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A block pushed up a wall

  1. Sep 25, 2009 #1
    1. The problem statement, all variables and given/known data

    A 2.0 kg block is pushed 3.0 m at a constant velocity up a vertical wall by a constant force applied at an angle of 27 degrees with the horizontal.

    The acceleration of gravity is 9.81 m / s2.

    If the coefficient of kinetic friction between the block and the wall is .30 ...

    A) Find the work done by the force on the block. Answer in units of J.

    B) Find the work done by gravity on the block. Answer in units of J.

    C) Find the magnitude of the normal force between the block and the wall. Answer in units of N.

    2. Relevant equations

    Wg = mgd

    w = Fdcos(theta)

    3. The attempt at a solution

    I actually figured out part 2 by using Wg = mgd

    I just cant seem to figure out how to set up parts one and three. I think I have to use w = Fdcos(theta) and set that equal to the force of gravity for part one. And for part three I have to use uN which will equal umg.
     
  2. jcsd
  3. Sep 25, 2009 #2

    kuruman

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    You cannot do this problem correctly unless you draw a free body diagram and put in all the forces. When you do that, you will see that the normal force is not mg. "Normal" means "perpendicular to the surface". Here the weight mg is parallel to the surface.
     
  4. Sep 25, 2009 #3
    Okay, I drew a free body diagram and I see what you mean which makes this question seem even more confusing. I believe I am trying to solve for the force between the block and the wall but the normal force is not even in that direction.

    Also, for part 1:

    I believe I need to use w = Fdcos(theta) but since I am not given a force and it is constant, can I assume that it is just zero? So I would then have w = dcos(theta)?
     
  5. Sep 25, 2009 #4
    The normal force is the force between two objects in contact, perpendicular to the surface of their contact. It is NOT the force that opposes gravity.

    Draw the free body diagram, and remember that the normal is between the mass and the wall.
     
  6. Sep 25, 2009 #5
    That is what I first thought. But before I answer the third part I would really like to figure out the first part. Is my assumption correct and would w = dcos(theta) actually work?
     
  7. Sep 25, 2009 #6
    Yes, that would be perfectly true. That's just the definition of work.
    You have to pay close attention to what the net force parallel to the wall is, though, or you'll end up making the same mistakes.
     
  8. Sep 25, 2009 #7
    So the net force parallel to the wall for the first problem would be uN. I am not sure whether N is mg - sin(theta) or mg - cos(theta).
     
  9. Sep 25, 2009 #8

    Doc Al

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    :confused: Why do you think you can just assume that the force is zero? Draw a free body diagram and figure out what the force is.
    No. If you assume that F = 0, then w = 0. (w = dcos(theta) assumes F = 1, not F = 0.)
     
  10. Sep 25, 2009 #9
    This is also false.
    First, the force parallel to the wall is not just [tex]\mu N[/tex]
    Second, your suggestions for [tex]N[/tex] are dimensionally inconsistent (Forces+pure numbers don't add up)

    Please, draw out the free body diagram.
    Find the net force in the [tex]x,y[/tex] axes. What you are allowed to assume is that the block doesn't move in the x axis (That wouldn't make much sense either)
    Use NSL to find the value of [tex]N[/tex] and the net force parallel to the wall.

    I'll give you a hint about the normal force, it doesn't take [tex]mg[/tex] in as a factor.
     
  11. Sep 25, 2009 #10
    Okay, I drew out my free body diagram again and came up with this:

    Forces in the horizontal:

    N - Fdsin27

    Forces in the vertical:

    Fdcos27 - mg - uN

    So N = Fsin27 and the forces in the vertical will equal:

    Fdcos27 - mg - uFdsin27

    I guess the next step would be to solve for F, but before I go on I want to make sure I am correct in my force equations.

    Thanks for the help.
     
    Last edited: Sep 25, 2009
  12. Sep 25, 2009 #11

    Doc Al

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    Looks good except for mixing up sine and cosine.
     
  13. Sep 26, 2009 #12
    Okay, I will try solving for F and post the equation.
     
  14. Sep 26, 2009 #13
    Fdsin27 - mg - uFdcos27 = 0
    -uFdcos27 = -Fdsin27 + mg
    F = utan27 + mg

    I am not sure if I can divide out the d or if the F's would end up on the left side. My algebra is a bit weak.
     
  15. Sep 26, 2009 #14

    Doc Al

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    I take it back. What's with the d's?
    In addition to the sine/cosine mixup, you have extraneous d's stuck in there. Get rid of them. (Sorry for not spotting that earlier.)

    The d should not be there. For example, the vertical component of F is Fsin27, not Fdsin27. Don't mix up force and work. Here you're just dealing with forces.
     
  16. Sep 27, 2009 #15
    The first part of the problem asks for work done by the force. I thought that in order to solve for work, a distance has to be taken into account?
     
  17. Sep 27, 2009 #16

    Doc Al

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    That's true, but first you need to solve for the force. Then you can worry about the work done.

    Also: in any physically meaningful equation, every term must have the same units. You can't have a work term (Fd) added to a force term (mg).
     
  18. Sep 27, 2009 #17
    Okay, so I plug my numbers into F = utan27 + mg and then use that F with W = Fd? Where d is equal to 3 m.
     
  19. Sep 27, 2009 #18

    Doc Al

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    That equation isn't correct; redo your algebra in solving for F. (Note that the "utan27" term has no units, thus cannot be added to a force term.)
    Don't forget to take the angle into account. W = Fd only if F and d are in the same direction.
     
  20. Sep 27, 2009 #19
    How does this look?

    Fsin27 - mg - uFcos27 = 0
    -uFcos27 = mg - Fsin27
    F = tan27/u + mg
     
  21. Sep 27, 2009 #20

    Doc Al

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    OK.
    :yuck:

    Just put all terms containing F on one side, then isolate F.

    Can you solve this equation?:
    ax -c -bx = 0
    If so, it's the same thing here.
     
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