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Calculating Force at an angle with kinetic friction coefficient

  1. Feb 2, 2010 #1
    1. The problem statement, all variables and given/known data
    A force, F, applied to a 12.1 kg crate at an angle (theta) of 41.0 degrees makes the crate move horizontally with a constant acceleration of 1.85 m/s^2. The coefficient of kinetic friction between the crate and the surface is uk=0.35. Calculate the magnitude of F.


    2. Relevant equations
    F = ma
    Fx = F + friction + N + W = ma
    Fy = F + friction + N + W = 0
    uk*N = friction
    so, F = mg(cos theta) - friction = ma


    3. The attempt at a solution
    I first found the force of kinetic friction, by multiplying the coefficient of kinetic friction (0.35) by the normal force (12.1 * 9.8), which equaled 41.5 N.
    Then, I found m*a = (12.1 * 1.85) = 22.385.
    I then set up the equation: F cos(theta) - friction = ma, or F(.7547) - 41.5 = 22.385
    This gave me F = 84.65 N, which is wrong.
    Can someone help me see where I'm going wrong?
    Thanks!!
     
  2. jcsd
  3. Feb 2, 2010 #2
    im not sure if i understand the question u have .. is the crate in a horizontal surface and the crate itself lifted to be at an angle 41 .. or the crate is moving in a plane which is at angle of 41 ?? and mention the direction as well is it upward or downward??
     
  4. Feb 2, 2010 #3
    Sorry, the crate is moving horizontally to the right, being pulled by a force F at an angle 41 degrees above the horizontal axis. (most likely a rope or something!!)
     
  5. Feb 2, 2010 #4
    The crate is not moving up an angle. it is moving across a flat surface, but being pulled by a rope at a 41 degree angle. Is there any way to paste diagrams in here? I could try to scan it...
     
  6. Feb 2, 2010 #5
    I think that the setup of your problem is something like the following:


    http://img52.imageshack.us/img52/8946/ed96dcopycopy.jpg [Broken]



    so, in that case u have calculated the friction force incorrectly ..

    it is true that the friction = (uk) * normal force .. in your case in order to find the normal force you should use Newton`s 2nd law which is :

    Fy(net force) = 0 , in your problem this is , normal force + Fy - weight = 0 .. normal force is not always mg as you suggested .. go on from this point and you should get the right answer ..
     
    Last edited by a moderator: May 4, 2017
  7. Feb 2, 2010 #6
    But if Fy = 0, and normal force + Fy (0) - weight (mg) = 0,
    then normal force = weight, so normal force = mg.

    Am I missing something??
     
  8. Feb 2, 2010 #7
    Oh, but you are saying that:
    normal + Fy (Fsin(41)) - weight (mg) = 0
    So normal + Fsin(41) = weight
    Can I use mg for the F of Fsin(41)?
    That would give me a normal force of 40.785
    and when I plug it into Fcos(41) - friction (40.785*0.35) = ma
    I get F to be 48.568
    Is this correct? Because I used mg for the F in the Fsin(41) but then I solved for the F in the Fcos(41)...I'm not sure that is right???
     
  9. Feb 2, 2010 #8
    noooooo you cant substitute mg for F .. Your questions is asking for the value of F ..

    I suggest you solve the following equations :

    Fx = Fcos41 - friction = ma
    Fy = Fsin41+normal force-mg = 0
    When you substitute you have friction = (ku) * normal ..

    I think this is clear enough .. Try again and tell me what you did
     
  10. Feb 2, 2010 #9
    sorry, i'm horrible at physics. here is what I think you are trying to explain:

    if I solve for normal force using Fy, I get:
    normal force = mg - Fsin41
    if I plug that into the Fx equation, I get:
    Fx = Fcos41 - (ku)*(mg - Fsin41) = ma
    If I solve for F, I get F = 122 N

    Am I going the right way now?
    Thanks for your help.
     
  11. Feb 3, 2010 #10
    what you are doing is right .. But i have solved your problem and the answer is not 122 ? .. I solved it twice so i dont give you a wrong answer .. You probably made simple mistake in your calculations .. Try again you should get some value between 60 and 68 ..
     
  12. Feb 3, 2010 #11
    I am so confused.
    When I do the equation, I get:
    Fcos41 - (ku)*(mg - Fsin41) = ma
    .7547F - .35(118.58 - .6561F) = 22.385
    .7547F - .2296F = 22.385 + 41.503
    F = 121.67

    Am I doing some trig function wrong here?
     
  13. Feb 4, 2010 #12
    ooooh u had made a simple mistake .. U have a sign mistake .. Ur equation should be :
    0.7547F + 0.2296F = 22.385 + 41.503
     
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