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Newtons Law Question

  1. Oct 4, 2006 #1
    Sorry people, not sure where im going wrong, im just not sure in which direction to take this question. I dont even know where to start.

    A box with mass of 22kg is at rest on a ramp inclined at 45 degrees to the horizontal. The coeffecient of friction between the box and ramp are (Static friction = 0.78, Kinetic Friction = 0.65)

    a) determine the magnitude of the largest force that can be applied upward, parallel to the ramp, if the box ix to remain at rest

    b) determine the magnitude of the smalled force that can be applied onto the top of the box, perpendicular to the ramp, if the box is to remain at rest
     
  2. jcsd
  3. Oct 4, 2006 #2
    for a start maybe u want to try drawing the free body diagram of the object and resolve the forces into 2 direction...
     
  4. Oct 4, 2006 #3

    arildno

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    a)
    i. Will an applied force parallell to the ramp influence the normal force?
    ii. If you apply a "strong" parallell force upwards along the incline, in which direction will the frictional force point? Upwards along the incline or downwards?
     
  5. Oct 4, 2006 #4
    i am doing this correspondence course and i dont have a teacher helping me so im not exactly sure where to take this question. i have drawn the free body diagram and i am trying to resolve it into two direction im just not sure what formulae i am supposed to be using to solve this question. As far as answering those questions you asked i have no idea thats exactly how its phrased in the book. As i said i have nobody else to ask for help
     
  6. Oct 4, 2006 #5

    arildno

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    Try to answer the questions I gave you, they are perfectly clear. Use your head for example.
     
  7. Oct 4, 2006 #6
    sorry about that .. i wasnt really paying attention
    ai) an applied force will not influence the normal force.. at least i dont think it should
    aii) frictional force will point downwards
     
  8. Oct 4, 2006 #7

    arildno

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    Correct!

    So, according to ai), you should be able to calculate the normal force without knowing what the applied force is. Do so!

    As for ii), how is MAXIMAL STATIC FRICTION related to the normal force?
     
  9. Oct 4, 2006 #8
    Coeffecient of static friction = Static friction max / force normal ??
     
  10. Oct 4, 2006 #9

    arildno

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    Quite so, meaning that Static friction max=?
     
  11. Oct 4, 2006 #10
    force normal / coeffecient of static friction.

    ok so here is what i have so far.. i calculated the force of gravity. 22 (9.8) then to calculate force norm do i do Fg cos 45?
     
  12. Oct 4, 2006 #11

    arildno

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    "force normal / coeffecient of static friction"
    TOTALLY WRONG! THINK AGAIN!

    Do not insert digits yet, keep the letters!
    So, the force of gravity is mg, where m is the mass and g is the acceleration due to gravity.
    Furthermore, as you've found out, the normal force N=mgcos(45).

    Now, what should you do further?
     
  13. Oct 4, 2006 #12
    its sin 45.. ok.. so what i got so far is that force normal = mass (acceleration) sin 45, which = 22 (9.8) sin 45 which = 152.45 N correct?
    now.. i take that and its force normal / coeffecient of static friction
    which = 152.45 / 0.78 which i equate to equal 194.87 N correct?
     
  14. Oct 4, 2006 #13

    arildno

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    its force normal / coeffecient of static friction

    THIS IS WRONG!
     
  15. Oct 4, 2006 #14
    sorry about that.. Fsmax = coeffecient of statfriction * Normal Force
     
  16. Oct 4, 2006 #15

    arildno

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    Yes, that's correct. Given the expression for the normal force in post 11, what is Fsmax?
     
  17. Oct 4, 2006 #16
    mgcos 45 * coeffecient Fs
     
  18. Oct 4, 2006 #17

    arildno

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    Correct!

    Now, set up Newton's 2.law PARALLELL to the incline, knowing that the box is at rest, and that the applied force is called A.
     
  19. Oct 4, 2006 #18
    mgcos45 * coeffecient of Fs
     
  20. Oct 4, 2006 #19

    arildno

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    Eeh, is this Newton's 2.law of motion??
     
  21. Oct 4, 2006 #20
    ok done.. let me take a crack at this one more time :
    force normal = mass (acceleration)cos 45, which = 22 (9.8) sin 45 which = 152.45 N correct?
    now.. Fsmax = force normal * coeffecient of static friction
    which = 152.45 * 0.78 which equals 118.9 N correct? when i did the diagram it only makes sense that the applied force can not exceed the normal force * coeffecient of static friction .. am i headed on the right track?
     
    Last edited: Oct 4, 2006
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