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Force Components, Friction, etc. Check my work?

  1. Oct 22, 2014 #1
    LqZ7Pwe.jpg

    Could someone credible please check my work on this one?

    1. The problem statement, all variables and given/known data
    This problem is "pulling a sled up a slope"
    All relevant data is in the picture. I wanted to find FP (The force one has to pull to start the sled moving up with slope).

    mass = 26.0kg
    μs = 0.096
    slope of hill = 12 degrees
    FP = 23 degrees (from horizontal)

    I did all my work in the rotated reference frame where the x direction is in line with the slope of the hill.

    2. Relevant equations
    I fully believe I got the right answer which is why I want this to be double and triple checked to make sure I really do understand everything here. I have my midterm on Friday so now is the time to learn!

    3. The attempt at a solution
    The solution I got was FP = 76.9N

    Thank you so much in advance! <3‿<3
     
  2. jcsd
  3. Oct 22, 2014 #2

    rude man

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    I got a slightly higher number. To wit, 78.4N.

    I notice a term μsFpsin(11) in your work which I did not have.
     
  4. Oct 22, 2014 #3
    Thanks for replying :) I see exactly what you did differently so I'll explain and can you tell me where/if I wrong in the following?

    To find the Ffriction it is dependent on FN which ordinarily would be equal to the y component of Fg. However, there is also a y component to FP (an applied force in the y direction) which reduces FN by FPY. In other words because you are pulling up on the object the normal force is reduced. If this were not the case there would be a net force in the y direction and it would accelerate off the slope. That is where μsFpsin(11) in my work comes from. It is part of 0.096[mgcos(12)-FPsin(11)] where FPsin(11) is the y component of the applied force FP.
     
  5. Oct 22, 2014 #4

    rude man

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    Well, guess what, I agree! Except one thing bothers me: the force Fpsin(11) is not applied at the c.g. of the sled, but at one edge. So the tendency of that vertical component of Fp is to tilt the sled upwards rather than lift it evenly off the ramp. So the part of the surface contact the sled makes with the ramp that is closest to the rope will get "lighter" but the part on the low end will hardly be affected. I guess you can say it averages out. Certainly I was remiss in neglecting that vertical component. So congrats! I'd say the apprentice beat the master except I'm no master!
     
  6. Oct 22, 2014 #5
    Thanks! That is an interesting point. Would it balance out perfectly or would that factor into a more accurate answer?
     
  7. Oct 22, 2014 #6

    rude man

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    I think it would balance out and your analysis was spot-on.
     
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