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A simple ladder problem (equilibrium)

  1. Nov 13, 2005 #1
    First test on equilibrium tomorrow, and I'm very sure there is going to be a ladder-type problem on it where I have to determine normal forces, frictional forces when someone is on the ladder up a certain distance, and how far can the person go up before it starts to slip. I don't know if you guys are familiar with masteringphysics.com or not, but it is an online homework system where it automates questions similar to that in the book. I've gotten pretty far in the problem, but for some reason I can't complete the 2nd two problems. Here is the exact problem on mastering physics:
    I drew a diagram and a free-body diagram of the problem, and was able to get the answer to A through the requirements for equalibrium. Here is my work:
    (Fnet)x: Fs-Fw=0
    (Fnet)y: Fn-Fp-Fg=0
    We have Fp and Fg, so Fn = Fp+Fg or Fn = 900
    Since Fs <= Us*Fn we have (Fnet)x: Us*Fn-Fw=0, or Fw=(.4)*(900), so Fw = 360. And since Fs<=Us*Fn, Fs = Fw, so Fsmax=360. This is the answer to A.
    I am confused as hell when it comes to B. It is difficult to explain because you cannot see my free-body diagram, but I will do my best to describe it.
    Don't know how to use tao, so I'll just use lowercase t.
    (tnet)=tn+ts+tp+tg+tw=0. I am choosing the origin to be at the lower end of the ladder where Fn and Fs are, so those would equal zero because they have no moment arm. So here is the new equation:
    (tnet): -(Fp)(R)Sin()-(Fg)(R)Sin()+(Fw)(R)Sin()=0. With the values we have
    (tnet): -(740n)(1m)sin(53.13)-(160n)(2.5m)sin(53.13)+(Fw)(5m)sin(36.87)=0.
    I found 53.13 and 36.87 through basic trig. Here are my questions: I am solving for a new Fw (normal force coming from the wall) in this equation right? i.e. I would NOT be using the value I obtained earlier. After I get that I plug it back in to (Fnet)X to get Fs, which would be my answer - right?

    My final answer was 301.3, which is NOT right - The correct answer is 171n. Why :confused:

    When it comes to part C I'm not sure how to go about it because I don't fully understand the answer to part B. I'm guessing you solve for the unknown moment arm in the tnet equilibrium equation.

    Thanks guys

    pink floyd says hi :cool:
  2. jcsd
  3. Nov 13, 2005 #2


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    You've got your angles the wrong way round!!

    Or, you should be taking cos instead of sin, and vice versa.

    The distance from the bottom of the ladder to the wall is 3 m.
    3m is not the height up the wall to the top of the ladder
  4. Nov 13, 2005 #3


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    Oh, btw, it's a 3-4-5 triangle, with cos@ = 3/5, sin@ = 4/5. It makes calculations a bit simpler if you recognise that :smile:
  5. Nov 13, 2005 #4
    Thanks Fermat :)

    btw for searching purposes the answer to B is 171n and C is 2.7m.
  6. Nov 13, 2005 #5


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    Yeah, I got those answers too.
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