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Maximum horizontal force F on ramp with friction?

  1. Dec 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Assume the ramp and block make the following coefficients of friction: (μs=0.9 μk=0.3)
    and the 2 kg mass is placed at rest on the ramp and a horizontal force F is also applied.

    http://img69.imageshack.us/img69/4097/filepq.jpg [Broken]

    What is the biggest magnitude force F that can be applied and the mass remains stationary?

    2. Relevant equations
    I know that FN = 20cos40 = 15.32N
    F|| = 20sin40= 12.85N

    However will I need to subtract any components from Fn and so on?

    3. The attempt at a solution

    No idea what to do, I did the problem without friction and for that F has to be 16.8N, but no idea what to do with friction incorporated into it.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Dec 2, 2012 #2
    Free-body diagram?
    Write expressions for the x and y components of all the forces.

    The largest value of F corresponds to the largest value for the static frictional force.

    If you can't solve it, show us your expressions for the components.
     
  4. Dec 2, 2012 #3

    TSny

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    Start with a carefully drawn free-body diagram for the block with all forces displayed and labeled. Decide on the orientation of your coordinate system - a good choice would be x-axis up along the slope and y-axis perpendicular to the slope. Use the diagram to find expressions for the x and y components of all the forces.

    [I see ap123 beat me. Good. Note that we both start with FREE BODY DIAGRAM! :smile:]
     
  5. Dec 2, 2012 #4

    haruspex

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    No, it will be increased because of the applied force F.
    List all the forces acting on the block, noting their directions, then write out the horizontal and vertical static equilibrium equations. Do the same for the ramp.
     
  6. Dec 2, 2012 #5
    Hmm, I think I get it. So acceleration is the F|| - ukFn right?
    Then a = m(sin 40 - .3 * cos 40)
    And i get acceleration =8.39 m/s/s
    Then force F = 8.39/cos40 = 10.95N, so max F = 10.95 N? Or is this wrong?
     
  7. Dec 2, 2012 #6

    TSny

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    Acceleration? The problem states that the block remains stationary.
    I suspect that you still haven't drawn a free-body diagram. :grumpy:
     
  8. Dec 2, 2012 #7
    What's happening here is that the block is stationary to start with.
    As you increase the force, then eventually the block will start to move.
    You need to calculate the value for F just before it starts moving.
    This happens when the static frictional force reaches its maximum value.
     
  9. Dec 2, 2012 #8
    I found the acceleration the block would have if there was no horizontal force, so basically the force down the incline. And then to balance it i divided by cos 40. I drew the diagram, but still don't get it..
     
  10. Dec 2, 2012 #9
    @ap123, so the block is not moving in the start. So I just have to find the max friction can be, and then I have my component of max force F?
     
  11. Dec 2, 2012 #10
    In the x-direction (along the ramp) you have the x-comp of F, the x-comp of the weight and the static frictional force. The sum of these will equal zero.
    You will also need the normal force.
    In the y-direction you have the y-comp of F and the y-comp of the weight and well as the normal force.
    If you put all these together you will get the answer :)

    Once you get all the components straight, it will be easier.
     
  12. Dec 2, 2012 #11

    haruspex

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    Yes.
     
  13. Dec 2, 2012 #12
    So, I think I got it.

    Fn = mg cos theta + F sin theta
    and then F = mg tan theta

    So F = 16.9 N

    and Fn = 26.10N, so max Ffric = 23.49 / cos 40 deg (component) = 30.7N for Max force F?
     
  14. Dec 2, 2012 #13
    Nearly there!

    Your first expression (for the y-components) is fine.
    The second expression ( for the x-components ) needs to include the frictional force :
    static friction + wsinθ - Fcosθ = 0

    You then need to combine both equations - the only unknown will be F which you can solve for.
     
  15. Dec 2, 2012 #14
    Wait, what? So this is okay?
    Fn = mg cos theta + F sin theta
    And how do I set up the second equation? I don't understand what you wrote for it...
     
  16. Dec 2, 2012 #15
    friction force down the ramp + x-comp of weight down the ramp = x-comp of F up the ramp
     
  17. Dec 2, 2012 #16
    So, does it become something like: Fn * Us + mg sin theta = F cos theta?
     
  18. Dec 2, 2012 #17
    Looks good.

    Take the value for the normal from the first equation and substitute it into this one.
    Then solve for F and put in all the values.

    Hope it works out!
     
  19. Dec 2, 2012 #18
    awesome, thanks!
     
  20. Dec 2, 2012 #19
    Hmm, I got -25N for the force of F? Is that right? Also can you explain the principle behind the second equation?
     
  21. Dec 2, 2012 #20

    haruspex

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    No, it should be positive.
    Your equations are Fn μs + mg sin(θ) = F cos(θ), and Fn = mg cos(θ) + F sin(θ), yes? If you can't find your mistake, please post your working from there.
     
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