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Moving Mass on a Moving Incline

  1. Jun 22, 2004 #1
    Here is another problem I'm struggling with: A mass m is sitting on an rough incline of mass M who in turn is resting on a frictionless surface. What minimum force F must be applied to the non-inclined side of the incline to make mass m move up the incline. The coefficient of static friction between m and M is [itex]\mu[/itex] and the incline has a slope that makes an angle [itex]\theta[/itex] with the horizontal.

    Let [itex]a_i[X][/itex] be the net acceleration of object [itex]X[/itex] along the [itex]i[/itex]-axis. Since m is moving up the incline while M is moving, then I must have
    [tex]
    0 < a_x[m] < a_x[M][/tex], and
    [tex]a_y[m] > 0[/tex].​
    Unfortunately, I don't get the right answer when I solve for F starting from these initial arguments. I don't know what else to do. Any help?
     
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  3. Jun 22, 2004 #2

    Gokul43201

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    Is the answer
    [tex]F = (M+m)g \frac {(sin \theta + \mu cos \theta)} {(cos \theta - \mu sin \theta)} ~[/tex] ?

    Or am I too screwing up somewhere ?
     
  4. Jun 22, 2004 #3
    Looks like the answer to me. How did you get that?

    Let me work this out so everyone can see what I'm doing. Let X be the magnitude of the force that M exerts on m and vice versa and let N be the magnitude of the force the surface exerts on M. Using Newton's second law yields
    [tex]\begin{align*}Ma_x[M] &= F - X\sin{\theta}\\
    Ma_y[M] &= N - Mg - X\cos{\theta}\\
    ma_x[m] &= X(\sin{\theta} - \mu\cos{\theta})\\
    ma_y[m] &= -mg + X(\mu\sin{\theta} + \cos{\theta})\end{align}[/tex]​
    Hmm...these four equations plus the two in my previous post give me six. Unfortunately I have seven unknowns. What equation am I missing? No wonder I got the answer wrong.
     
    Last edited: Jun 22, 2004
  5. Jun 22, 2004 #4

    Doc Al

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    I agree with Gokul43201's answer.

    The way to get that answer is to consider forces on m in the y direction to just balance and forces in the x direction to give an acceleration a. You can combine these relations to find out what the minimum "a" must be. (If you exceed "a" then there will be a net upward force on m.) Then find the force on the incline needed to produce that acceleration.
     
  6. Jun 22, 2004 #5

    Doc Al

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    Waaay too many equations. All you need is two to find "a" (forces in the x-direction; forces in the y-direction), then one more to find F.
     
  7. Jun 23, 2004 #6
    So you're saying that [itex]a_x[m] > 0[/itex] and that the sum of the forces on m in the y-direction is zero? And why is it "a" (I don't get the reason for the double-quotes)? I still don't get the right answer.
     
  8. Jun 23, 2004 #7

    arildno

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    Rather than use Newton's 2.law in "the absolute" horizontal and vertical reference frame for mass m, you should use Newton's 2.law in the accelerated frame of reference moving with the incline, and require balance of forces in the tangential and normal directions (to the incline), respectively.
    (While balance of forces in the normal direction is necessary in order for the particle m to remain on the incline, a greater than zero tangential acceleration might be allowed for)

    Note first, that in the inertial reference frame, this entails that the force F fullfil in the positive x-direction:
    [tex]F=(M+m)a[/tex]
    We have the following equations for particle m in the accelerated reference frame:
    [tex]N-ma\sin\theta-mg\cos\theta=0[/tex]
    [tex]-\mu{N}+ma\cos\theta-mg\sin\theta=0[/tex]

    Solving for a and N, and solving for F yields Gokul's answer.

    However, since m moves relative to M, it should be the coefficient of kinetic friction that comes into play, and not the coefficient of static friction (even if the surface of M has the property that these coefficients are equal)
     
  9. Jun 23, 2004 #8

    Doc Al

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    I look at the problem like this. Mass m has these forces acting on it: normal force from the incline, friction force from the incline, and its weight.

    In the vertical direction, I want to describe the point just before slipping begins: so that means equilibrium (net force = 0) with the friction acting down the incline. (Static friction.) That gives you one equation.

    In the horizontal direction, assume that there is an acceleration "a" (quotes just make it easier to read :smile:). Thus the net force Fx = ma. That's your second equation. You can now solve for "a".

    Once you have "a", what applied force will produce it? F = (M + m)a.

    Note that I interpreted the problem as what force just starts the block sliding upward, so I use static friction. And I saw no need to use an accelerated frame. (But, as usual, arildno's analysis is perfectly fine as well. :smile: )
     
  10. Jun 23, 2004 #9
    I'm really confused now. You say I can use Newton's laws in an accelerated reference frame but from what I've learned (particularly from my book), Newton's laws are invalid in such frames. So which is right here?
     
  11. Jun 23, 2004 #10
    I didn't use F = (M + m)a because it implies (at least to me) that both m and M have the same acceleration which further implies that if m is in equilibrium in the vertical direction, then m remains in the same place on the incline. The impression I'm getting is that F is the force needed so that m remains in the same location on the incline and so any force greater than this would cause m to move up the incline. Hence, this F is the minimum force the problem asks for. Does this make any sense?
     
  12. Jun 23, 2004 #11

    Gokul43201

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    Okay, I solved this in the rest frame of the floor.

    Remember, you want the limiting case where the bloch just starts to slide up the incline. So the relative acceleration between the block and the incline = 0. Or, the block and the incline have the same acceleration, a - as you seem to have figured. The normal force on the block is

    [tex]N=mgcos\theta + masin\theta [/tex]

    The second term in the above expression is clearly the normal component of the acceleration of the incline. This effect is the same as the increase in your weight in an elevator that's starting to go up.

    So the forces on the block along the incline are

    [tex]mgsin\theta + \mu N = mgsin\theta + \mu (mgcos \theta + masin \theta) [/tex]

    The component of the acceleration of the block along the incline is [tex] acos\theta[/tex]

    So, F(incl) = ma(incl) from the above 2 equations gives you 'a'.

    And finally use [tex]F = (M+m)a[/tex] to get F.
     
    Last edited: Jun 23, 2004
  13. Jun 23, 2004 #12

    Doc Al

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    But m and M do have the same acceleration! (Just before m starts to slip.) And mass m does not slide up the incline--yet!
    Exactly.

    Also: The way I would solve this is exactly as Gokul43201 explained it.
     
  14. Jun 23, 2004 #13

    arildno

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    Let's make a careful examination of the concept of "accelerated reference frame" and inertial reference frame.

    Suppose we have an inertial frame I and an object O, whose acceleration in I satisfies Newton's 2.law:
    [tex]F=ma_{I}[/tex]
    I have put the subscript I to signify that this is the acceleration of O as measured in I.

    Now, let us first consider a reference frame I' whose origin moves with velocity [tex]v_{0'}[/tex] as measured in I (axes in I' remains parallell to axes in I).
    Let us first consider the case where an observer in I measures
    [tex]v_{0'}[/tex] to be a constant in time.

    Now, by the Galilean transformation of velocities, the velocity of object O as measured in I' is related to object O's velocity as measured in I by:
    [tex]v_{I'}=v_{I}-v_{0'}[/tex]

    We want now to find the acceleration of O as measured in I':
    [tex]a_{I'}=\frac{dv_{I'}}{dt}=\frac{dv_{I}}{dt}-\frac{dv_{0'}}{dt}[/tex]
    But [tex]v_{0'}[/tex] was constant in time; hence we have:
    [tex]a_{I'}=a_{I}[/tex]

    That is, an observer in I' measures O's acceleration to be the same value as the one an observer in I gets!

    But, therefore, we gain:
    [tex]F=ma_{I'}[/tex]
    That is, Newton's 2.law is valid even though we look at O from I' rather than from I
    Hence, I' is also an inertial frame.

    Now, let's take the case with a reference frame I'', whose origin's velocity is meaused as non-constant in I (axes still parallell, for simplicity)
    Doing the same computations, we get:
    [tex]a_{I''}=a_{I}-a_{0''}[/tex]
    Substituting for [tex]a_{I}[/tex] in Newton's 2.law and rearranging, we get:
    [tex]F-ma_{0''}=ma_{I''}[/tex]

    That is, the set of forces an observer in I'' can glimpse from measuring O's acceleration is not the same set of forces that an observer in I measures!
    For an observer in I'' it seems there is a new force present (we call it a fictitious force) which is equal to [tex]-ma_{0''}[/tex]

    This is what is meant by using Newton's 2.law in an accelerated frame of reference:
    To add the requisite fictitious forces to the left-hand side of the equation.
    (Often, this simplifies problems)
     
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