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How do I find the forces acting on a ladder?

  1. Mar 18, 2017 #1
    1. The problem statement, all variables and given/known data
    "To get up on the roof, a person (mass 70.0 kg) places a 6.00-m aluminum ladder (mass 10.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be friction-less. The center of mass of the ladder is 2.00 m from the bottom. The person is standing 3.00 m from the bottom. Find the normal reaction and friction forces on the ladder at its base."
    hLY9Yvc.png

    2. Relevant equations
    ##\theta=cos^{-1}(\frac{1}{3})=tan^{-1}(2\sqrt{2})##
    ##m_0=10kg##
    ##s_0=\frac{2 m}{tan\theta}##
    ##m_1=70kg##
    ##s_1=\frac{3 m}{tan\theta}##
    ##||s||=6m##

    3. The attempt at a solution
    I'm choosing my lever arm to be the point at which the ladder touches the ground.

    ##∑|τ|=-g(m_0s_0+m_1s_1)+||s||cos\theta⋅F_r=0##
    ##∑F_x=-gcos\theta(m_0+m_1)-F_r+f_s=0##

    ##F_r=f_s-gcos\theta(m_0+m_1)##
    ##F_r⋅||s||⋅cos\theta=g(m_0s_0+m_1s_1)##
    ##F_r=\frac{g}{||s||}sec\theta(m_0s_0+m_1s_1)##
    ##f_s-gcos\theta(m_0+m_1)=\frac{g}{||s||}sec\theta(m_0s_0+m_1s_1)##
    ##f_s=gcos\theta(m_0+m_1)+\frac{g}{||s||}sec\theta(m_0s_0+m_1s_1)##

    Can anyone tell me if I'm going in the right direction? Thanks.
     
    Last edited: Mar 18, 2017
  2. jcsd
  3. Mar 18, 2017 #2

    TSny

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    You have the correct approach. But I noticed a few things:

    (1) It is always helpful to define the symbols that you are using. For example, I had to try to decipher your equations to decide what angle you are calling ##\theta##. Likewise for the symbols ##s_1##, ##s##, and ##s_0##.

    (2) The problem states that "the center of mass of the ladder is 2.00 m from the bottom". I think this probably means that the center of mass is 2.00 m along the ladder starting from the bottom of the ladder. If they meant it the way you interpreted it, I think they would have said that the center of mass is 2.00 m above the ground. Similarly for the position of the person.

    (2) You wrote the equation ##∑|τ|=-g(m_0s_0+m_1s_1)+||s||cos\theta⋅F_r=0##. Check to make sure that the expression for the lever arm for ##F_r## is correct.
     
  4. Mar 18, 2017 #3

    TSny

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    On more thing, it looks like you have assumed that the forces of gravity have a horizontal component. Is this right?
     
  5. Mar 18, 2017 #4
    Oh, it's ##||s||sin\theta⋅F_r##, right? ##F_r## acts horizontally, so the lever arm is the vertical component of ##||s||##.

    Forgive me. I think I should treat the ladder as a point mass for the summation of forces expression.

    ##∑F_x=f_s-F_r=0##
    ##∑F_y=-(m_0+m_1)g+F_N=0##
     
  6. Mar 18, 2017 #5

    TSny

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    That all looks good.
     
  7. Mar 19, 2017 #6
    ##\theta=cos^{-1}(\frac{1}{3})##
    ##m_0=70 kg##
    ##s_0=3 m##
    ##m_1=10kg##
    ##s_1=32 m##

    ##∑|τ|=-gcos\theta(m_0s_0+m_1s_1)+F_r(||s||sin\theta)=0##
    ##F_r=(\frac{g}{||s||})(cot\theta)(m_0s_0+m_1s_1)=(\frac{9.8\frac{m}{s^2}}{6m})(\frac{1}{2\sqrt{2}})((70kg)(3m)+(10kg)(2m))=133 N##
    ##∑F_x=f_s-F_r=0##
    ##f_s=F_r=133N##

    ##∑F_y=F_N-g(m_0+m_1)=0##
    ##F_N=g(m_0+m_1)=784N##
     
  8. Mar 19, 2017 #7
    I tried to do a quick check and it looks like your results for Fr and FN are correct.
     
  9. Mar 19, 2017 #8
    Is that right? I probably need a new physics problem book because it says that ##F_N=784N## but ##f_s=376N##.
     
  10. Mar 19, 2017 #9
    I just repeated the sum of the moment equation and once again calculated 132.8 N for Fr, which is equal to -fs, since those are the only 2 horizontal forces. So I think that is the correct answer unless you and I are both doing something fundamentally wrong.
     
  11. Mar 19, 2017 #10
    It looks like the book incorrectly used (Fr)(||s||)(cosθ), instead of (Fr)(||s||)(sinθ) when calculating the moment from force Fr.
     
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