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Difficulty with force problems involving friction

  1. Feb 22, 2006 #1
    I finally managed to work through my difficulty with frictionless force problems involving Newton's second law, but now what's throwing me off is how to throw friction into the mix.

    For this problem, for example:

    A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 24° above the horizontal. (a) If the coefficient of static friction is 0.47, what minimum force magnitude is required from the rope to start the crate moving? (b) If the coefficient of kinetic friction = 0.29, what is the magnitude of the initial acceleration (m/s^2) of the crate?

    I know that I first need to draw a freebody diagram, and when I do, I can perceive 4 forces acting on the crate: Normal Force, Tension, Weight, and of course Friction.

    However, my problem is where to go from there.
  2. jcsd
  3. Feb 22, 2006 #2
    The coefficient of friction can be expressed as


    where [tex]F_f[/tex] is the force excerted by the friction. This should give you a start.
  4. Feb 22, 2006 #3

    Well.. as assyrian_77 suggested u do indeed need to know the force of friction acting against the pull on the rope. But u must remember that this force is related to the normal force of the box on the ground. And the normal force will be lowered because your actually pulling UP on the box a little...

    Once u have set up the correct resolving equations it should just be a matter of Cranking the Handle.

    Good luck
  5. Feb 22, 2006 #4
    Okay, I've got part a resolved but am confused as to how part b works.

    For part a, I approached it like this:

    After drawing a free body diagram, identifying the 4 forces involved and their relation to the object being acted upon, and seeing that an angle is involved, I know that I need to resolve it into components.


    x component: Tcos(24°) - f = 0
    y component: N + Tsin(24°) - mg = 0

    The two equations I get from this are:

    f = Tcos(24°)
    N = mg - Tsin(24°)

    In order for the crate to move, the friction force has to be greater than the normal force multiplied by the static coefficient, so...

    Tcos(24°) = 0.47 * (mg - Tsin(24°))

    From all that I can get the equation I need for the tension, which, after some algebraic manipulation, yields:

    T = (mg * static coefficient) / (cos(24°) + sin(24°) * static coefficient)

    Then plugging in the values...

    T = 283.52.

    However I'm uncertain as to how to proceed for part b.
  6. Feb 22, 2006 #5


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    Go back to the x equation and set [itex] f = \mu_k N [/itex] and set the whole x equation equal to [itex] m a_x [/itex]. T is the same as what you just found. also, the normal force will be the same as in part (a) because nothing has changed for the equation along y. So, from part a you know T and N. Just plug in the x equation to find [itex] a_x [/itex].
  7. Feb 22, 2006 #6
    Ok, now you have T. What other force can you calculate? The kinetic friction is given by the same expression I stated above. Is the friction force the same as earlier? And using Newtons 2nd law....... Those hints should do it. :smile:
  8. Feb 22, 2006 #7

    Doc Al

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    In part a, the forces produced equilibrium. In part b, once movement has begun, the static friction is replaced by the lower kinetic friction: Now the forces are no longer in equilibrium. Find the net force and then the acceleration.
  9. Feb 22, 2006 #8
    Okay, the concepts are starting to click into place for me, but I'm stuck on the reason for a certain thing present in the solution for part b shown in my book. Basically, I have the two (X and Y) component equations figured out for part b, but my Y component is off.

    I have it set to Tsin(theta) + Fn - mg = ma.

    However, the book's solution sets the ma to 0.

    But why is the y equation set to 0 still?
  10. Feb 22, 2006 #9

    Doc Al

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    What direction does the friction act? Why would changing the friction (from static to kinetic) affect forces in the y-direction?
  11. Feb 22, 2006 #10
    There is no acceleration in the y-direction (up). The crate is dragged along the floor, never lifted.
  12. Feb 22, 2006 #11
    That's a good point. I suppose that was my main problem when approaching Newton's 2nd Law for the first time. (Nervousness about a future exam didn't help either, but besides that..) I wouldn't realize what was happening or how to represent it and as such would be stumped on a problem. But little by little I've been noticing that Physics problems, at least at this level, are usually just needing a slight adjustment or consideration of an overlooked detail when I get them incorrect.
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