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Force that the wall exerts on the wedge

  1. Oct 26, 2007 #1

    klm

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    A wedge with an inclination of angle theta rests next to a wall. A block of mass m is sliding down the plane, as shown. There is no friction between the wedge and the block or between the wedge and the horizontal surface.

    Find the magnitude, F_ww, of the force that the wall exerts on the wedge.
    Express F_ww in terms of theta and m, along with any necessary constants.

    [​IMG]

    ok i tried it and i got mgcos(theta)^2 . but it saying that is incorrect. i drew my free body diagram and cant figure out where i am going wrong. i found Fnet = mg sin theta as well
     
  2. jcsd
  3. Oct 26, 2007 #2
    I think what you need to do is calculate the horizontal force exerted by the block on the wedge. And that would be the force that the wedge exerts on the wall and hence the force the wall exerts back on the wedge.

    Did you do that?
     
  4. Oct 26, 2007 #3

    klm

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    isn't the only force the box acting on the wedge is weight
     
  5. Oct 26, 2007 #4

    klm

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    no i mean, like okay we know n=mg cos theta right. and that is the force the box is acting on the wedge and the same the wedge is acting on the box right?
     
  6. Oct 26, 2007 #5
    Hmm, it's been a while since I did mechanics, but since this is in an incline, there has to be a force in the horizontal vector as well as the vertical doesn't it?

    So the horizontal vector would be found by F = mg sin theta
     
  7. Oct 26, 2007 #6
    I am really not sure though, best if someone else answers this.
     
  8. Oct 26, 2007 #7

    klm

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    right i found mgsintheta for my fney x. but i dont understand how that is going to help...sorry!
     
  9. Oct 26, 2007 #8

    klm

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    thanks for trying =)!
     
  10. Oct 26, 2007 #9
    Well what is the answer given?
     
  11. Oct 26, 2007 #10

    klm

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    i dont know! i have to try to figure what the force of the wall exerts on the wedge.
     
  12. Oct 26, 2007 #11
    Well try putting in mg sin theta as the answer, and cya.
     
  13. Oct 26, 2007 #12

    klm

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    i dont think it should be mg sin theta only
     
  14. Oct 26, 2007 #13

    klm

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    nvm i think i figured it out. thanks guys
     
  15. Oct 26, 2007 #14
    What was it? I am interested to know :)
     
  16. Oct 26, 2007 #15

    Kurdt

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    Staff Emeritus
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    Gold Member

    What you are doing here is resolving components of forces twice. So you need to find the component of weight acting perpendicular to the incling which you have as [itex]mg\cos(\theta)[/itex]. Then you have to resolve this force into the component that pushes the block against the wall.
     
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