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Force and motion: box on a ramp question

  1. Feb 28, 2009 #1
    1. The problem statement, all variables and given/known data
    A box with a mass of 22kg is at rest on a ramp inclined at 45 degrees to the horizontal. The coefficient of friction between the box and the ramp are mu static = 0.78 mu kinetic = 0.65.

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

    b) Determine the magnitude of the smallest force that can be applied onto the top of the box, perpendicular to the ramp, if the box is to remain at rest.


    2. Relevant equations

    Fa = ma
    Fs = Fn mu static
    Fk = Fk mu kinetic


    3. The attempt at a solution


    Given:
    m = 22kg
    Incline angle = 45 degrees
    mu static = 0.78
    mu kinetic = 0.65

    a)

    The box is at rest, so Fg downward parallel to ramp = Fs upward parallel to ramp.
    The largest force that can be applied upward parallel to ramp must be > Fs to accelerate from 0 m/s to some velocity to move.
    But if I push the box from a lower altitude (upward) then would Fs be amplified by Fg parallel to ramp? Or does Fs already accommodate for gravity?

    Fs = Fn mu static

    Does Fn = Fg? Since there is no vertical movement? Does Fg = mg? Or is it modified by the inclination of ramp? Would it be Fg = mg cos 45?

    Assuming Fg = mg sin 45
    Assuming Fg = Fn = mg sin 45 = (22kg)(9.8N/kg)(cos45) = 152.46N

    Fs = Fn mu static
    Fs = (152.46N)(0.78) = 118.91N


    Assuming Fs has gravity parallel to ramp from up to down taken into consideration:
    Fa must be greater than 118.91N upward parallel to hill to accelerate the box.


    b)
    I'm not sure how to do this. Adding a force perpendicular to ramp is like increasing the Fg perpendicular to hill, hence increasing the Fn, which affects Fs.
    Remain at rest needs all forces to be balanced, so Fg parallel to ramp = Fs parallel to ramp, Fg perpendicular to ramp = Fn.

    I don't see how changing Fg perpendicular to ramp can move the box, unless it becomes greater than the stability of the surface of the hill and crushes it so the box collapses into the hill? :/

    Fs = Fn mu static
    Fn = Fg cos 45
    Fg = mg

    So the m is being questioned here.
    If Fgx parallel to ramp is > Fs then it will move, so Fgx = Fs
    So mg sin45 = mg cos45 mu static. But it doesn't.
    Why is the Fs less than the Fgx but the object is still at rest?

    I'm so confused at this point I don't think I've done anything right since the beginning.
     
    Last edited: Feb 28, 2009
  2. jcsd
  3. Mar 1, 2009 #2

    Doc Al

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    Staff: Mentor

    You've correctly caculated the maximum force of static friction, but that's not all there is to this problem.

    Do this: Draw yourself a diagram of the box showing all forces acting on it at the point just before the box begins to slide. What's the direction of the friction force?

    Do the same thing for part b. How does the normal force depend on the applied force? How does that affect the maximum value of static friction?
     
  4. Mar 1, 2009 #3
    Ok I have done this. The direction of the friction force is upwards, since gravity parallel to ramp is acting downwards. But they are balanced since the box is at rest. So at this point force of gravity acting downwards should be the same as static friction acting upwards.

    There is also Fg perpendicular to ramp that acts perpendicular downward and Fn that acts perpendicular upwards. These are also balanced since the box is at rest, so Fn = Fg, but I'm not sure how these affect the horizontal forces.

    I know that Fs = (Fn)(mu static)
    So does this mean Fn does not = Fgy if the object is on a plane? Since it is affected by (mu static)? And also I think Fgy is affected by the elevation of ramp?
    Fgy = (mg)(cos degree of ramp)? Am I making this more complicated than it needs to be? :/

    So if I were to push the box upwards, then friction would be acting downwards, along with Fgx? So I would have to push harder than Fs + Fgx together?

    So after I overcome the Fs = 118.91N [downwards] I have to also overcome the Fgx = 152.46N [downwards] to accelerate from 0 to some velocity to move it yes?
    So my totally applied has to be greater than 118.91N + 152.46N?

    But this is where I am confused since 118.91N doesn't = 152.46N to begin with, and if you say 118.91N is correct then 152.46N must not be the Fgx.

    Fgx must be 118.91N at rest.

    Does Fn perpendicular to ramp = 152.46N?
    I think I'm getting something really confused here.
     
    Last edited: Mar 1, 2009
  5. Mar 1, 2009 #4

    Doc Al

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    Staff: Mentor

    If friction and gravity were the only forces acting, then friction would point up the ramp. But that's not the case here. Remember there is an applied force up the ramp. Hint: Which way will the box slide if the applied force is too much? Static friction must oppose that.

    If by Fg you mean the weight, then it acts vertically downward. Fn = the component of Fg perpendicular to the ramp = mg cosθ.
     
  6. Mar 1, 2009 #5
    Ok yes, if I was pushing the box upwards, then static friction would act downwards.
    It would slide upwards if I applied too much force.

    So now there is
    me vs. static friction + gravity downwards yes?
     
  7. Mar 1, 2009 #6

    Doc Al

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    Staff: Mentor

    Yes.
     
  8. Mar 1, 2009 #7
    Ok cool beans now I can solve it.
    So I have

    Fa > Fs + Fgx
    Fa > (m)(g)(cos45)(mu static) + (m)(g)(sin45)
    Fa > (22kg)(9.8N/kg)(0.707106781)(0.78) + (22kg)(9.8N/kg)(0.707106781)
    Fa > 118.9127332N + 152.452222N
    Fa > 271.3649552N

    So I must push harder than 271.3649552N yes?
     
  9. Mar 1, 2009 #8

    Doc Al

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    Yes, that's the largest force you can exert without the box moving. (But round your final answer to a reasonable number of significant figures.)
     
  10. Mar 1, 2009 #9
    The normal force depend on the applied force in that normal force is the same as (but going up) the total force going down perpendicularly, which is mass x gravity x cos angle of ramp.
    So If I increase the mass, I would increase the total force going down and thus increase the normal force going up.

    And the maximum value of static friction depends on the normal force. So if mass is increased, total force going down is increased and normal force going up is increased and static friction is increased.

    But how do I know at what mass the static friction will be overcome by the gravity going down parallel to ramp? Is that what I need to find?

    So
    Fn[up] = Fg(weight)[down] = (m)(g)(cos45)
    Fn[up] is also = Fs / mu static coefficient
    Fs is (m)(g)(sin45) / mu static?

    I don't really understand.

    What is the difference between mg sin(degree) and mg cos(degree)?
    Are these to find the x and y component of normal force? sin for x and cos for y?
     
  11. Mar 1, 2009 #10

    Doc Al

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    Staff: Mentor

    mg cosθ is just the component of the weight perpendicular to the ramp; it's not the total force going down.
    True, but we're not talking about changing the mass of the box.

    Identify all force components perpendicular to the ramp. (Hint: There are three such force components.) They must sum to zero. Use that to deduce the normal force and how it depends upon the applied force. (Call the applied force "F" if you like.)

    For part b, which way is the static friction force pointing?

    Those are the components of the weight parallel and perpendicular to the ramp.
     
  12. Mar 1, 2009 #11
    Ok, the three forces perpendicular to the ramp are:
    1) The weight of the box
    2) The normal force
    3) The extra applied force asked in the question

    So normal force is the only force going up, the two other forces go down.
    So normal force = weight + extra applied force

    The static friction for part b is pointing up the ramp. It is equal to the weight of the box going down the ramp.

    I don't really know how to make the connection between forces going perpendicular and parallel.
     
  13. Mar 2, 2009 #12

    Doc Al

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    Staff: Mentor

    Good, as long as you realize that you should be talking about the component of the weight perpendicular to the ramp.

    Good. Here realize that you are talking about the component of the weight parallel to the ramp.

    Friction provides the connection. Friction depends on the normal force.
     
  14. Mar 2, 2009 #13
    Ok, after some thought I have come up with this:
    At rest:
    Fs = Fwx
    Fn = Fwy + Fa
    Then
    Fa = Fn - Fwy
    Fa = (Fs/mu static) - Fwy
    Fa = (Fwx/mu static) - Fwy

    Is this right?

    Fa = (mg sin45 / mu static) - mg cos45
    Fa = [(22kg)(9.8N/kg)(sin45)/(0.78)] - [(22kg)(9.8N/kg)(cos45)]
    Fa = 43N

    The largest force that can be applied perpendicular is 43N.
     
  15. Mar 2, 2009 #14

    Doc Al

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    Looks good!
     
  16. Mar 2, 2009 #15
    Oh man finally..
    Thanks a lot, I think I learned quite a bit from this.
     
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