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Find the acceleration of the wedge (see picture)

  1. Oct 14, 2015 #1

    Nathanael

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    1. The problem statement, all variables and given/known data
    In the arrangement shown in fig 1.24, the masses of the bar, m, and of the wedge, M, and also the wedge angle, ##\alpha##, are known. The masses of the thread and pulley are negligible, and friction is absent. Find the acceleration of the wedge M.
    irodov1.82.JPG

    3. The attempt at a solution
    Let the positive x direction be towards the wall, and the positive y direction be the direction of gravity.

    Let ##a_x## and ##a_y## be the respective components of mass m, and let ##a## be the magnitude of the acceleration of M. (Note: we expect ax to be negative.)

    We can write three net-force-equations for the three accelerations:
    ##ma_y=mg-F_N\cos\alpha##
    ##ma_x=T\cos\alpha-F_N\sin\alpha##
    ##Ma=T(1-\cos\alpha)+F_N\sin\alpha##

    Of course this introduces two more unknowns (the normal force FN between m and M, and the tension T in the thread).
    So we need two more equations for the problem to be solved.

    Conceptually, the two constraints that lead to these two equations are that m stays in contact with M, and that the length of the string is constant.

    We can describe the fact that m stays in contact with M with the following equation:
    ##\frac{a_y}{a-a_x}=\tan\alpha##

    Now about the constraint that the string length is constant... Well we can see the distance M moves towards the wall must be the distance m moves down the wedge. If we let ##\hat u## be the unit vector in the direction down the wedge's slope, and ##\hat x## and ##\hat y## be the unit vectors for the x and y directions respectively, then we can write this constraint as follows:
    ##a=(a_x\hat x+a_y\hat y)\cdot \hat u = a_x \hat x \cdot \hat u + a_y \hat y \cdot \hat u= a_y\sin\alpha-a_x\cos\alpha##


    So now we have five equations and five unknowns:
    ##ma_y=mg-F_N\cos\alpha##
    ##ma_x=T\cos\alpha-F_N\sin\alpha##
    ##Ma=T(1-\cos\alpha)+F_N\sin\alpha##
    ##\frac{a_y}{a-a_x}=\tan\alpha##
    ##a= a_y\sin\alpha-a_x\cos\alpha##

    But upon solving I don't get the correct answer. I thought I was careful with the algebra.
    Can anyone find any mistakes in these equations? (Particularly in the last two.) I can try to explain how I got any of them if requested.
     
  2. jcsd
  3. Oct 14, 2015 #2

    haruspex

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    In your ay equation, what about the tension?
     
  4. Oct 14, 2015 #3

    Nathanael

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    Of course o:) I'll give it another shot now.
     
  5. Oct 14, 2015 #4

    Nathanael

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    My answer: ##a=\frac{mg\sin\alpha}{M+2m-m\cos\alpha(1+\cos\alpha)}##

    The correct answer: ##a=\frac{mg\sin\alpha}{M+2m(1-\cos\alpha)}##

    It hurts to be so close :frown:
    I just want to know if my equations are wrong or if I'm making a careless mistake somewhere (5 equations takes a lot of algebra!).

    edit:
    and yes I fixed the ay equation properly:
    ##ma_y=mg-F_N\cos\alpha-T\sin\alpha##
     
  6. Oct 14, 2015 #5

    haruspex

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    Not sure about the equation you derived from the string length constancy. For the downslope acceleration, shouldn't you be using the acceleration relative to the wedge, so a-ax rather than -ax?
    [Sanity check: if alpha is zero then ax is zero.]
     
  7. Oct 14, 2015 #6

    Nathanael

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    You are correct, thank you for resolving this. Luckily that was the last equation I eliminated so I can check at once that it is now correct.

    My thinking was that the radial component of the motion of m (w.r.t. the point where the thread bends) would be along the direction of the wedge's slope. What I failed to consider is that the radial component of m's acceleration would appear different in a frame where the origin (the pulley) is not accelerating.

    Thanks again, Haruspex.
     
  8. Oct 14, 2015 #7

    haruspex

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    You're welcome. I do recommend that you develop the habit of doing extreme case sanity checks on equations.
     
  9. Oct 14, 2015 #8

    ehild

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    Try to use conservation of energy. You get the result much easier.
     
  10. Oct 14, 2015 #9

    Nathanael

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    Ah right.
    ##mga_y(0.5dt^2)=0.5M(adt)^2+0.5m(a_x^2+a_y^2)dt^2##
    In general, though, this method can only be used to find the initial acceleration, right?
    In this problem of course the accelerations are constant so the initial acceleration is always the acceleration.
     
  11. Oct 14, 2015 #10

    ehild

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    You can express both the KE and PE of the system at any time. PE+KE= const. Differentiating with respect to time and using the constraint equation, you get the acceleration of the wedge.
    ds/dt is the speed of the block downward along the slope, with respect to the slope. Because of the constraint, the velocity of the wedge is V=dX/dt=ds/dt.
    The components of velocity of the block in the rest frame of reference are ##v_x= V-\dot s \cos(\theta)=V(1-\cos(\theta))## and ##v_y=\dot s \sin(\theta)=V\sin(\theta)##

    ##KE+PE=0.5\left(mV^2(1-\cos(\theta))^2+V^2\sin^2(\theta)+MV^2\right)-mgs\sin(theta) =const##
    Simplify, take the derivative with respect to time, replace ##\dot s = V ##, divide the whole equation by V. Isolate ##\dot V##.
     
  12. Oct 14, 2015 #11

    Nathanael

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    Ahh, I see.

    The energy expression simplifies to ##0.5V^2(2m(1-\cos\alpha)+M)-mgs\sin\alpha=\text{constant}##
    Differentiating and dividing out the factor of V gives ##\dot V(2m(1-\cos\alpha)+M)-mg\sin\alpha=0## which gives ##\dot V##

    That is indeed a much easier method, nice! Thanks, Ehild.
     
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