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Torque, equilibrium and calculating angle of forces

  1. Dec 20, 2011 #1
    1. The problem statement, all variables and given/known data

    I just did two equilibrium/torque problems and it seems like they contradict each other. I understand the basic steps here, but I'm baffled about what I perceive to be a contradiction. Here are the problems:

    A 75 kg window cleaner uses a 10 kg ladder that is 5 m long. He places one end on the ground 2.5 m from a wall, rests the upper end against a cracked window and climbs the ladder. He is 3 m up along the ladder when the window breaks. Neglect friction between the ladder and window and assume that the base of the ladder does not slip. When the window is on the verge of breaking, what is the angle (relative to the horizontal) of the force of the ground on the ladder?

    The second question entails a horizontal beam 3 m long, of weight 500 N, suspended horizontally. On the left end, it is hinged to a wall. The right end has a cable that runs to the wall, a distance of D above the hinge point. (The wall, cable and beam form three sides of a triangle). The least tension that will snap the cable is 1200 N. It asks what value of D corresponds to that tension.

    2. Relevant equations

    Horizontal forces are in equilibrium, as are vertical forces and torques.


    3. The attempt at a solution

    I'm confused here because in the first problem, the angle of the force of the ground on the ladder is NOT the same as the angle of the ladder and the ground. The angle between the ladder and ground is 60 degrees, whereas the answer to number 1 (and therefore the angle of the force relative to the ground) is 71 degrees.

    HOWEVER, in the second problem, it seems like you find the answer by calculating the vertical component of the force of the cable on the beam (it ends up being 250 N), then using the sine trig relationship to find the angle of the force of the cable on the beam (it ends up being 12 degrees). You then assume that this angle is the same as the angle between the beam and the cable, and use the tangent trig relationship to relate 12 degrees, 3 meters (the length of the beam) and D. This yields a D of 0.64 m, which is the answer. My question is, why do we assume that the angle of the force and the angle between the ladder and ground is different whereas, in the second case, we assume that that the angle of force and the angle between the beam and the cable are the same.
     
  2. jcsd
  3. Dec 20, 2011 #2
    The first thing you need to do is calculate the force of the ladder on the window at the instant it breaks.
    You are told to ignore friction with the ladder and the window which means that the force on ladder/window is at right angles to the window (ie horizontal)
    Take moments about the base of the ladder to find the force on the window..... I got the force to be 284N
    Hope this helps... get in touch if you need more guidance.
     
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