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Friction: Static Cylinder And Ramp Problem: Again!

  1. May 31, 2005 #1
    Thanks for the help,

    i think i'm getting it, but i still have doubts, anyway, here's the problem:

    A wedge of 15° is pushed under a tube of 50 kg as shown in fig 1. [tex] \mu_S = 0.20 [/tex] in all the surfaces. Determine the required force P to move the wedge.

    answer is [tex] P &= 283 N\leftarrow [/tex]
    but i get 377.73 N, please give some hints of what i'm doing wrong

    what i'm doing is (please refer to attachments to see the fbd's)

    [tex] \arctan\mu_S = \theta_S[/tex]

    [tex] \arctan0.20 = \theta_S[/tex]

    [tex] 11.3 &= \theta_S [/tex]

    [tex] R &_B=\frac{W\sin101.3}{\sin52.4} [/tex]

    [tex] R &_B=607.09 N [/tex]

    and in the second system:

    [tex] P &=\frac{607.09N\sin37.6}{sin78.7} [/tex]

    which yields:

    [tex] P &=377.3N\leftarrow [/tex]


    please help!!

    thanks in advance for any help you can give me
     

    Attached Files:

    Last edited: May 31, 2005
  2. jcsd
  3. May 31, 2005 #2

    OlderDan

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    I have an idea on this, but I have not yet gotten it to work out to the given answer. See how this strikes you. I think your calculation is assuming that the maximum frictional force is applied at all surfaces. That is not likely the case here. For a very small angle, the normal force at the plane is going to be much greater than the normal force at the wall. The cylinder is going to slide up the wall, and roll up the plane. The torques have to be equal and opposite, so the vertical component of the force at the wall, which will be the maximaum friction force, must equal the friction force parallel to the plane, which will be less than the maximum friction force. This is going to reduce the force at B, and therefore reduce the force at C, resulting in less force at P. I believe the same will be true even at 15 degrees. See what you can do with this.
     
  4. May 31, 2005 #3
    uhmm, i have a couple of questions

    ok, let's say [tex] F &_A=F &_B^' [/tex]

    where [tex] F &_B^' [/tex] is the frictional force parallel to the plane

    if my calculations are right [tex] F &_A = R &_A\sin 11.3 [/tex]

    [tex] F &_A = 274.3 N\sin 11.3 [/tex]

    [tex] F &_A = 53.74 [/tex]

    ok, so if [tex] \frac{F &_B^'}{\mu_S} = 268.7 = N &_B^' [/tex]

    and [tex] R &_B^{'} = 274.06 [/tex]

    because [tex] R &_B^{'} = \sqrt{N &_B^{'}^{2} + F &_B^{'}^{2} [/tex]

    that would make [tex] P &= 171.72 N [/tex]

    plz, any help!!!!

    the other question, are the fbd right?

    thanks in advance for any help you can give me

    edit:

    another thing i tried was to make the angle between [tex] W \ \mbox{and} \ R_B [/tex] 15 degrees in the first fbd, so that instead of having [tex] R_B [/tex] you have [tex] N &_B [/tex] directly and you can obtain [tex] F_B [/tex] and eventually [tex] R_B [/tex], but that didn't do it either.
     
    Last edited: Jun 1, 2005
  5. Jun 1, 2005 #4

    OlderDan

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    Those numbers look too high to me. It works as I described earlier. I got the given answer. Your original free body diagrams are not correct because they assume maximum friction at all three surfaces. The direction of your B force is not correct. You need to equate the frictional force at A to the frictional force at B. The frictional force at A is 20% of the normal force at A. Here are my starting equations, where I have broken the normal and frictional forces at B into horizontal and vertical components. These are the forces acting on the cylinder.

    [tex]
    \overrightarrow {F_A } = F_{AN} \widehat i - \mu F_{AN} \widehat j
    [/tex]

    [tex]
    \overrightarrow {F_B } = F_{BN} \left( { - \sin 15\widehat i + \cos 15\widehat j} \right) + \mu F_{AN} \left( { - \cos 15\widehat i - \sin 15\widehat j} \right)
    [/tex]

    Add the weight to these to get zero and solve for the normal force components. You should get

    [tex]F_{AN} = 177.76[/tex]

    [tex]F_{BN} = 554.13[/tex]

    Plug those in to get the total force at B. The normal force at C must counter the vertical force on the wedge at B. The horizontal force at C is 20% of the normal force because the wedge is moving. Add the horizontal components at B and C, and you get the magnitude of P.
     
  6. Jun 1, 2005 #5
    Thanks, thank you very much

    once again, you saved me....thank you very much

    [tex] F_B^{'} [/tex] which is parallel to the plane is 35.55, from that, everything was 'easy'.....

    [tex] N_C = 526.05 N [/tex]

    [tex] \sum F_x = 0 [/tex]

    [tex] F_N^{'}\cos15° + F_{BN}\sin15° + 0.2N_C - P = 0 [/tex]

    [tex] P = 283.6 N [/tex]

    see you next problem....

    :wink:
     
  7. Jan 29, 2010 #6
    Here's one that I just saw on a Canadian First Class Operating Engineer Applied Mechanics exam:

    A solid cylinder rests at the top of a ramp 1.2 meters high. Neglecting friction what is the coefficient of friction? How fast does the cylinder move down the ramp (in meters per second)?
    Note: The radius of gyration is the same as the diameter of the solid cylinder.

    I have no problem admitting that I had no idea where to start on this one (though I knew the note was a good place to start).
    Any help with this question would be greatly appreciated!
     
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