gcosθ.
     
  5. May 5, 2016 #4
    I believe that is the frictional force acting on m2. Looks right to me.
     
  6. May 5, 2016 #5
    Seems to me that it should be u (M2 g - T2 sin theta) cos theta.
     
  7. May 5, 2016 #6
    I still don't understand why the cosine is used because when I draw out all the force vectors the mgcostheta cancels out with the normal force. :/ Unless I am drawing my force diagrams wrong.
     
  8. May 5, 2016 #7

    haruspex

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    Please explain your reasoning.
     
  9. May 5, 2016 #8

    haruspex

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    Yes, the normal force is equal and opposite to the component of gravity normal to the plane, i.e. mg cos(θ). Why is that a problem? How do you obtain the kinetic friction force from that?
     
  10. May 5, 2016 #9
    This is the answer for the tension from my solution manual: T2 = m2gsinθ−μm2gcosθ−m2a
    So if the cosine cancels out, my question is why does it still appear in the equation?
     
  11. May 5, 2016 #10

    haruspex

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    It only cancels out in the sense that there is no net force normal to the plane. But the friction is not related to the net force. It depends on the normal force, which is not zero.
     
  12. May 6, 2016 #11
    The normal force is the resultant vector of T1 and the weight of M2.
     
  13. May 6, 2016 #12

    haruspex

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    T2 (I think you mean) is parallel to the plane. It has no component normal to the plane.
     
  14. May 6, 2016 #13
    Yes, I meant T2.
    Try putting the forces T2 and W2 on a force table.
    What vector in the direction of N is required to balance the those 2 vectors?
    T2 is perpendicular to N, but it is not perpendicular to the weight W2.
    You can't use u W2 cos theta as the frictional force because T2 has a component
    that lies along the vector W2 and that component is T2 sin theta.
    I don't see any way around that.
    (I didn't see your previous post where you mentioned that N is the resultant of W2 and T2)
     
  15. May 6, 2016 #14

    haruspex

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    W2cos(θ).
    you can get yourself confused analysing things that way, but if you insist:
    The system is accelerating, but to make the discussion simpler here suppose it is static. I'll drop the 2 suffix.
    T has a vertical component Tsin(θ) (opposing) W. T=Wsin(θ), so that component is Wsin2(θ). Take that from W leaves Wcos2(θ). Likewise, N has a vertical component Ncos(θ) opposing W. Conclusion, N=Wcos(θ).

    Since there is acceleration, it is much easier to analyse in the directions parallel and normal to the plane.
     
  16. May 7, 2016 #15
    Well, I disagree with the statement T = Wsin(θ) and retract my statement that (M2 g - T2 sin theta) cos theta.
    T = Wsin(θ) is true for the direction along the plane, but then you use Ncos(θ) which has no component along the plane.
    If I were to balance forces in the vertical direction I would get N cos theta + T sin theta = W.
     
  17. May 7, 2016 #16

    haruspex

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    Which is entirely consistent with T=Wsin(θ) and N=Wcos(θ).
     
  18. May 8, 2016 #17
    Well, my statement is OK for static forces only, but as you pointed out earlier there is an acceleration component in the
    vertical direction, so that acceleration would need to be included in any dynamic solution.
    However, there is no component of acceleration in the N direction, but both T and W have components
    in the N direction so the Normal force must include both of these vectors when calculating the frictional force
    and that seems to involve considerable "arithmetic" in solving the problem.
     
  19. May 8, 2016 #18

    haruspex

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    T has no component in the N direction.
     
  20. May 9, 2016 #19
    I agree (finally) with that and that the value of N (the normal force) has to be N = M2 g cos theta.
     
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