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Maximal theta given the coefficient of kinetic friction

  1. Jul 7, 2009 #1
    1. The problem statement, all variables and given/known data

    While moving in, a new homeowner is pushing a box across the floor at a constant velocity. The coefficient of kinetic friction between the box and the floor is 0.410. The pushing force is directed downward at an angle theta below the horizontal. When theta is greater than a certain value, it is not possible to move the box, no matter how large the pushing force is. Find that value of theta.

    2. Relevant equations

    FKinetic = [tex]\mu[/tex] FN

    [tex]\Sigma[/tex]Fx = 0

    3. The attempt at a solution


    FN = mg + F sin [tex]\theta[/tex]

    FKinetic = [tex]\mu[/tex]FN

    [tex]\Sigma[/tex]Fx = -FKinetic + F cos [tex]\theta[/tex] = 0



    This is as far as I can get. Can you please explain how theta is found?

    Thanks so much!
     
  2. jcsd
  3. Jul 7, 2009 #2

    alphysicist

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    Hi tkahn6,

    Plug your first equation into the second to get rid of the normal force variable, and the second equation into the third to get rid of the kinetic frictional force variable.

    You can then determine the value of theta by examining the equation you get. (What is F at the angle that you are looking for?)
     
  4. Jul 8, 2009 #3
    Would F be any arbitrary magnitude or would it increase linearly with theta?


    Fcos(theta) is always going to equal mu(mg + F sin(theta)) because the box is in equilibrium right?

    Am I looking for where FKinetic > F cos (theta) ?

    I am really lost.
     
  5. Jul 8, 2009 #4

    alphysicist

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    When theta=0, F has to equal the frictional force, and after that it grows larger with theta (but not linearly), until you get to an angle where no force is large enough to push it.

    That's right. You can set them equal, and then solve the equation for F. (What do you get?) Now you are looking for an angle so that no force is large enough to push the box. What then would F be? (this is the important part of the problem)

    Once you know what F has to do, what angle does that correspond to?
     
  6. Jul 8, 2009 #5
    Thank you for your help!

    OK.

    F = [tex]\frac{(mu)mg}{(cos(theta) - sin(theta)(mu))}[/tex]

    When I solve the denominator for 0 I get 67.723o which is the answer that is in the book.

    Is this because at all theta values below 67.723 F is positive? But I thought F represents the magnitude of the force and not necessarily its direction.

    Thanks!
     
  7. Jul 8, 2009 #6
    Yes, that is exactly the correct explanation. [tex]\vec F[/tex], according to the work premises you worked with, is by definition positive. Any value of [tex]\theta[/tex] that provides you with a non-positive value for [tex]\vec F[/tex] would contradict your work premises and therefore be impossible.

    At times, a fun exercise is to see what is the physical meaning of an impossible result, and under what terms it would be possible. :)

    Oh, and on a side-note, use the forward slash (\) before names of greek letters to produce the symbol in LaTeX.
     
    Last edited: Jul 8, 2009
  8. Jul 8, 2009 #7

    alphysicist

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    I think it would be more to the point to say that when theta is 67.723 degrees, then the the right side is going to infinity (in the limit), so (in the limit) F is going to infinity. That matches what the problem indicates, in that it is not possible to push it any more at that angle.

    (In other words, that angle would require an infinite F to push it at constant speed, which of course is not going to happen.)
     
  9. Jul 8, 2009 #8
    Thank you both for your help - especially alphysicist.

    I need to wrap my head around the fact physics is applied math. I totally did not connect the fact that F is inversely related to [tex]\theta[/tex] and therefore prone to asymptotic behaviour as (cos([tex]\theta[/tex]) - [tex]\mu[/tex] sin([tex]\theta[/tex])) approaches 0!

    Thanks guys!!
     
  10. Jul 8, 2009 #9
    Hey I have another question.

    A 225kg crate rests on a surface that is inclined above the horizontal at an angle of 20o. A horizontal force (magnitude 535 N and parallel to the ground, not the incline) is required to start the crate moving down the incline. What is the coefficient of static friction between the crate and the incline?


    [tex]F_{Normal} = cos(20)W + sin(20)535N [/tex]
    [tex]F_{Static} = \mu F_{Normal} = 535N [/tex]

    For the coefficient of static friction I get .2373 but the answer in the book is .665

    Thanks again.
     
  11. Jul 8, 2009 #10
    Try and adopt a different system of axes, one that's aligned with the incline, rather than with the floor. It should simplify things considerably.

    The index y in my calculations is used for the forces perpendicular to the incline, and the index x, for those in the direction of the incline.
    The positive directions are up from the incline, and down the incline, for the y and x axes respectively.

    [tex]\Sigma \vec F_y=F_{ext}\sin {20^o}+N-mg\cos {20^o}=0[/tex]
    Because the mass is at rest in this direction.

    [tex]\Sigma \vec F_x=F_{ext}\cos{20^o}+mg\sin {20^o}-\mu mg\cos{20^o}=0[/tex]
    Since we're told [tex]\vec F_{ext}[/tex] is the minimal force required to set the body in motion, we know that it means that the mass is on the verge of slipping, and that it would move at a constant velocity if it were acted on by [tex]\vec F_{ext}[/tex]
     
  12. Jul 8, 2009 #11
    Thanks! I got it.

    But why is Fexty positive? Is is not pushing in the same direction as W?
     
  13. Jul 8, 2009 #12
    You're very welcome. :)

    You quoted the problem as saying that [tex]F_{ext}[/tex] is the minimum force required for the mass to start moving down the incline, so I could only assume that it was pointed down the incline, which would make its vertical component point perpendicular to the incline, and going out of the incline.
    Were the external force to come from the other side (Making its vertical component go into the incline, rather than out of the incline, the force would also end up being larger, because it would then need to overcome a bigger bit of friction, as its own action would increase the normal. I hope I'm making sense.
     
  14. Jul 9, 2009 #13
    Oh duh!

    Thanks! Yeah, that makes perfect sense.

    Thank you for all your help!!
     
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