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Kinetic frictional force question

  1. Jun 14, 2006 #1
    Hi there,

    I'm doing a problem in my book ("Check your understanding" 4.3 in Cutnell/Johnson, 7th ed.). The book says the answer is x=43 degrees, but I don't know how to find it exactly. Given that it's an in-chapter problem, I'm guessing there's something wrong with my brain, but I'd appreciate your help.

    Here's the problem:
    A box has weight 150 N and is being pulled across a horizontal floor by a force of 110 N. The pulling force can be applied to the box either horizontally or can point above the horizontal at an angle, x. When the pulling force is applied horizontally, the kinetic frictional force acting on the box is 2x as large as when applied at angle x. What's x?

    I know that fk = uk(Fn).
    I started by writing
    F_{x} #1 = 110 - fk = ma = 0
    F_{x} #2 = 110cos(x) - fk = ma = 0
    and making those equal to each other, but I end up with multiple variables, and I'm stuck.

    Thanks! :D
  2. jcsd
  3. Jun 14, 2006 #2
    Don't you need to change the weight of 150N to mass in kilograms?
    To say that you have a weight of 150N is incorrect. Newtons are a measure of force.
  4. Jun 14, 2006 #3


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    well, if these were the right equations, you should have no problem solving! From the first equation, fk=110 N. Plug this in the second equation and solve for cos(x), you get cos(x)=0 so x= 0 degrees! That is telling you that the only solution for your two equations is x=0 degrees.

    So what is wrong?

    The point is that you must realize that the kinetic friction forces in the two situations are NOT the same (*unless* x=0 degrees which is not the case we want here).

    fk is given by [itex] \mu_k N [/itex] (where N is the normal force here, not the unit Newtons). So what is N? It is not necessarily the weight of the object, right? If there are no other vertical forces other than gravity and the normal of the surface, then N=mg = 150 Newtons. So in your first equation, N is 110 Newtons. Therefore your first equation allows you to find [itex] \mu_k[/itex].

    Now, your second equation is correct except that there, the normal force
    is no longer 150 Newtons. How do you find it? The best thing to do is to draw the free-body diagram. Notice that the external force of 150 N is at an angle "x" so it has both vertical and horizontal components.
    Now, write an equation in the y direction [itex] \sum F_y =0 [/itex] and [itex] \sum F_x=0 [/itex] . Notice that those two equations contain *two* unknowns, namely the normal force N and the angle x (recall that you know by now the value of [itex] \mu_k [/tex] from the horizontal force case). Then you can solve for the angle x.

    Last edited: Jun 14, 2006
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