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Simple Torque problem. Help please!

  1. Mar 15, 2009 #1
    A uniform ladder of a length 5m and a mass 15 kg leans against a wall, making a 45 degree angle with the ground. The wall is capable of supporting a maximum of 500 N.

    A fireman must climb the ladder to it's midpoint, what is the max mass that the fireman can be without collapsing the wall.


    Ok...So I know that for a uniform problem we know that this means that the Center of Mass will be halfway along the lengh of the board.

    Again since this is a center of mass problem the sum of the torques must equal zero.

    Et=0

    I assumed that I would need to take the (max force)(cos 45) + (Weight of ladder)(distance of ladder) then divide that by the (total distance of the ladder)


    The answer should be 86.95kg

    But I got..

    (500N)cos45 + (15kg)(5m)

    = 353.33N + 75 kg.m
    = 428.33 N.kg.m / 5.0m = 85.66 = 86 kg

    I am doing something wrong. I am not able to cancel out some of my units. Its seems like a simple mistake but any advice would help!!

    Thanks!
     
    Last edited: Mar 15, 2009
  2. jcsd
  3. Mar 16, 2009 #2
    Force X Perpendicular distance = Torque.
    Net torque = 0
    500 X 5 cos45 = (15+m) g X 2.5 cos 45
    m = (1000/9.8) - 15 = 87 Kg
     
  4. Mar 16, 2009 #3

    tiny-tim

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    Welcome to PF!

    Hi Rha1828! Welcome to PF! :smile:
    I don't understand this at all :confused:

    you should be taking moments (torques) about the base of the ladder …

    that's ∑(force x perpendicular distance)

    try again :smile:

    sArGe99, on this forum, please don't try to give complete answers​
     
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