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Calculate the force exerted by the wall on the ladder

  1. Dec 30, 2005 #1
    A 1.29-m long ladder weighing 200N rests against a vertical wall so that the top of the ladder is at a height of .590m. A 603-N man stands on the ladder at a distance of .390m along the ladder. Calculate the force exerted by the wall on the ladder. Assume that the wall is perfectly smooth.

    Well.. i've never had to deal with a problem like this, so do not really know where to go from here. Do I need the equation Fnet=Fa+F+(-Fg) and solve for Fa? It seems like i need to find the torques? I really don't know what to do here.
     
  2. jcsd
  3. Dec 30, 2005 #2

    Doc Al

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    Yes, you'll need to examine the torques: For the ladder to be in equilibrium, the sum of the torques on the ladder (about any point) must be zero. What point do you think you should use as your pivot for computing torques?
     
  4. Dec 30, 2005 #3
    I would think either the ground level or the place where the man is on the ladder.
     
  5. Dec 30, 2005 #4

    Doc Al

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    OK, but one of those points will get you the answer you need with the least amount of effort. Which one?

    The trick is to pick a point (if possible) where you can identify all the forces exerting a torque and where the only unknown force is the one you are trying to find.
     
    Last edited: Dec 30, 2005
  6. Dec 30, 2005 #5
    I would have first tried the man on the ladder. Would that be the simplest way to go?
     
  7. Dec 30, 2005 #6

    Doc Al

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    Try it and see for yourself. Start by identifying all the forces on the ladder. (Draw a diagram of the ladder and mark the forces on it.)
     
  8. Dec 30, 2005 #7
    Ok. I've drawn the picture
    [​IMG]
     
    Last edited: Dec 30, 2005
  9. Dec 30, 2005 #8

    Doc Al

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    You'll need to revise your picture to show each force, exactly where it acts, and the direction that it points.

    To help you out, I'll list all the forces that I see acting on the ladder. It's your job to draw them on your diagram accurately.

    (1) Weight of the ladder
    (2) Weight of the man
    (3) Force that the wall exerts
    (4) Upward (normal) force that the floor exerts
    (5) Frictional force that the floor exerts
     
  10. Dec 30, 2005 #9
    Alright, i redid the image. I was sure about the frictional force that the floor exerts, so i have not added that in. Which way is the force for the friction the floor exerts?

    [​IMG]
     
  11. Dec 30, 2005 #10
    careful- the weight of the ladder would be at the ladder's center of mass.
     
  12. Dec 30, 2005 #11

    Doc Al

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    OK, you have the directions of the forces correct, but you need to accurately show where each force acts. As andrewchang points out, the weight of the ladder acts at the ladder's center. But what about the forces exerted on the ladder by the floor and the wall? Where do they act? (If you don't show where they act, you'll have a hard time calculating the torque they generate.)

    If the friction force disappeared, which way would the ladder slip? The friction force points so as to oppose that slipping.
     
  13. Dec 30, 2005 #12
    Ok. I think i've got it drawn out.
    [​IMG]
     
  14. Dec 30, 2005 #13

    Doc Al

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    Looks good. Now pick your pivot point and start calculating torques about it. If you pick the right point, the only unknown force in your torque equation will be the force you're trying to find: The force of the wall on the ladder. (If you pick the "wrong" pivot point, you'll have multiple unknowns and will need additional equations to solve the problem.)
     
  15. Dec 30, 2005 #14
    Ok, well, i think i'll pick where the man is on the ladder as the pivot point, since i'm trying to set both sides equal to eachother, correct?(equilibrium?). The torque equation is t=Frsinθ... i'm not exactly sure how to go about finding the torques on a point.
     
    Last edited: Dec 30, 2005
  16. Dec 31, 2005 #15

    Doc Al

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    No matter where you pick your pivot point you will be setting clockwise torques equal to counter-clockwise torques. So that's no reason to pick any particular point as your pivot. I'd pick the point that eliminates as many forces from the torque equation as possible. (Realize that if a force acts directly at the pivot point, it will produce zero torque since "r" would equal zero.)

    To find the torques about a pivot point, apply that equation:
    F is the force

    r is the distance between the pivot point and the point of application of the force

    θ is the angle that the force makes (measured from the "radial line" that connects the pivot point and the point of application)​
    Clockwise torques are considered negative; counter-clockwise torques, positive.
     
  17. Jan 1, 2006 #16
    I've decided that i'll try to pick the point where the ladder is on the floor, since it elimates the friction of floor and force of floor, since r=0. I guess i'm confused about the forces 603-N and 200-N acting on the ladder... should i add those together? Or should the equation look like this: t=603-N x .390m x sin(27.2172)? And: t=200-N x .645-N x sin(27.2172). By the way.. I got the 27.2172 from solving for the sides and angles that that the ladder makes with the wall and ground. If this is the way to go with the torques.. then how would I go about the force exerted by the wall?
     
  18. Jan 1, 2006 #17

    Doc Al

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    Good thinking!

    Careful with the angle: 27.2 degrees is the angle that the ladder makes with the ground, not the angle that the forces make with the radial line.

    But you have the right idea: find the torque exerted by each force. Note that the torques due to the weight of the ladder and the weight of the man are clockwise torques.

    To find the torque exerted by the wall force, do the same thing. Since you don't know the force (after all, that's what you're trying to find!) just call it F. Now find the torque it exerts in the same manner as with the other forces. (Taking care with the angle.) This torque is counter-clockwise.
     
  19. Jan 1, 2006 #18
    So would the angle be 62.8 instead of 27.2? I'm not exactly sure on what you mean by the radial line.

    T=F(1.29)(sinθ) is what you mean to find the force caused by the wall, right? Will i end up with an answer that contains both T and F? Or will it be a number?
     
  20. Jan 1, 2006 #19

    Doc Al

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    That's correct.
    The radial line is the line between the pivot and the point of application of the force.

    Right. Be sure to use the correct angle.
    T just stands for torque. In your torque equation, you'll put in "F(1.29)(sinθ)" not "T". You'll get an equation with only one unknown, the wall force F. When you solve the equation, you'll find out what that force is: Some number of Newtons.
     
  21. Jan 1, 2006 #20
    T of man on ladder=(603N)(.390m)(sin62.7828)=-209.132
    T of ladder=(200N)(.645m)(sin62.7828)=-114.717
    T of friction on floor=0
    T of force exerted by floor=0

    So. T=F(1.29)(sin27.2172)? Would T be the -209.132 and -114.717 added?
     
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