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Forces and Friction Homework

  1. Jan 19, 2017 #1
    1. The problem statement, all variables and given/known data
    A block is placed on a inclined plane. The incline of the plane is 21° .A 24 N force is required to push the block up the incline with constant velocity. The force is applied at an angle of 15° as shown in the picture. The coefficient of kinetic friction (μk) is 0.26. What is the weight of the block? Answer in units of N. The acceleration of gravity is 9.8 m/s/s.

    2. Relevant equations
    Force Kinetic Friction = μk * Normal Force

    3. The attempt at a solution
    The farthest I got in solving the problem was to come up with the force of gravity being 186.5076 N, however I am not entirely sure that this is correct.
    To get this I listed the components for each force:
    Fn= < 0, Fn >
    Fg= < -sin(21)Fg, -cos(21)Fg >
    Fpush= < cos(15)Fp, -sin(15)Fp >
    Fk= < -μkFn, 0 >
    Then is added all the x and y parts together:
    Fx:
    -sin(21)Fg + cos(15)Fp - μkFn = 0
    Fy:
    Fn - cos(21)Fg - sin(15)Fp = 0
    Then I solved for Fn for each:
    Fx:
    (cos(15)Fp - sin(21)Fg) / μk = Fn
    Fy:
    sin(15)Fp + cos(21)Fg = 0
    Then I plugged in one Fn for the other:
    (cos(15)Fp - sin(21)Fg) / μk = sin(15)Fp + cos(21)Fg
    Then I solved for Fg and I am not sure where I messed up doing so...
    Physics.png
     
    Last edited: Jan 19, 2017
  2. jcsd
  3. Jan 19, 2017 #2

    gneill

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    Staff: Mentor

    Please show the details of what you've tried.
     
  4. Jan 20, 2017 #3

    gneill

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    Staff: Mentor

    What you've done so far is good. I spotted one typo (red comment) which didn't affect the later step. I hope that you understand that equating the two expressions for the normal force is really a subtle way of enforcing the given constraint that the acceleration is zero. A less subtle way would be to declare that the net force in the x-direction (along the slope) is zero and write the sum accordingly. You'll arrive at the same relationship that you did in your last step shown.

    (cos(15)Fp - sin(21)Fg) / μk = sin(15)Fp + cos(21)Fg

    which looks good to me. So it seems that your problem now is solving the above for Fg as you've surmised and is a matter of algebra. Can you show us how you've tried to isolate Fg in the above?
     
  5. Jan 20, 2017 #4
    Thanks but I already figured it out. But again thank you and the site for the help! Should I delete the thread or post the answer?
     
  6. Jan 20, 2017 #5

    gneill

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    Staff: Mentor

    Threads are never deleted unless there's some extraordinary reason. You can post your answer if you wish, but if you're satisfied with the way things are it's not a requirement. You might wish to mark the thread "solved" to indicate that you're done here (find the "MARK SOLVED" button at the top right of the page).
     
  7. Jan 20, 2017 #6
    I have solved the problem and the answer is 35.87958 N for anyone that wishes to know.
     
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