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Statics Ladder Problem

  1. Aug 6, 2006 #1
    I am having a little trouble with this statics problem:
    "A 15.0 m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 60 degree angle with the horizontal. (a) Find the horizontal and vertical forces the ground exerts on the base of the ladder when an 800-N firefighter is 4.00 m from the bottom."

    In solving this problem, the first thing I did was draw a free-body diagram of the scenario and label all the exerting forces. The forces that I came upon were the force of the wall exerting on the ladder (Fw), the weight of the ladder (WL), the weight of the firefighter (WF), and the x and y components of the force of the ground exerting on the ladder (Fxg and Fyg).

    The next thing I told myself was: because the ladder is at rest, it is in equilibrium, so the net force and torque will be zero.

    I go ahead and start my sum of forces in the x and y directions:
    Fx = Fw - Fgx = 0.
    Fx = Fgy - WL - WF = 0.
    Because the weight of the ladder and firefighter are given, I was easily able to solve for the Fgy component which came out to be 1300 N.

    My problem lies in solving for the Fgx component. This is my attempt:
    I realized that because torque (t) is equal to force multiplied by distance (fd), I would need to use it to solve for the Fgx component. Therefore, I created my sum of torques:
    t = -800(4)(sin60) -500(7.5)(sin60) + Fw(15)(cos30), and I solved for Fw which came out to be 4.63 N.

    In my above equation, the 4 comes from the distance the firefighter is standing from the base of the ladder, the 7.5 is the center of gravity of the ladder (halfway point of the ladder), and the 15 comes from the total length of the ladder because the force of the wall is at the top of the ladder. All of these values are in meters.

    I think my problem lies in me possibly using the wrong trig function and/or angle value. It looks like if I used 30 degrees instead of 60 for the above first two trig function values, I get to the right answer. Why is it mandatory to use 30 degrees instead of 60 when the problem specifically states the angle is "60 degrees with the horizontal" - which is right at the base of the ladder and the ground. I understand that the other angle of the formed triangle would be 30 degrees, but why use that angle instead of 60 degrees?

    I hope my question/analysis isn't confusing. Thank you for your time and help.
  2. jcsd
  3. Aug 6, 2006 #2


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    Everything up to here is correct.
    The first 2 terms in that expression have the wrong trig functions in them. Do it again - draw the triangles, label the correct sides, write out the ratios carefully and find the required unknowns.

    Remember that the 'd' in the torque equation is the perpendicular distance from the line of action of the force to the chosen point.

    PS: Note that cos30=sin60
    Last edited: Aug 6, 2006
  4. Aug 6, 2006 #3


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    The reason you must use sin(30) (or cos(60)) comes from the definition of the torque.

    Look at the weight of the ladder, say. The component of that force that contributes to the torque is the component that is *perpendicular* to the ladder. And that component is WL sin(30).
    In general, you must take the component of the force that is perpendicular to the line joining the axis of rotation to the point where the force is applied.

    If you look at the force exerted by the wall on the top of the ladder, you will see that the part perpendicular to the ladder is FWFW cos(30) so that part was right.
  5. Aug 6, 2006 #4
    I understand. Both of you explained it very well. Thanks.
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