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    tiffanymessias's Avatar
    tiffanymessias Posts: 2, Reputation: 1
    New Member
     
    #1

    Aug 12, 2015, 12:31 PM
    Urgent
    A ladder whose length L = 18 m and whose mass, m, is 40 kg rests against a wall. Its upper end is adistance h = 12 m above the ground. The centre of mass of the ladder is one-third of the way up theladder. A fire fighter whose mass M is 75 kg climbs the ladder until his centre of mass is halfway up.Assume the wall but not the ground is frictionless. What forces are exerted on the ladder by thewall (Fw) and by the ground (Fg)?
    Curlyben's Avatar
    Curlyben Posts: 18,514, Reputation: 1860
    BossMan
     
    #2

    Aug 12, 2015, 12:35 PM
    What do YOU think ?
    While we're happy to HELP we wont do all the work for you.
    Show us what you have done and where you are having problems.
    tiffanymessias's Avatar
    tiffanymessias Posts: 2, Reputation: 1
    New Member
     
    #3

    Aug 12, 2015, 12:46 PM
    Quote Originally Posted by Curlyben View Post
    What do YOU think ?
    While we're happy to HELP we wont do all the work for you.
    Show us what you have done and where you are having problems.
    the length of the ladder from wall= 13.42
    cos alpa= 12/18
    ala=48.19 degrees

    sin beta= 12/18
    beta=41.81 degrees
    fighter fighter force= 735N
    Ladder force= 392N

    I don't know where I go from here or if I'm on the right track?
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
    Expert
     
    #4

    Aug 13, 2015, 08:18 AM
    Hello Tiffany, so far so good. It's always a good idea to draw a free-body diagram, including the unknown reaction force of the wall against the top of the ladder and the ground against the bottom of the ladder. Note that the reaction force at the top is normal to the wall, whereas at the bottom because of friction there are both normal and tangential forces. Then the next step is to consider that the sum of forces in both the vertical and horizontal directions must be equal to zero - this allows you to solve for one of those reaction forces. Finally, consider that the sum of torques (or moments) about any point must be zero - in general it's best to pick a point that has two unkown forces, so that you can ignore them in this calculation. Take a stab at this, and post back with what you get.

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