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Find forces in ladder

  1. Nov 21, 2006 #1
    Each side of a stepladder is 8m long and there is a hinge at the top. An 87kg man is 6m up one side and there is a 2.5m long rope connecting the two sides. The rope is exactly halfway up the ladder. Assume that the floor is frictionless and neglect the weight of the ladder. Find the tension in the rope, the forces exerted at the points A and B and the forces at the hinge C. It will be easiest if you separate the ladder into two halves and continue from there.

    The answers in the answer key are 210N, 533N, 320N, 320N, 210N

    I split the ladder into two.

    On the left between A and C, I know that there is a tension on the rope pulling right. There is a gravitational force on the person pushing down, and there is a normal force pushing up on A.

    On the right, I have Tension pulling right and that's it. The ladder is massless.

    I don't know what to do though. I suspect that some of the stuff I relate looking at the free body diagram might have been wrong. I don't know what to do to solve for these forces.
  2. jcsd
  3. Nov 21, 2006 #2


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    Do you have an image? i am not getting the geometry.
  4. Nov 21, 2006 #3
  5. Nov 21, 2006 #4


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    Ultimately you want to split the ladder in 2 and analyze the FBD for each. However, FIRST find the vertical reactions at A and B before you split up the ladder. The only forces acting are the weight, and the vert. reactions at A and B. Everyting else is internal to the system. So just sum torques = 0 about A and solve for the vert force at B. Then the vertical force at A plus the vert force at B must equal the weight of the person. That's most of the battle, then take it from their with your FBD"s.
  6. Nov 21, 2006 #5


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    Ok, start off by not breaking at the hinge, yet!. Do torque to find the normal forces on A and B, first, then break apart and find the rest of the forces.
  7. Nov 22, 2006 #6
    I know that net torque is zero

    0 = -T_na + T_nb -T_mg + T_t

    I got

    0 = -1.25N_a + 1.25N_b - .3123*mg + T_t

    I know that T_t is there, but I don't know the direction. Is it pushing right or left?

    After I get this, do I plug into those equations to get the rest. The answers are 210N, 533N,320N,320N, 210N
  8. Nov 22, 2006 #7


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    Did you find the normal forces on A and B first?

    In case you didn't understand, if you do not break apart, the only forces you will have are Weight of person, normal force A and normal force B, the rest are internal forces for the whole as your system.
  9. Nov 22, 2006 #8
    I am having problems with this part.

    I know that N_a + N_b = Mg, and I know that N_a is greater than N_b because the person is closer to A.

    N_a + N_b = 852.6

    Since tension is internal as you said, I should not include it in the torque equation - right?


    0 = -T_na + T_nb - T_mg
    0 = -1.25*N_a + 1.25*N_b - .3123*mg
    0 = -1.25*N_a + 1.25* (mg - N_a) -.3123mg

    -2.5N_a = -.9377mg
    N_a = 319.8

    and N_b = 532.8

    Did I do something wrong? Why is N_b>N_a when the person is closer to A?

    Now, how do I find the forces at the hinge at the tension of the rope? I know the rope is resisting rotation (ladder falling) but how do I know how much?
  10. Nov 22, 2006 #9


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    Your problem seems to be in your lever arms, right no tension, to find the rest of the forces, break apart the bodies.
  11. Feb 4, 2008 #10
    Is there any possibility you could repost the image I am confused as to where B is located with only the words to go by.
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