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Coefficient of friction and forces proof

  1. Oct 17, 2007 #1
    1. The problem statement, all variables and given/known data

    A box of weight W is pushed by a force F on a horizontal floor (the diagram below shows the force on the box is at an angle theta to the horizontal)
    If the coefficient of static friction is mu, show that the minimum value of F that will move the crate is given by:

    F=(mu*W*sec theta)/(1-mu*tan theta)

    2. Relevant equations

    The only maths needed i think is trig functions such as sec=1/cos, 1=sin^2+cos^2 etc.
    For physics, I have called the frictional force acting against the box fr and the reaction against the horizontal plane R. fr=mu*R is the only equation I could find with the exception of the equations that are results of Newton's thid law (which are listed in my attempted solution)


    3. The attempt at a solution

    Horizontally I worked out F*cos theta=fr=mu*R
    Vertically, F*sin theta=W=n (not sure which values to use so I plugged them all in there)

    1st attempt: R is rective force and W=mg so R=W.
    fr=mu*W
    F*cos theta=mu*W
    F=(mu*W)/cos theta

    But obviously the result is more complicated than that so I then tried:

    fr=R*tan theta (taken from previous class notes but I don't know if this value is suitable for this problem...)

    F*sin theta=W=n

    F*sin theta*tan theta=mu*W=(F*sin^2 theta)/cos theta

    F=(mu*W*cos theta)/sin^2 theta (from rearranging the above line)

    F=(mu*W*cos theta)/(1-cos^2 theta) then I just trailed off...
    I have made many more attempts but have rubbed them out as they were yet more unsuccessful...

    Please help!
     
  2. jcsd
  3. Oct 17, 2007 #2

    nrqed

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    Is it above the horizontal or below?
    :confused: Iguess that by R you mean the normal force. This is not standard notation.
    No. You must set the net force along y equal to zero. You can't set the three forces eqaul to one another!
    the normal force is certainly not equal to the weight here since there is another force with a y component.
     
  4. Oct 17, 2007 #3
    Yeah for R I do mean the normal force, I've heard it as both R and n depending on the teacher... I prefer n to be honest but my teacher today is an R person.
    Looking at y now I'm thinking F*sin theta - n = 0 (hence F*sin theta = n) but that doesn't seem to get me any further than my previous attempts...
     
  5. Oct 17, 2007 #4
    Oh and the angle theta is above the horizontal.
     
  6. Oct 17, 2007 #5
    Oh hang on, F*sin theta + W = n doesnt it... sorry, I don't know why I didn't write that before. Blonde moment.... I'll have another go.
     
  7. Oct 17, 2007 #6
    Right, I'm now getting:

    For y: F*sin theta+W - n=0
    so F*sin theta+W=n

    For x: F*cos theta - mu*n=0
    so F*cos theta = mu*n

    Substituting in the first equation for n:
    F*cos theta=mu*(F*sin theta+W)
    F*cos theta=mu*F*sin theta+mu*W
    F*(cos theta - mu*sin theta)=mu*W

    so F=(mu*W*cos theta)/sin^2 theta
    =(mu*W*cos theta)/(1-cos^2 theta)

    and then the trail goes cold again...
     
  8. Oct 17, 2007 #7

    nrqed

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    This does not look quite right (unless some of your quantities there are negative)

    The way to do it is to start with [tex] \sum F_y = m a_y [/tex]
    In your case a_y is zero. Therefore the net force along y is zero. Therefore
    [tex] F \sin \theta + n - mg = 0 [/tex]
    I had never heard of "R" for a normal. In what country are you, just out of curiosity?
     
  9. Oct 17, 2007 #8
    I'm from England, there's just one teacher I have who calls it the "reaction" to the plane.

    So if we take the direction of n to be positive, why is Fsin theta also positive... isn't that acting in the opposite direction? i.e. pushing down, therefore adding to the weight?
     
    Last edited: Oct 17, 2007
  10. Oct 17, 2007 #9

    nrqed

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    I obtain the equation you quoted except that I have a division by [itex]1+ \mu \tan \theta [/itex] instead of with a minus sign like you gave. Are you sure you typed the rigt formula?
     
  11. Oct 17, 2007 #10
    You mean the right result?
    Yes, it's definitely supposed to be 1-mu*tan theta on the bottom...
     
    Last edited: Oct 17, 2007
  12. Oct 17, 2007 #11

    Doc Al

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    The force must be acting downward, so that should be:
    [tex] -F \sin \theta + n - mg = 0 [/tex]

    [tex]n = F \sin \theta + mg[/tex]
     
  13. Oct 17, 2007 #12
    Yup, thought so, but I'm still not getting the right result...
    I used n=Fsintheta + mg earlier in post #6... which shows my latest (unsuccessful) attempt. I can't really see how to carry it on.
     
  14. Oct 17, 2007 #13

    Doc Al

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    OK.

    OK.

    Good.

    Huh???

    Just divide by (cos theta - mu*sin theta) and you're done. (Almost.)
     
  15. Oct 17, 2007 #14
    Sorry! My notes are rather messy and I typed out the wrong bit.
    I meant:

    F(cos theta - mu*sin theta)=mu*W
    so F=(mu*W)/(cos theta-mu*sin theta). which is where I got stuck but if that's right I'll have another look first.
    Thanks!
     
  16. Oct 17, 2007 #15

    Doc Al

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    You are one small step away from the final answer. Hint: What do you have to do the denominator to make it the way you want?
     
  17. Oct 17, 2007 #16
    Oh yeah you just divide throughout by cos theta!

    Right, got it, thanks again

    (Edit: and for the record, I did that before I saw your post...)
     
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