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Equilibrium of a Rod

  1. Sep 14, 2013 #1
    Problem Statement: As shown in the figure below, a uniform beam is supported by a cable at one end and the force of friction at the other end. The cable makes an angle of θ = 30°, the length of the beam is L = 1.50 m, the coefficient of static friction between the wall and the beam is μs = 0.420, and the weight of the beam is represented by w. Determine the minimum distance x from point A at which an additional weight 2w (twice the weight of the rod) can be hung without causing the rod to slip at point A.
    9-p-017.gif

    Relevant Equations:
    Ff = μN
    Ʃτ = 0
    ƩF = 0

    Solution Attempt:
    I'm not really sure where to start. Does Fn (normal force) come from the weight of the rod multiplied by cos(30) due to the angle of the cable holding it against the wall? And once I find Fn, I multiply that by μ, giving the upward force of friction. Then, placing the rotational axis at the end of the rod that the cable is attached to (because it's not moving, I can place the axis anywehere) and set Fn equal to the weight of the object being attached multiplied by the distance from the wall. Would that work?
     
  2. jcsd
  3. Sep 14, 2013 #2
    You just need to go about the problem systematically.

    Write down the equation for x-components of the forces.

    Write down the one for y-components.

    Then for moments/torques.

    You are going to have a bunch of known and unknown forces in those equations. Of those, the normal force at A and friction at A will be related with the coefficient of friction.
     
  4. Sep 16, 2013 #3
    Okay so:

    ƩFy: Ff + FBsin30-2w-w = 0
    ƩFx: FN - FBcos30 = 0
    Ʃτ: -w(0.75) - 2wxm + FBsin30(1.5)

    Is that correct?
     
  5. Sep 16, 2013 #4
    Assuming the third formula ends with " = 0 ", that looks good.
     
  6. Sep 16, 2013 #5
    Alright so all of that is correct, but now what? We don't have most of those variables and the only ones we can solve for are FF/FN. How do we relate w to FN? Everything has to be in terms of "w" as it stands.

    What am I missing?
     
  7. Sep 16, 2013 #6
    Observer that once you relate friction and the normal force, the first to equations can be solved, giving you all the forces in the system. But you still have the third one to satisfy.
     
  8. Sep 17, 2013 #7
    So rewriting ƩFy we get:

    μFN - 2w - w = -FBsin30 → -(μFN - 2w - w) = FBsin30

    And rewriting ƩFx we get:

    FN = FBsin30

    We set these equal which gives:

    -(μFN - 2w - w) = FN

    then

    -μFN + 2w + w = FN

    then

    (-0.420)FN + 2w + w = FN

    then

    What? Nothing cancels so we can't solve for anything.
     
  9. Sep 17, 2013 #8
    It was FBcos 30 originally.

    How so? Assuming it is correct (which it is not), this is the same as 3w = 1.420 FN. Surely this can be solved for FN.
     
  10. Sep 17, 2013 #9
    Is the answer 0.9275m?
     
  11. Sep 17, 2013 #10
    Oh yeah, so it really doesn't work then.

    Now I feel really dumb. How do you solve that for anything, but an answer in terms of w?

    But back to the problem, I'm at a loss of what to even try. You say relate FN and μ, but I can't.

    If I try setting it equal with FN rather than FB...

    I get:
    FBcos30 = (FBsin30 - 3w) / -μ

    Does cos30/sin30 cancel somehow that I'm unaware of? That would help with one of the variables. Actually that would help everything! Waitwaitwaitwaitwait...

    [FBcos30] / [FBsin30] = -3w/ -μ

    substitute in for μ:

    [FBcos30] / [FBsin30] = -3w/ -0.420

    Cancel FB and multiply by -0.420:

    -0.420(sin30/cos30) = -3w

    Divide by -3:

    (0.420/3)(sin30/cos30) = w

    RIGHT???
     
  12. Sep 17, 2013 #11
    I don't have the answers. It's a Web Assign, if you'v heard of it. Basically online physics homework that allows for 5 solution attempts.
     
  13. Sep 17, 2013 #12
    The way I did mine was:

    We need to find the Fb in terms of w then we can eliminate it from the final answer as all the terms in the torques will have w. So I found 0.42Fn=Ff and that Fn=Fbcos30. From there i found Ff in terms of Fb and then put this into Fy to find Fb in terms of w. From here i substituted this into the torque equation and simplified
     
  14. Sep 17, 2013 #13
    @voko ?
     
  15. Sep 17, 2013 #14
    How many more answer attempts do you have?
     
  16. Sep 17, 2013 #15
    Is there any help available on this thread?
     
  17. Sep 17, 2013 #16
    I think you need to brush up on your algebra skills. When you have ## a F_B = (b F_B + c)/d ##, you should have no problem doing this: $$

    a F_B = (b/d) F_B + (c/d)

    \\

    a F_B - (b/d) F_B = c/d

    \\

    (a - b/d)F_B = c/d

    \\

    F_B = \frac {c/d} {a - b/d}

    \\

    F_B = \frac {c} {ad - b}
    $$
     
  18. Sep 17, 2013 #17
    Burnst14 got any answers?
     
  19. Sep 17, 2013 #18
    This approach should work.
     
  20. Sep 17, 2013 #19
    Cheers Voko. Sorry for being quite brash - I get really into problem solving
     
  21. Sep 17, 2013 #20
    There is also another approach, somewhat more complex but "more correct". Note that by letting ## F_f = \mu F_n ## we assume that the friction is maxed out. But we do not really know if it is going to be maxed out at a minimum or at a maximum of ##x##.

    So the more correct approach would involve not letting ## F_f = \mu F_n ## just yet, and instead solve the system of three equations in three variables ##F_B, \ F_f, \ F_n ## and one parameter ##x##. Then find the range of ##x## where ## F_f \le \mu F_n ##. You could try that.
     
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