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Weight/Normal Force of a block

  1. May 30, 2008 #1
    [SOLVED] Weight/Normal Force of a block

    1. The problem statement, all variables and given/known data

    If P = 1.98, M = 1, Theta = 45 degrees what is the weight of m in newtons?

    [​IMG]

    2. Relevant equations



    3. The attempt at a solution

    I'm apparently doing this wrong. I was trying to solve for the normal force given that N = Mg/cos theta but this is just 9.8 so I'm doing something wrong. Any advice would be appreciated.
     
    Last edited: May 30, 2008
  2. jcsd
  3. May 30, 2008 #2

    Doc Al

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    Any further description of the problem? Is the block sliding down the incline?
     
  4. May 30, 2008 #3
    P (1.98) is such that the block is not moving.

    Thats all I'm given :(
     
  5. May 30, 2008 #4

    Doc Al

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    Start by identifying the forces acting on each mass. Draw a free body diagram for each.

    What can you say about the net force on the block?
     
  6. May 30, 2008 #5

    Doc Al

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    How did you deduce this?
     
  7. May 30, 2008 #6
    N cosθ – mg = ma(y). a(y) = 0 so N = Mg/cos theta?

    Forces on triangle block are W (Mass pointed down), P (->), Normal force up mgcos(theta) opposite the angle of the block?

    Small block is w pointed straight down and Normal force at an angle opposite (90d?)
     
  8. May 30, 2008 #7

    Doc Al

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    Good. But don't mix M and m: N = mg/cos(theta).
    Only worry about horizontal forces on the triangle.
    OK

    Keep going. Apply Newton's 2nd law to the horizontal direction and see what you can deduce. Hint: What net horizontal force acts on the block? On both masses?
     
  9. May 30, 2008 #8
    P = (M + m) g tanθ?

    But this gives me m = .2020408163 so * 9.8 just = 1.98 again which is wrong.

    N = Nx i + Ny j = N sinθ i + N cosθ j

    –Nsinθ = –mgtanθ

    P – mg tanθ = Max

    N sinθ = mg tanθ = max

    P = (M + m) g tanθ


    I guess I'm just having a hard time seeing what I need to get out of this.

    F = Mgcos(45) - mgsin(45)? That doesn't even look right...
     
  10. May 30, 2008 #9

    Doc Al

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    No. Just apply Newton's 2nd law to M + m. (Where did you get the g tanθ?)

    After you do that, apply it to the block alone.
     
  11. May 30, 2008 #10
    Ah its equal to 1! Sweet! I dunno if I did it right or not though. I did N = Mg/cos(45) to find N then N = mg/sin(45) so m = 1!
     
  12. May 30, 2008 #11

    Doc Al

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    Actually, I take it back. This seems correct to me. (I just didn't see how you got. You gave your conclusion first. :wink:)

    But your data doesn't seem OK.
    I don't see how you deduced this value for m.
     
  13. May 30, 2008 #12

    Doc Al

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    Where did you get N = Mg/cos(45) = mg/sin(45)? I thought we had established that N = mg/cos(45).
     
  14. May 30, 2008 #13
    I was just trying to find a value for N without having m. Like I said I got the right answer but I'm sure I did it wrong
     
  15. May 31, 2008 #14

    Doc Al

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    What makes you think you got the right answer?
     
  16. May 31, 2008 #15

    tiny-tim

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    Hi AnkhUNC! :smile:

    waaah … you've left out the y-component of the poor little friction force. :cry:

    When you draw a diagram, you should always mark in all the forces

    You need to take components in the normal direction, so that the friction component will be 0.

    Then N = ? :smile:
     
  17. May 31, 2008 #16

    Doc Al

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    Don't cry, tiny-tim! Now I'm starting to cry... :cry:

    I would assume, lacking any statement to the contrary, that the surfaces are frictionless.
     
  18. May 31, 2008 #17
    The answer checked as right but like I said I was sure I did it wrong. If its not moving then N is going to be equal to P?

    Any websites you can recommend to help with these types of problems? :(
     
  19. May 31, 2008 #18

    Doc Al

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    How about the site you're on right now? :wink:

    Just attack it systematically, as I suggest in post #7. (I assume the surfaces are frictionless, correct?)

    (I'd say that given the data you supplied, there is no correct answer. But you can solve for m symbolically in terms of P and M.)
     
  20. May 31, 2008 #19
    Yeah there isn't a frictional force. I think the problem was just to get me thinking about the process and I spent way too much time on it :P
     
  21. May 31, 2008 #20

    Doc Al

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    Just do it step by step. Analyze the forces and apply Newton's 2nd law:
    (1) To the block (vertical direction)
    (2) To the block (horizontal direction)
    (3) To the entire system (horizontal direction)
     
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