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Friction Force Help

  1. Feb 23, 2006 #1
    Hello everyone,

    I have have no clue on how to go about this problem here, if someone can explain how to go about these it would be great! THANK YOU ALL!

    Problem 1: :grumpy:

    A 62 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15° above the horizontal.
    (a) If the coefficient of static friction is 0.51, what minimum force magnitude is required from the rope to start the crate moving?
    (b) If µk = 0.36, what is the magnitude of the initial acceleration of the crate?
    m/s2
     
    Last edited: Feb 23, 2006
  2. jcsd
  3. Feb 23, 2006 #2
    Ok so i found a) by using T = mg*u / cos x + sin x * u

    yielding => 282.23 Newtons,

    trying to find B) can anyone help?

    EDIT: Just a little help to get me moving is good enough, PLEASE SOMEONE HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP
     
    Last edited: Feb 23, 2006
  4. Feb 23, 2006 #3
    still no post ?
     
    Last edited: Feb 23, 2006
  5. Feb 23, 2006 #4
    So no ones gonna help me, thanks alot
     
  6. Feb 23, 2006 #5
    I think your part (a) is wrong.

    Where did you get this?

    T = mg*u / cos x + sin x * u
     
  7. Feb 24, 2006 #6
    No question a is correct i already put it in webassign, (a online physics homework thing for college). for b i just asked my professor, it was pretty simple. Using F = ma
    F = Fnet while in motion

    Thanks anyway dude :smile:
     
  8. Feb 24, 2006 #7
    It strikes me as somewhat odd that you found the answer to a), but can't get b). Can I assume the T equation was something from your book? (Otherwise I can't figure out how you got it!)


    b) The object has a normal force (N) and a weight (w) in the vertical direction. I took the applied force (F) to be acting up and to the right, so the friction force (f) will be acting to the left. I choose a coordinate system with +x to the right and +y upward.

    As is typical with a friction problem, you will need to do the net force in the y direction, so do that first:
    [tex]\sum F_y=Fsin \theta +N-w=0[/tex]
    So you can find N.
    In the x direction we note that the box is presumed to be sliding, so we are using kinetic friction. [tex]f_k= \mu_kN[/tex] so we have:
    [tex]\sum F_x=Fcos \theta-f_k=ma[/tex].

    Plug in for kinetic friction and the normal force and solve for a.

    To do a) you basically do the same thing. The difference is that we are looking at a stationary situation, so we use static friction. Since we are looking for the minimum force required, we are using maximum static friction and a=0 in the limit. This will allow you to derive the equation you used for part a).

    -Dan
     
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