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Frictional force on Inclined Plane

  1. Oct 6, 2011 #1
    1. The problem statement, all variables and given/known data

    A box with a mass of 6.00 kg is at rest on a ramp that is at an angle of 20 degrees with the horizontal. The coefficient of static friction between the box and the ramp is 0.800. Use g = 9.80 m/s2. You now want to make the box move by applying a force. To start the box moving, what is the minimum force you need to apply if your force is directed ...


    (a) parallel to the slope?
    N
    (b) perpendicular to the slope?
    N
    (c) horizontally?
    N

    2. Relevant equations

    Force static = μs * Fn

    3. The attempt at a solution

    I calculated part a by realizing that the minunum force would be if the box is moving down the incline. The difference of the static force and x-directional value of force gravity is the answer for part a, 24.092.

    However, for part b and c, I attempted to solve the values for solving the x and y components of the force needed to to move the box parallel to the slope in part a and did not arrive at the right answer.
     
  2. jcsd
  3. Oct 6, 2011 #2

    Doc Al

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    Staff: Mentor

    Sounds good to me.

    Not sure what you mean. Realize that a force applied at an angle will change the normal force and thus the friction. Show exactly what you did.
     
  4. Oct 6, 2011 #3
    Thanks for the confirmation. Using the force vector needed to push the box parallel to the slope, I solved for the directional components of that force vector.

    If I define the x-direction as parallel to the slope, then the horizontal force required (part c) should be cos 20 = 24.092/force horizontal = 25.69. The perp. force to slope should be the sqrt of the 25.69^2-24.092^2= 8.77.

    Can you clarify why this is wrong and how to approach this problem correctly?
     
  5. Oct 6, 2011 #4

    Doc Al

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    Staff: Mentor

    You cannot assume that the parallel component of the force is the same as for part a. Again, Fn and thus the friction force will change as the angle of the force changes.

    If the applied force is F applied horizontally, what's the new Fn? What's the new max friction force? Use that to figure out the minimum value of F.
     
  6. Oct 6, 2011 #5
    If the applied force is F applied horizontally, would the new Fn=mgcos0? And for the F applied vertically be Fn=mgsin90? Wouldn't these two values be the same? Where does the 20 degrees inclination go? Sorry, my mind is really flustered.
     
  7. Oct 6, 2011 #6
    I tried to take the x and y components of the opposing forces, Fstatic and mgsin20 and added the values to each component and got the same values as before. I do not follow your instructions.
     
  8. Oct 6, 2011 #7

    Doc Al

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    Staff: Mentor

    No. To start with, what would be the y-component of that force? To find the new Fn, set ƩFy = 0 and solve for it.
    No. You'll need a similar analysis as above for that case.
     
  9. Oct 7, 2011 #8
    If a perpendicular to incline force is added, the new Fn would be (mg+Fy)(cos 20). This will cause the static force to increase; I theorize that no matter how much Fy force is added, the static force will always increase and have no effect on moving the box down the incline.

    For the horizontal force, how would you find the y-component to that force if you do not know that force?
     
  10. Oct 8, 2011 #9

    Doc Al

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    How did you deduce this?
    The perpendicular force may push in against the block, in which case that will just add to the amount of friction to be over come. But what if it pulled up out of the box?
    Just call the force F and use some trig. Remember that you are solving for the force, you don't have to know it up front.
     
  11. Oct 9, 2011 #10
    I have a similar problem.

    What I did was that to find the new force of friction I first found the normal force algebraically.

    Fn= (cos θ)(mg) - (sin angle) horizontal force

    Then I did this to find the force of friction

    Ff = (coefficient of static friction) [(cos θ)(mg) - (sin angle) (horizontal force)]

    Since there is no acceleration the force of friction and the force of gravity acting on the slope should add up to be 0.

    So by setting the force of friction equal to the force of gravity acting on the slope you should be able to solve to F, the horizontal force, which is the only unknown at this point.

    However when I did this, I got an answer that does not seem reasonable. My given numbers were
    m=5kg
    θ=20°
    coefficiant of static friction=.8
    gravity=9.8/m/s^2

    The final answer I got when solving for F is 73.376N which I see as ridiculous seeing as the answers I got for the first two parts were in the 20s.
     
  12. Oct 9, 2011 #11

    Doc Al

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    Staff: Mentor

    This is incorrect. The applied force also has a component parallel to the slope.
     
  13. Oct 9, 2011 #12
    So would I be right if I do this

    Ff - (cos20 F + sin 20 mg) = 0

    This would be the force of friction that acts on the box, with the x component of the force acting on the box as well as the component of gravity on the slope. All this added up would have to equal to 0 to show that the box moves.

    I get a more reasonable answer, but its still incorrect.
     
  14. Oct 9, 2011 #13
    Never mind. I figured it out. I didn't get the right answer because I forgot to multiply the one of the components of the normal force by the coefficient of static friction.

    Now I got the right answer. Thank you for the help!
     
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