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Box on an incline with friction, min and max forces involved

  1. Jul 10, 2012 #1
    μSomeone is pushing a packing case of mass 30 kg up an inclined slope of angle 11.3deg.
    as shown in the diagram below. The coefficient of kinetic friction between the case and the slope is µk = 0.100 and the coefficient of static friction is µ = 0.200.

    1. The minimum force the person must apply parallel the slope to
    keep the case from sliding down the slope (i.e., just to keep it in
    equilibrium, at rest) is:

    A) 59 N B) 120 N C) 152 N D) 32 N E) Zero

    2. The maximum force that the person can apply parallel to the slope without causing the crate to move is:
    A) 59 N B) 118 N C) 29 N D) 136 N E) 88 N

    My prof discussed this in class a bit, but he went through it so fast, I'm not sure I have this right.

    1. The F applied plus F static friction must equal the F parallel to keep it from slipping. I think then:

    Fapplied + μ(mg cos theta)= (mg sin theta) So,
    Fapplied min. = (mg sin theta) - μ(mg cos theta)= Zero???? because:
    30 x 10 (sin 11.3) - .2(30 x 10 cos 11.3) = -.05

    I feel like conceptually I'm missing something. I don't really understand what the minimum applied force to keep the box stationary means vs. the max applied force to keep the box stationary.

    2. For the max applied force to keep box stationary, then would this be Fapp = (mg sin theta) + μ(mg cos theta)=
    30 x 10 (sin 11.3) + .2(30 x 10 cos 11.3) = 118N??

    At first I thought the answer to this was just the coeff of static friction x FN(mg cos theta), which I thought gives the max static frictional force which would be 59N, but now I just don't really know because they are asking for the max force applied by the person which I think is certainly different then just the max Fstatic.

    There is another component to the question also asking what the min and max force to keep the box moving at constant velocity is and I assume to figure these out I would use the same formulas above but substitute μ for kinetic friction.

    I'm getting really confused though about this conceptually. Can someone please help explain the "why" behind this? For one, I don't quite understand the min. and max forces to apply to keep a box stationary or moving. Do I even have the formulas above correct?

    Thanks so much! It seems the more I look at this the less I get it! I seemed to "get it" in class, but now I feel like my understanding went out the window!
     
    Last edited: Jul 10, 2012
  2. jcsd
  3. Jul 10, 2012 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    If you apply the minimum necessary, then you will be barely stopping the box from sliding DOWN the slope. If you apply the maximum necessary, then you will be on the verge of moving the box UP the slope.
     
  4. Jul 10, 2012 #3
    Thank you for the reply!

    I think I'm starting to get it: For the maximum applied force:
    the magnitude of this applied force should be the same as the Fstatic plus the Fparallel. In this scenario, the free body diagram would show Fapplied going up the slope, with the static friction going down the slope to oppose this force AND the Fparallel going down the slope for the x axis components. Initially I wasn't thinking this way as I was still stuck on the Fapp and the Fstatic going up the slope, but that is clearly not correct for this part of the problem as the frictional force must oppose the applied force! So, the max force necessary is Fapp= Fstatic friction + Fparallel = μmgcos theta + mgsintheta which equals approx 118N. Do I have this clear now?

    I'm still a bit stumped on the minimum force, though. My professor indicated Fapp minimum= Fparallel minus Fstatic friction= mgsin theta - μ (mg cos theta). When I calculate that for this problem I get zero, so by your description, zero force, or rather no force is needed to even keep the box from slipping at all in this scenario? I guess it seems the FBD (showing Fapp up the slope and Fstatic up the slope and Fparallel down for the x-axis components) and the formulas are correct, but when I got zero as an answer I started to question my evaluation of the problem.

    For the second part of the problem, which I did not copy, I am to find the minimum and maximum forces to keep the same object moving at constant velocity. So, I assume that this is essentially the same problem as no net force is involved. It will just require substituting the coefficient of kinetic friction in for the static friction.

    Am I finally on the right path?

    Thanks so much!
     
  5. Jul 10, 2012 #4
    The value of static given is the maximum value.
    It varies from zero to μs.
    The static frictional force is always equal to applied force until it moves.
     
  6. Jul 10, 2012 #5
    Thanks again for the help. I'm starting to grasp this concept much better now. Plus, I can report that my answers are correct. The min. force for this problem to keep the object from moving is in fact zero. My question also was not whether or not it is zero in all cases, as I know this number varies, I just wanted to make sure I was on the right path with this. So, I was/am! And, 118N for the max force that can be applied w/o causing the crate to move is also right so I guess my understanding of the concepts is at least close enough to correct for me to be able to solve other problems. The wording for the second half of the problem that I did not include here involving an object with constant velocity was a little different so I had to apply this same understanding to those questions and I am also happy to report that my answers to those were correct also, so I'm getting it; slowly maybe, but better than not at all!
     
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