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Dynamics: Ramp and Friction

  1. Jan 22, 2006 #1
    Hi everyone,

    I'm having a problem in solving dynamics problem involving ramp and friction forces. I'm wondering if anyone could give me some help to check my answers and give some hints on how to solve the problem :P

    Anyways, here is the description of the problem. If you find it unclear, please let me know:

    A box with a mass of 22 kg is at rest on a ramp inclined at 45 degrees to the horizontal. The coefficients of friction between the box and the ramp are [tex]\mu_{s}=0.78[/tex] and [tex]\mu_{k}=0.65[/tex].

    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.

    This is how I solved (a):

    [tex]F_{app}[/tex] is the magnitude of the force applied to the box upward and parallel to the ramp.
    [tex]F_{g}[/tex] is the magnitude of the gravity force.
    Fsmax
    [tex]F_{s, max}[/tex] is the maximum static friction force.
    [tex]\theta[/tex] is the inclination of the ramp.

    If [tex]F_{app} > F_{s, max} + F_{g} sin \theta [/tex], the box will move upward parallel to the ramp. Therefore, the maximum [tex]F_{app}[/tex] if the box is to remain at rest:

    [tex]F_{app} = F_{s} + F_{g} sin \theta[/tex]
    [tex]F_{app} = \mu_{s} * m * g * cos \theta + m * g * sin \theta [/tex]
    [tex]F_{app} = m * g ( \mu * cos \theta + sin \theta ) [/tex]
    [tex]F_{app} = 22 kg * 9.8 \frac{m}{s^{2}} * (0.78 * cos 45 + sin 45) [/tex]
    [tex]F_{app} = 270N[/tex] (correct to two significant digits).

    Am I doing it correctly? I'm suspicious that I'd need to plug in the kinetic friction force somewhere.. Plus when I do some calculation, it seems that the downward force parallel to the ramp ([tex]F_{g} sin \theta[/tex]) is greater than the [tex]F_{s, max} = F_{N} * \mu_{s}[/tex] which means the object should be experiencing a downward acceleration parallel to the ramp. However, the question explicitly mentioned that the box is at rest. Is this caused by the kinetic friction force?

    About question (b), I'm not sure how could a force perpendicular to the ramp affects the motion of an object on the ramp since the applied force is perpendicular to the possible direction of the boxes motion (down the slope or up the slope). And I don't know whether the motion will be upward or downward parallel to the ramp.

    However, I have a partial solution. But I don't know how to insert my free body diagram to this post. If anyone could help me, I'd greatly appreciate that :smile:.

    But basically I assumed that the net force will be going down parallel to the ramp since any object will naturally fall to ground. If the downward force parallel to the ramp > (Fs, max) the box will move from rest. As described: [tex]F_{down} > F_{s, max}[/tex]

    So I need to calculate:
    a) the downward force parallel to the ramp
    b) the [tex]F_{s, max}[/tex] opposing the motion

    I can find the force mentioned in (a) by considering that the resultant force perpendicular to the ramp can be calculated by adding the applied force to the component of gravity perpendicular to the ramp: [tex]F_{resultant} = F_{app} + F_{g} cos \theta[/tex].

    From there, I could find [tex]F_{down} = F_{resultant} * tan \theta[/tex].

    Since [tex]F_{resultant} = F_{app} + F_{g} cos \theta[/tex]. Through subsitution, if the object is to move from rest:

    [tex]( F_{app} + F_{g} cos \theta ) * tan \theta > F_{s, max}[/tex]

    Therefore, the minimum magnitude of Fapp is:

    [tex]F_{app} = (F_{s, max} / tan \theta) - F_{g} cos \theta[/tex]

    Is it correct?

    Thanks very much!!
     
    Last edited: Jan 22, 2006
  2. jcsd
  3. Jan 22, 2006 #2
    your part a looks right to me

    for part b, when you apply the perpindicular force, the normal force changes, correct? so then the friction force will also change.
     
  4. Jan 22, 2006 #3
    So if the object is to stay at rest, I need to find the Fs where: [tex]F_{s} = F_{s, max}[/tex]. From there, I could find Fn through: [tex]F_{N} = F{s} * \mu_{s} [/tex]. Then I could find The Fapp since: [tex]F_{app} = F_{N} - F_{g} cos \theta[/tex]

    Is that correct?
     
    Last edited: Jan 22, 2006
  5. Jan 22, 2006 #4
    Also, does it mean kinetic friction force has no effect here?
     
  6. Jan 22, 2006 #5
    yes, seems like it
     
  7. Jan 22, 2006 #6

    If the object is to stay at rest, you would need to find Fs where Fs = Fgsin theta.
     
  8. Jan 22, 2006 #7
    ah that's correct. Thanks a lot andrew :)
     
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