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Calculate force required to tip beam on a rope w/ friction

  1. Apr 29, 2015 #1
    1. The problem statement, all variables and given/known data
    The uniform I-beam has a mass of 74 kg per meter of length and is supported by the rope over the fixed 300-mm drum. If the coefficient of friction between the rope and drum is 0.50, calculate the least value of the force P which will cause the beam to tip from its horizontal position.

    2rxtjpj.jpg

    2. Relevant equations

    T1/T2 = e^(βμ)

    Where β = the angle of the rope on the drum, μ = 0.50, T1 = the bigger tension, and T2 = the smaller tension

    3. The attempt at a solution

    So to start off, I added the lengths up to see that the beam is 1.8 m long, I then used that to determine the weight = 1.8*74*9.81 = 1306.7 N. Also the rope on the drum suggests the angle β = π rads and μ is known to be 0.50.

    I began by making my FBD, which I believe is probably where I've made my mistake. I broke it into two separate FBD, one with just the beam and one with just the drum.

    FBD of beam

    The FBD of the beam I saw 4 forces acting on it. the weight (1306.7 N) acting downward in the middle of the beam, the force P acting downward (as indicated in the problem), and the two ropes which I called tension 1 (left rope) and tension 2 (right rope) both with forces acting upward due to them being tension ropes (therefore they cannot be downward, I believe).

    I used the moment equation ∑M = 0 at T1 and T2.

    ∑M =0 at T2 gives me: (-.75)(-P) + (.15)(-1306.7) + (.3)(T1) ⇒⇒ T1 = (1.05P + 196)/0.3

    ∑M =0 at T1 gives me: (-1.05)(-P) + (-.15)(-1306.7) + (-.3)(T1) ⇒⇒ T2 = (-.75P + 196)/0.3

    FBD of drum

    For this FBD I just had T1 going downward (due to the tipping) and T2 going upward. I used the T1/T2 equation to have (1.05P+196)/(-.75+196) = e^(.5π), which gave me P = 244.5 N.

    This is incorrect, as the book says the answer should be P = 160.3 N.
     
  2. jcsd
  3. Apr 29, 2015 #2

    haruspex

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    Are you mixing up T1 and T2 somewhere? Isn't P 1.05m from T2?
     
  4. Apr 29, 2015 #3
    Ah, sorry, I mistyped what I had written down, so it should have been:

    ∑M =0 at T2 gives me: (-1.05)(-P) + (-.15)(-1306.7) + (-.3)(T1) ⇒⇒ T1 = (1.05P + 196)/0.3

    ∑M =0 at T1 gives me: (-.75)(-P) + (.15)(-1306.7) + (.3)(T2) ⇒⇒ T2 = (-.75P + 196)/0.3

    But I think the T1 and T2 values I got from that are correct (or at least correct based on my incorrect assumptions)
     
  5. Apr 29, 2015 #4

    TSny

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    You left out a factor of P in the denominator. Maybe just a typo. Otherwise, I think your equation gives the correct answer.
     
  6. Apr 29, 2015 #5
    Yeah, sorry, that's another typo. I had the -.75P in the equation while working it out, and I got P =244.5. Unless I'm messing up algebraically, but I don't think I did.
     
  7. Apr 29, 2015 #6

    TSny

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    I thought it was probably just a typo. But when I solve your equation, I get the correct answer!
     
  8. Apr 29, 2015 #7
    Actually, according to wolframalpha I spoke too soon, I must be making an algebra error, because wolfram says the answer is 160.

    I guess I had the correct method, I'm just screwing up the algebra part lol.

    Thanks you guys!
     
  9. Apr 29, 2015 #8

    haruspex

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    I think it's not just a typo in the post. Reinstating the P leads to the book answer.
     
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