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This is so annoying me

  1. May 28, 2005 #1
    Another problem I cant solve

    A box of mass 6kg lies on a rough plane inclined at an angle of 30 to the horizontal. The box is held in equilibrium by means of a horizontal force of magnitude P newtons.
    The coefficent of friction between the box and the plane is 0.4.

    Find the normal reaction exerted on the body by the plane.

    The box is in limiting equilibrium and on the point of moving up the plane. :confused:
  2. jcsd
  3. May 28, 2005 #2
    You know that in limiting equilibrium, the frictional force is [itex]F = \mu R[/itex], where R is the normal reaction.

    The box is in equilibrium, but if it were to move, it would move up the plane. That should tell you in which direction the forces on the box are (or rather, if they're in the same direction, or in opposite directions).

    Draw a diagram, those help loads in A Level Maths!

    Edit: you don't even need all that info to find the normal reaction. Resolve the weight into components parallel and perpendicular to the slope and you can get the normal reaction from that.
    Last edited: May 28, 2005
  4. May 28, 2005 #3
    but the force P (which is unknown at the moment) will also need to be taken into consideration. If I resolved into perpendicular components it becomes Pcos60 +6gcos30 = R?
  5. May 28, 2005 #4
    Ahh, sorry. I didn't see that horizontal force bit. Yes, you're right about that.
  6. May 28, 2005 #5
    I included P in my equations and then elinated it. But I dont get the right answer. What would be your suggested method?
  7. May 28, 2005 #6
    Post your working, then we can see where you've gone wrong (if anywhere!). You're not using Edexcel books are you? Sometimes they get the answers wrong.
  8. May 28, 2005 #7
    Resolving perpendicualr to the plane:

    R = Pcos60 + 6gcos30

    Resolving parallel to the plane:

    Fr (friction) = Pcos30 - 6gcos60

    is this right so far?
  9. May 28, 2005 #8

    Doc Al

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    Staff: Mentor

    Write the equations representing the equilibrium condition for forces parallel and perpendicular to the plane. You'll get two equations and two unknowns (P and the Normal reaction); solve for N.
  10. May 28, 2005 #9
    Yep, looks ok to me.
  11. May 28, 2005 #10
    and Fr (the friction will also be in the equation which is the problem.
  12. May 28, 2005 #11
    Why is the frictional force a problem?
  13. May 28, 2005 #12

    Doc Al

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    Staff: Mentor

    But the friction can be written in terms of R. Since it is just about to move, the static friction will equal its maximum value [itex]F_r = \mu R[/itex], as was pointed out by Nylex in his first post in this thread.
  14. May 28, 2005 #13
    ok. I just did everything again. And I found out that I went wrong with the sins and cos during the calcualtions. My value for R comes out to be 88.3N which is correct.
  15. May 28, 2005 #14
    Yes I used Fr = uR
  16. May 28, 2005 #15

    Andrew Mason

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    Science Advisor
    Homework Helper

    The last point tells you that the actual friction force is the maximum static friction force (ie. exactly the coefficient of static friction multiplied by the normal force). The components of force along and perpendicular to the plane are:

    Parallel to plane:

    (1) [tex]mgsin(\theta) + Pcos(\theta) + \mu_s N = 0[/tex]


    (2) [tex]mgcos(\theta) + Psin(\theta) = N[/tex]

    Substitute N from (2) into (1) to solve for P in terms of mg and [itex]\theta[/itex]:

    [tex]mgsin(\theta) + Pcos(\theta) + \mu_s (mgcos(\theta) + Psin(\theta)) = 0[/tex]

    [tex]mgsin(\theta) + \mu_smgcos(\theta) + P(cos(\theta) + \mu_ssin(\theta)) = 0[/tex]

    [tex]P = \frac{-(mgsin(\theta) + \mu_smgcos(\theta))}{cos(\theta) + \mu_ssin(\theta)}[/tex]

    Substitute that into (2) to get N.

  17. May 28, 2005 #16
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