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Homework Help: Masses slide on a smooth wedge

  1. Sep 10, 2016 #1
    1. The problem statement, all variables and given/known data
    Snap1.jpg
    Mass m lies on a Weighing scale which is on Wagon M. the inclined surface is smooth, between m and M there is enough friction to prevent m from moving.
    1) What does the weigh show?
    2) What is the minimum coefficient of friction between m and M to prevent slipping?
    3) What is M's acceleration parallel to the slope if there is no friction between m and M? can it be bigger than ##g\sin\alpha##?
    4) What does the Weighing scale show in this case?

    2. Relevant equations
    Mass-acceleration: F=ma

    3. The attempt at a solution
    1) F=ma. ##a_{\alpha}## is parallel to the slope:
    $$(M+m)g\sin\alpha=(M+m)a_{\alpha}~~\rightarrow~~a_\alpha=g\sin\alpha,~~a_y=a_\alpha\sin\alpha=g\sin^2\alpha$$
    W is the weight, what the scale shows:
    $$W=mg\sin\alpha$$
    Snap1.jpg 2)
    $$a_x=a_\alpha\cos\alpha=g\sin\alpha\cos\alpha$$
    $$f=ma_x:~mg\mu=mg\sin\alpha\cos\alpha~~\rightarrow~~\mu_{\rm min}=\sin\alpha\cos\alpha$$
    3)
    $$(M+m)g\sin\alpha=Ma_\alpha~~\rightarrow~~a_\alpha=\frac{M+m}{M}g\sin\alpha$$
    Yes, it's bigger than ##g\sin\alpha##
    4)
    $$a_y=a_\alpha\sin\alpha=\frac{M+m}{M}g\sin^2\alpha$$
    $$W=ma_y=\frac{m}{M}(M+m)g\sin^2\alpha$$
     
  2. jcsd
  3. Sep 10, 2016 #2

    TSny

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    One way to check an answer is to consider limiting cases. Does your answer for (1) give you a reasonable result for ##\alpha## going to ##0## or ##\pi/2##?
     
    Last edited: Sep 10, 2016
  4. Sep 10, 2016 #3

    haruspex

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    Justify that claim. Consider the FBD of m.
     
  5. Sep 11, 2016 #4
    ##a_\alpha=g\sin\alpha## seems reasonable since when α→0 aα→0
    The weight, as the scale shows:
    $$mg-W=ma_y=mg\sin^2\alpha~~\rightarrow~~W=(1-\sin^2\alpha)mg$$
     
  6. Sep 11, 2016 #5

    haruspex

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    Right. Which simplifies to....?
     
  7. Sep 11, 2016 #6
    1)
    $$mg-W=ma_y=mg\sin^2\alpha~~\rightarrow~~W=(1-\sin^2\alpha)mg=mg\cos^2\alpha$$
    4)
    $$mg-W=ma_y=m\frac{M+m}{M}g\sin^2\alpha~~\rightarrow~~W=\left[ 1-\frac{M+m}{M}g\sin^2\alpha \right]mg$$
     
    Last edited: Sep 11, 2016
  8. Sep 11, 2016 #7

    TSny

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    OK
    You have ##a_y=\frac{M+m}{M}g\sin^2\alpha##. Does this behave properly as ##\alpha \rightarrow \pi/2##?

    You need to go back and rework parts (2), (3), and (4). Use separate FBD's for m and M.
     
  9. Sep 12, 2016 #8
    Snap1.jpg
    The first equation is for M and the second for m. W is m's weight as M sees it, i.e. the force applied on M by m:
    $$\left\{\begin{array}{l} (W+mg)\sin\alpha-W\mu\cos\alpha=Ma_\alpha \\ ma_\alpha\cos\alpha =W\mu \end{array}\right. ~~\rightarrow~~W=\frac{mMg\sin\alpha\cos\alpha}{(\mu\cos^2\alpha-\sin\alpha\cos\alpha)m+M\mu}$$
    W becomes 0 when α=0, wrong. i expect it to be mg, am i right?
     
    Last edited: Sep 12, 2016
  10. Sep 12, 2016 #9

    PeroK

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    Why not try part 2). Once you get that, move on to parts 3 & 4.

    Also, for 3 & 4, the equations of motion for m and M are coupled, so you may not be able to solve them independently.
     
    Last edited: Sep 12, 2016
  11. Sep 12, 2016 #10
    Post #8 is for part (2), but the answer doesn't comply with the boundary condition ##\alpha=0##. i expect, for ##\alpha=0## that W=mg, but if i substitute ##\alpha=0## in my W, i get W=0
     
  12. Sep 12, 2016 #11

    PeroK

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    I don't really understand post #8. From part 1 you have ##W_1 = mgcos^2(\alpha)##

    So:

    a) What is the maximum frictional force between ##m## and ##M##?

    b) What horizontal frictional force between ##m## and ##M## is needed to sustain the motion without slipping?
     
  13. Sep 12, 2016 #12
    The maximum friction force between m and M is ##f=mg\mu\cos^2\alpha##. the horizontal friction force needed to sustain the motion is what i wrote in the OP:
    $$a_x=a_\alpha\cos\alpha=g\sin\alpha\cos\alpha$$
    $$f=ma_x,~~W\mu=ma_x,~~mg\mu\sin^2\alpha=mg\sin\alpha\cos\alpha~~\rightarrow~~\mu_{\rm min}=\frac{1}{\tan\alpha}$$
     
  14. Sep 12, 2016 #13

    PeroK

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    You've got the ##sin^2## and ##cos^2## mixed up. It's ##\mu \ge tan \alpha##.

    Anyway, I think 3 & 4 are quite tricky. You can see by energy considerations that ##M## has more energy in case 3 after both masses have fallen a given vertical height, as ##m## has no horizontal velocity in case 3, unlike case 1. Or, to put it another way:

    The larger ##m## becomes, the more force it exerts on ##M## and the greater the acceleration of ##M##. But, the greater the vertical acceleration, the less proportion of the normal weight of ##m## is felt by ##M##.

    Can you see how to use this to set up some equations for ##a## and ##W_3##?
     
  15. Sep 13, 2016 #14
    I made in post #1 for case 3:
    $$(M+m)g\sin\alpha=Ma_\alpha~~\rightarrow~~a_\alpha=\frac{M+m}{M}g\sin\alpha$$
    $$a_y=a_\alpha\sin\alpha=\frac{M+m}{M}g\sin^2\alpha$$
    $$mg-W_3=ma_y~~\rightarrow~~W=(g-a_y)m=\left[ 1-\frac{M+m}{M}\sin^2\alpha \right]mg$$
     
    Last edited: Sep 13, 2016
  16. Sep 13, 2016 #15

    TSny

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    In post #8 you assume correctly that both blocks have the same component of acceleration, ##a_{\alpha}##, along the incline . Is this true for parts 3 and 4?

    For these parts, in what direction must the two blocks have the same component of acceleration?

    EDIT: Backing up to your first post, you wrote for part 3 that ##(M+m)g\sin\alpha=Ma_\alpha##. Can you explain how you got the left side?
     
  17. Sep 13, 2016 #16
    ##a_y## is the same for M and m. did you see my new, completed, post #14?
     
  18. Sep 13, 2016 #17

    TSny

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    I don't see how you got the left side of the first equation.

    Yes.
     
  19. Sep 13, 2016 #18

    PeroK

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    I see how you got this, but it's not right. You imagine that ##m## is pushing with its full, normal weight down on ##M##. But, because they are both accelerating downwards, it should be:

    ##(M + W_3)## instead of ##(M + m)##
     
  20. Sep 15, 2016 #19
    3) The first is FBD for M and the second is for m:
    $$\left\{ \begin{array}{l} [W+Mg]\sin\alpha=Ma_\alpha \\ mg-W=ma_y=ma_\alpha\sin\alpha \end{array}\right.~~\rightarrow~~a_\alpha=\frac{(M+m)\sin\alpha}{M+m\sin^2\alpha}$$
     
  21. Sep 15, 2016 #20

    PeroK

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    If you add "g" somewhere, you've got it!
     
  22. Sep 15, 2016 #21
    $$\left\{ \begin{array}{l} [W+Mg]\sin\alpha=Ma_\alpha \\ mg-W=ma_y=ma_\alpha\sin\alpha \end{array}\right.~~\rightarrow~~a_\alpha=\frac{(M+m)\sin\alpha}{M+m\sin^2\alpha}g$$
     
  23. Sep 15, 2016 #22

    TSny

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    Looks right. It's interesting to consider the limiting case where m >> M and α → 0.
    EDIT: The behavior of ##a_\alpha## for ##\alpha##→ 0 is very different for M = 0 compared to M ≠ 0 with M << m. Graphing ##a_\alpha## vs ##\alpha## for various M/m values is useful. I think this has something to do with childhood memories of squeezing a slippery watermelon seed between thumb and forefinger to launch the seed across the room (not at anybody of course :oldeyes:).
     
    Last edited: Sep 15, 2016
  24. Sep 16, 2016 #23
    I will try and post here the graphs
     
  25. Sep 16, 2016 #24
    17.9.jpg all the graph's results must be multiplied by g, the gravity constant.
    I don't understand why, for M=0, aα tends to infinity. it must be a mistake of mine in the graph.
     
    Last edited: Sep 16, 2016
  26. Sep 16, 2016 #25

    PeroK

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    ##M \rightarrow 0## is equivalent to ##m >> M##
     
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