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Ladder problem

  1. Jun 6, 2005 #1
    A uniform ladder 5.0 m long rests against a frictionless, vertical wall with its lower end 3.0 m from the wall. The ladder weighs 160 N. The coefficient of static friction between the foot of the ladder and the ground is 0.40. A man weighing 740 N climbs slowly up the ladder.

    a) What is the maximum frictional force that the ground can exert on the ladder at its lower end?
    **This is what I did: (160N + 740N) x 0.4 = 360 N

    b) What is the actual frictional force when the man has climbed 1.0 m along the ladder?
    **For x = 1.0 m: {(160)(1.0) + (740)(1)(1/2.5)}/ (5)(2/2.5) = ? NB

    c) How far along the ladder can the man climb before the ladder starts to slip?
    **(160)(2.5)(1.5/5) + (740)(1.5/2.5) x_max = 360 and solve for x_max

    Can you please check my work?
     
  2. jcsd
  3. Jun 6, 2005 #2

    OlderDan

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    Part a) looks OK. I'm having trouble deciphering the other parts, but I'm not seeing some things I expect to see. What principle is involved in writing the equation for those parts? Please either give an answer to the questions based on what you have posted, or rewrite it in terms of identifiable parts, like forces and moments (torques). Including units would be helpful.
     
  4. Jun 6, 2005 #3
    net torque = 0

    W_L = weight of ladder; W_m = weight of man; L = length of ladder;
    x = distance the man has climbed up the ladder from the base;

    W_L(L/2) cos(theta) + W_m x cos(theta) - fs * L sin(theta) = 0
    fs = {(160 N)(5 m/2)(1.0 m/2 m) + (740 N)(1.0 m)(1.0 m/2.0 m)}/ (5 m)(2 m/2.5 m)

    Please help
     
  5. Jun 7, 2005 #4

    OlderDan

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    Your starting equation looks good. I assume (theta) is the angle between the ladder and the floor at the base. Your trig ratios are not correct. The ladder is 5m long and is 3m from the wall at the base. This makes a 3-4-5 right triangle. The cosine should be 3/5 and the sine should be 4/5. Other than that, it looks OK
     
    Last edited: Jun 7, 2005
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