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Force needed to make a disk climb a step

  1. Apr 19, 2008 #1
    [SOLVED] Force needed to make a disk climb a step

    Howdy ho. First off, let me say that I'm a newbie here, and I'm not sure if this is the correct place to put this doubt. Here's the deal:

    1. The problem statement, all variables and given/known data

    We have a disk, and we need to know the force F needed to have it climb the step, with 50mm. The force is applied at the top of the disk.

    ............F <---- .....
    ........50mm | |........|

    Sorry for that draw, it's the best I could do.
    The only known data is:
    the step has 50mm;
    the disk is uniform and weights 40N;
    no friction is to be considered.

    2. Relevant equations


    3. The attempt at a solution

    I've tried solving it in several ways. I've considered a normal vertical reaction 1 between the ground and the disk, and considered a normal diagonal reaction 2 between the disk and the tip of the step. I also considered that the angle this reaction 2 makes is 45º. So, I concluded that, in the vertical axis, we have: 40 = N1 + N2.sin45º.
    After this, we would have in the horizontal axis the force F and N2.cos45º. I thought that one condition for the disk to climb the step would be: F > N2.cos45º.
    However, this wouldn't be enough. So I used the moment. The moment the force F causes in the center of the disk is F.r . We do not know r, so this would lead to another variable in another equation. So, unless there is missing some data here, I don't know how to solve it. I have the solution, and it says the force should be F=17,88N .

    Sorry for any language error, english is not my native language, and is not the language I'm using in classes, so I may have mistaken some technical terms.
  2. jcsd
  3. Apr 19, 2008 #2

    Doc Al

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    Staff: Mentor

    Taking moments is the right approach. Hint: You can choose any pivot point you like, so choose wisely.
  4. Apr 19, 2008 #3
    I can't seem to get a solution. I tried using the tip of the step as the pivot. But the distance between this pivot and the 3 forces (weight, reaction 1 and force F) all depend on R. Besides, the sum of all moments is depending on reaction 1, which is unknown.
    Thanks for any help
  5. Apr 19, 2008 #4


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    Homework Helper

    Hi Towelie,

    Was this a book problem? This answer does depend on R so they would have to give it to you somehow. (The answer you have of F=17.88 N corresponds to a radius of 15 cm.)
  6. Apr 20, 2008 #5
    No, it's in a work sheet, made by my professor. It is indeed possible some data is missing, it wouldn't be the first time. But thanks for all your help!
  7. Apr 20, 2008 #6

    Doc Al

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    Staff: Mentor

    You need to be given R.
    At the instant the disk begins to rise up, what is that reaction force?
  8. Apr 20, 2008 #7
    That would be 0, am I correct?
  9. Apr 20, 2008 #8

    Doc Al

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    Staff: Mentor

    You are correct.
  10. Apr 20, 2008 #9
    Ok. So I assumed the instant the disk is rising up, that is, N1=0. I also assumed R=0,15 , like alphysicist said. In this scenario, all we need is the moment caused by force F to be bigger than the moment caused by weight:

    F . 0,25 > 40 . 0,1118 (=) F > 17,888N

    The same result from my solutions. So I think it is correct! Thank you very much for your help, Doc Al and alphysicist.
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