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A sliding block - what's wrong with my arguement

  1. Aug 16, 2008 #1
    I'm trying to solve the http://www.geocities.com/kemboja_4a/problem.JPG".

    I have done the following, using Newton's Second Law

    [tex]\frac{N}{\sqrt{2}}i + \frac{N}{\sqrt{2}}j - mgj = m(\ddot{x} i + \ddot{y}j) [/tex]

    where N is the normal force on the block m. Comparing the coefficients where [tex]\ddot{x}=A [/tex],

    [tex]\frac{N}{\sqrt{2}} = mA[/tex] and [tex]\frac{N}{\sqrt{2}} - mg= m \ddot{y}[/tex].

    This gives [tex]\ddot{y}=A - g[/tex].

    But the clue in the book gives [tex]\ddot{y}=g[/tex] if A=3g. I don't think the answer given in the book is wrong (because I never encounter with incorrect answer before). What could possibly be wrong with my arguement ?
    Last edited by a moderator: Apr 23, 2017
  2. jcsd
  3. Aug 16, 2008 #2

    Doc Al

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    Staff: Mentor

    What makes you think that's true? That's the acceleration of the wedge, not the block.
  4. Aug 16, 2008 #3
    The block is always in contact with the wedge, which moved horizontally with acceleration A. This must implies that the horizontal acceleration of the block is also A.
  5. Aug 16, 2008 #4

    Doc Al

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    This is true.
    This is not true. Note that the block slides down the wedge.

    Hint: It may be easiest to analyze the problem from an accelerating frame in which the wedge is at rest.
  6. Aug 16, 2008 #5
    Yes, the block slides down. That's why the acceleration of the block is not the same as the acceleration of the wedge. The difference between the two is only in the acceleration vertical component, I presume. Hence the [tex]\ddot{x}=A[/tex].

    Thanks Doc Al for the hint. But I do not know how to analyse the problem in an accelerating frame. Need to do some reading then.
  7. Aug 16, 2008 #6

    Doc Al

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    That presumption is incorrect. Imagine you were riding along with the wedge. Since the block slides down the wedge (at 45 degrees), its acceleration with respect to the wedge has both a vertical and horizontal component.
    Reading is good. But if you haven't worked with accelerating frames, you can still do the problem from an inertial frame. The trick is to incorporate the constraint that the block maintains contact with the wedge. Try this: Call the acceleration of the block with respect to the wedge "a", then set up the equations of motion for horizontal and vertical components.
  8. Aug 16, 2008 #7
    Tricky problem. Took me awhile to figure it out. Couldn't have done it without Doc Al's hints.

    To the OP, the force equation you started with is correct for a block on a wedge which is not moving. That is a good start. This gives you two equations with three unknowns, ax, ay, N. There are two things you need to realize to finish the problem. One is that forces and thus acceleration (via F=ma) can simply be vectorally added. You need to account for the acceleration of the Wedge in your equation of motion by adding the Wedge acceleration to your block acceleration (which assumes the Wedge is not moving).

    The second thing to realize is the fact that the block stays in contact with the wedge gives you the third equation you need to solve for the three unknowns. I found it difficult to incorporate the constraint using the reference frame provided. I found it easier to make my own reference frame x' and y' that had x' pointing down the ramp and y' normal to the ramp surface. Of course once I solved for a in x',y' I had to do a little trig to get the final answer in x, y. You might try that unless this just causes more confusion than it helps.
  9. Aug 16, 2008 #8
    I have done some reading on accelerating frame and also gone through again the note in the same book. Really appreciate all the responds before. There is a flaw in my previous arguement.
    Just checking whether I have done correctly this time.

    Method 1
    The equations of motion parallel to the slope http://www.geocities.com/kemboja_4a/fig.JPG"is

    [tex]\frac{mg}{\sqrt{2}} - \frac{F}{\sqrt{2}}=ma, \\\ F=mA[/tex]

    [tex]\Rightarrow\\\ \ddot{y}=-\frac{a}{\sqrt{2}}=-\frac{1}{2}(g - A)[/tex]

    Hence if A=3g, we obtain [tex]\ddot{y}=g[/tex]. (agree with the clue in the book)
    My concern: Why is the vertical component accerelation positive? Is it because the block is slowing down?

    We should have a different set of coordinate system for the wedge and the block.
    Using similar triangle (h is the height and base of the wedge).
    Differentiate twice wrt time,
    This is the relation that relate [tex]\ddot{x}[/tex] to A that I was wrong previously.
    Last edited by a moderator: Apr 23, 2017
  10. Aug 16, 2008 #9
    If A is 3g, then that is a very hard pull. Basically the wedge is being pulled so hard that it causes the block to slide up the wedge instead of down. If the block slides backwards and up, then y acceleration is up which is positive.
  11. Aug 17, 2008 #10

    Doc Al

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    To add to what Chrisas already explained: The greater the acceleration of the wedge, the greater the non-inertial force on the block (what you call F), which has a component up the incline. If it's great enough, the block is pushed up the wedge.

    Viewed from the inertial frame, compare the vertical component of the normal force--which increases with A--to the weight.


    Another way to view it is to call the acceleration of the block with respect to the wedge "a" (down), thus the acceleration of the block with respect to the inertial frame will have components:
    [tex]\ddot{x} = A + a \sqrt{2}/2[/tex]
    [tex]\ddot{y} = -a \sqrt{2}/2[/tex]
    Last edited by a moderator: Apr 23, 2017
  12. Aug 17, 2008 #11
    Chrisas and Doc Al, both of you are :cool:

    Thank you.
  13. Nov 14, 2008 #12
    hey I just tried the question.. But I m stuck on the first equation.. anyone would like to enlightene on the working? Thx
  14. Nov 15, 2008 #13
    Just forget my not so good solution. Try to understand the solution by Doc Al.

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