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Homework Help: Help with hard problem about friction

  1. Oct 27, 2004 #1
    Hi, I need help with the following problem:

    You are pushing a mop of mass m with a force P at an angle theta. The coefficiant of friction is [tex]U_k [/tex].

    Find P so that the mop will start moving in terms in [tex]\theta,U_k ,m,g [/tex].

    I solved this part and got:

    P = \frac {(U_k *mg )}{( \sin\theta- U_k*\cos\theta)} [/tex]

    Now, for part 2 I have to find the minimum angle [tex] \theta [/tex] for which it will be impossible for me to push the mop in terms of [tex]\theta,U_k ,m,g,P [/tex]..

    Like at 90 degrees It is impossible to push the mop. Any help on part 2 would be appreciated.

  2. jcsd
  3. Oct 27, 2004 #2
    is theta relative to to the vertical or to the horizontal ?
    if it is relative to the horizontal, i got 45 degree. Pcos(theta) = miu*N
  4. Oct 27, 2004 #3
    Theta is relative to the Vertical. Can you explain how you got 45?

  5. Oct 27, 2004 #4
    45 degree is wrong, sorry. i think you have got the right anwer ninety degree.
  6. Oct 27, 2004 #5
    The answer is below 90 degrees. From theta to 90 it will not move.
  7. Oct 27, 2004 #6
    if theta is below ninety degree if it won't move, i don't know how the p found in the first part can move the mop.
  8. Oct 27, 2004 #7
    P is the force required for it to move at the angle Theta. The 2nd part of the problem is to find an angle theta such that any amount of P would not be able to move the mop.

    My guess is that theta is somewhere around 80 degrees, however I am looking for the true solution.
  9. Oct 27, 2004 #8
    sorry, can't help you with that !

    i found a similar question in my book. it introdueces both [tex]\mu_s [/tex] for coefficient of static friction and [tex]\mu_k [/tex] as coeffiecient of kinetic friction.

    if the force P can't move the mop, then [tex]Psin\theta<\mu_s*N[/tex]; then
    so [tex]\theta_{min}=sin^{-1}\frac{\mu_s*mg}{P*\sqrt{1+\mu_s^2}}+tan^{-1}\mu_s[/tex] so that the mop will move.
    Last edited: Oct 27, 2004
  10. Oct 28, 2004 #9
    thank you Leong, for the help
    Unfortunately, it it a little too complex and there is no [tex]U_s[/tex] in the problem.

    The correct answer is that [tex]\tan\theta > U_k [/tex]

    I am still looking for how to arrive at this answer.
  11. Oct 28, 2004 #10
    to move a thing in a frictional surface, don't we need to overcome the static friction first to get started to move, we use uk to relate the friction when it is in motion, a little weird i think for the answer to have uk.
    Last edited: Oct 28, 2004
  12. Oct 28, 2004 #11
    I found the answer. It was actually very simple

    From the answer I got for part 1:

    P = \frac {(U_k *mg )}{( \sin\theta- U_k*\cos\theta)} [/tex]

    We can say that [tex] \sin\theta- U_k*\cos\theta > 0[/tex]

    Working it out you get that [tex]\frac{\sin\theta}{\cos\theta} > U_k[/tex]

    And then you get that [tex]\tan\theta > U_k[/tex]
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