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Equillibrium ladder problem

  1. Sep 22, 2010 #1
    1. The problem statement, all variables and given/known data
    Untitled-3.jpg


    2. Relevant equations
    Torque formulas, equilibrium formulae


    3. The attempt at a solution

    I tried to find the force of gravity on both the ladder and pale. The I thought that the wall would have to excrete that exact same force to hold it up, but that wasn't the answer.

    Thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 23, 2010 #2
    What made you think that the force exerted by the wall would be same as that of the force exerted by gravity on the ladder and the bucket??
    Are the directions of force of gravity and normal force by the wall same?
     
  4. Sep 23, 2010 #3
    Yes you're right. Now that I think of it, that did not make any sense.

    How would we go about solving this one? I guess we would first establish the foot of the ladder as the pivot point, then calculate the torques? Where I get screwed up is calculating the torques.

    Thanks for the help! :)
     
  5. Sep 23, 2010 #4
    You are always welcome!!

    This time you have got the correct path. Carefully calculate the torques ( be careful about the signs) due to all forces taking the foot of the ladder as the pivot point, then and equate it to zero (which is the necessary condition for rotational equilibrium)!
     
  6. Sep 23, 2010 #5
    How does this sound?

    Torque of Ladder: (12kg)(9.80)(2.5m)(sin50)=-225.2(negative because clockwise)
    Torque of pale: (2.5kg)(9.80)(4.0m)(sin50)=-75.1(negative because clockwise)

    -225.2+-75.1=(F)(5.0)(sin50)

    I have made a mistake somewhere but can't tell where. Can you please tell me where I have made a mistake? Thanks.
     
  7. Sep 23, 2010 #6
    Look once again at the equations you have written.
    For the expressions of the torque due to the ladder's and bucket's weight, which component of their weights are you taking? Should they be the sin or the cos component?? (Use a little geometry and that shall help you figure out your mistake)
     
  8. Sep 23, 2010 #7
    I don't understand how to find the correct angle you use. Why would you use cos(50), or would you use cos(40) -its supplimentary angle.
     
  9. Sep 23, 2010 #8
    I saw someone solve a problem like this online.

    They went about it like this:

    (12)(9.80)(2.5)(3.21)/5+(2.5)(9.80)(4.0)(3.21)/5-F(5)(sin50)=0

    I don't understand how this works. It gets the right answer-66N, but for example why divide by 5?
     
  10. Sep 24, 2010 #9
    To get the correct angle do the following:
    Draw a right triangle ABC (rt angled at B, base AB)with angle CAB=50. Now if you draw a line perpendicular to AC and passing through C, then what angle does the line make with BC?? (You know that angle ABC is 40)
    In this problem, the line AC is somewhat like the rod and the vector along BC is the line of action of gravity, so the torque you must take its component that is perpendicular to AC, which is the cos component!

    Hope you got it now!

    So far as the solution you have put up is concerned, look at it a bit more carefully, the term (3.21/5) is indeed the value of cos 50!! That's why its correct and you get the correct answer!
     
  11. Sep 24, 2010 #10
    123.jpg

    like this?
     
  12. Sep 26, 2010 #11
    so the component for the wall would be sine? How would you know that?
     
  13. Sep 26, 2010 #12
    I know for the bucket and ladder it's cose. but what about the wall.
     
  14. Sep 27, 2010 #13
    you dont have too break up the force from the wall into components, the normal from the wall is completely horizontal and it points in the negative x direction
     
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