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Proving tension in rope of equilibrium system

  1. Jul 28, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-7-29_0-26-50.png

    2. Relevant equations
    principle of moments

    3. The attempt at a solution
    Okay so when resolving the forces horizontally and vertically, I get RA= T cosθ for horizontal component and RB= W + T sinθ for vertical component. To get my T, I chose to take moments about B, though I'm not sure if it's the fastest way. So 2 RA = ¾W + T sinθ( 1.5 - OC cosθ).

    I get the ¾ value by using similar triangles. Now, the problem I have is eliminating OC from my equation, because OC is not perpendicular to AB, which leaves me with many unknowns that I don't know how to eliminate.
    So, can someone help me find a way to eliminate OC?
     
  2. jcsd
  3. Jul 28, 2015 #2

    SammyS

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    You are missing a contribution to the torque. The horizontal component of T also exerts torque about point B.

    Also: You will need to do some trig stuff to eliminate length OC .
     
  4. Jul 28, 2015 #3
    Okay, so I rewrite my equation as 2 RA = ¾W + Tsinθ (1.5-OC cosθ) + Tcosθ (2-OC sinθ)
    Since RA= T cosθ, I substituted this value into the equation above.
    Then. after some algebraic manipulation, I still did not manage to eliminate my OC, I get T= 3W / ( 8 OC cosθsinθ- 6 sinθ), which looks a bit like the answer, but isn't.
    Must I form another equation to eliminate OC?
     
  5. Jul 28, 2015 #4

    andrewkirk

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    You are given all angles in the diagram, plus ##W##, so you have three unknowns: the forces of the wall on the ladder (horizontal), the floor on the ladder (vertical) and the rope on the ladder (at the angle ##\theta##).

    So you need three equations to solve it. The torque gives you one equation that is required in order for the ladder not to rotate. You have written that (I have not checked it). You can get the other two equations by requiring the vertical and horizontal components of force on the ladder to be zero, so that it does not move.
     
  6. Jul 28, 2015 #5
    I have the three equations already, but I still can't manage to eliminate OC from my equation. That's my main problem. Now I'm not sure if it's necessary to include the OC in my moment equation, since I cannot eliminate it. But I can't find a better way, since its hard to find the perpendicular distance of tension from B, with OC not perpendicular to AB
     
  7. Jul 28, 2015 #6

    andrewkirk

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    I notice a couple of things about your equation:

    1. I think you should have ##\frac{3}{5}W## not ##\frac{3}{4}W## for the torque about B provided by the ladder's weight.
    2. I don't think there's any need to put OC in your equation. You have resolved the rope's tension into horizontal and vertical components. Now resolve each of those into components parallel and perpendicular to the ladder, using the 3:4:5 triangle geometry. Add the two perpendicular ones, multiply by BC and you have the contribution to torque from the rope. To get BC, drop a perp from C to OB to divide OCB into two triangles. The geom of one of those is given by ##\theta##, the other is 3:4:5, and you know OB=1.5m, so you can express BC in terms of ##\theta##.
     
  8. Jul 29, 2015 #7

    SammyS

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    (3/4)W is correct.
     
  9. Jul 29, 2015 #8

    andrewkirk

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    Ah yes. The force that generates the torque is ##\frac{3}{5}W##. I neglected to multiply by the distance of ##\frac{5}{4}m## from ladder's CoM to ##B##, which makes the torque ##\frac{3}{4}W##.
     
  10. Jul 29, 2015 #9
    I still can't manage to express BC in terms of θ only. Here are my steps:

    tanθ = x/ (1.5-y), where x and y are two unknowns.
    tan α = x/y , where α is part of the 3:4:5 triangle.

    Since both x are the same, tanα = tanθ (1.5-y)/ y
    which gives 4/3 = tanθ (1.5-y)/ y

    cosα=y/ BC
    BC = ¾ tanθ (1.5-y) x 5/3
    =5/4 tanθ (1.5-y)

    I've tried for quite some time, and I can't eliminate the y from my equation. This question has exasperated me for hours.
     
  11. Jul 29, 2015 #10
    I still can't manage to express BC in terms of θ only. Here are my steps:

    tanθ = x/ (1.5-y), where x and y are two unknowns.
    tan α = x/y , where α is part of the 3:4:5 triangle.

    Since both x are the same, tanα = tanθ (1.5-y)/ y
    which gives 4/3 = tanθ (1.5-y)/ y

    cosα=y/ BC
    BC = ¾ tanθ (1.5-y) x 5/3
    =5/4 tanθ (1.5-y)

    I've tried for quite some time, and I can't eliminate the y from my equation. This question has exasperated me for hours.
     
  12. Jul 29, 2015 #11
    I still can't manage to express BC in terms of θ only. Here are my steps:

    tanθ = x/ (1.5-y), where x and y are two unknowns.
    tan α = x/y , where α is part of the 3:4:5 triangle.

    Since both x are the same, tanα = tanθ (1.5-y)/ y
    which gives 4/3 = tanθ (1.5-y)/ y

    cosα=y/ BC
    BC = ¾ tanθ (1.5-y) x 5/3
    =5/4 tanθ (1.5-y)

    I've tried for quite some time, and I can't eliminate the y from my equation. This question has exasperated me for hours.
     
  13. Jul 29, 2015 #12
    Sorry about the multiple posts...internet was laggy, but I can't delete them.
     
  14. Jul 29, 2015 #13
    Why don't you try balancing torque from point O ?

    Now you wouldn't have to worry about OC or anything else .

    Hope this helps .
     
  15. Jul 29, 2015 #14
    I don't see how balancing torque from point O eliminates the need to use OC. How about the perpendicular distance from horizontal component of T to O, and also for the vertical component?
     
  16. Jul 29, 2015 #15
    Tension will not produce a torque about point O . What is the line of action of tension ?
     
  17. Jul 29, 2015 #16
    So if O lies in the line of action of tension, then there would be no moment. Hmm, I see that I'm wrong. Thanks for pointing it out.
     
  18. Jul 29, 2015 #17
    So , can you now solve this question ?
     
  19. Jul 29, 2015 #18
    Wow, just wow. I'm speechless. After trying for so long and wasting so much time, you made it look so easy, and I feel relieved and rather dumb at the same time. Many many thanks to you, this is one of the few questions that have managed to stump me for a long time.

    By the way, I just started to learn this topic in the syllabus, which is equilibrium of rigid bodies. Questions from this topic often involves many unknowns, and I find myself having to form many simultaneous equations to solve them. So, I'm wondering if you have any tips on how to best approach these kind of equilibrium questions?
     
  20. Jul 29, 2015 #19
    Well , usually , the main part of the solution involves balancing torque .
    Choose a point from which torque balancing is easy - in many cases , it is from where minimum torques are visible , in many it is not .

    Force balancing always forms part of the equations .

    Practice really helps .
     
  21. Jul 29, 2015 #20

    haruspex

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    See if the tip at the end of section 1 of https://www.physicsforums.com/insights/frequently-made-errors-mechanics-moments/ helps.
     
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