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Homework Help: Student on Ladder (Equilibrium)

  1. Jun 16, 2010 #1
    1. The problem statement, all variables and given/known data

    A student is standing on a ladder as shown in the figure below Each leg of the ladder is 2.7 m long and is hidged at point C. The tie-rod (BD) attached halfway up and is 0.76 m long. The student is standing at a spot 2.025 m along the leg and her weight is 478 N. (You may ignore the weight of the ladder and any minor friction between the floor and the legs.)

    5u1out.gif

    a) What is the tension in the tie-rod?

    b) What is the vertical component of the force of the ground on the ladder leg at point A?

    c) What is the horizontal component of the force of the ground on the ladder leg at point A?

    d) What is the vertical component of the force of the ground on the ladder leg at point E?

    2. Relevant equations

    F=ma
    Fg=mg
    c^2 = a^2 + b^2 -2abcos(angle c)

    3. The attempt at a solution

    I used law of cosines to find the angles in the diagram.
    I tried finding torques but I'm pretty sure I did those completely wrong.

    Help me out please! I'm stuck!
     
  2. jcsd
  3. Jun 17, 2010 #2
    Consider two separated rods. How many forces exerted on each rod?
    Use summation F(x)=0
    summation F(y)=0
    summation torque (C)=0
     
  4. Jun 17, 2010 #3
    For the forces acting on the feet of the ladder, would that just be the normal force since it said to ignore friction? Are there others?
     
  5. Jun 17, 2010 #4
    No, there is only normal force, since friction is negligible as stated in the problem.
     
  6. Jun 17, 2010 #5
    What other forces are in the x direction then other than the tension?
    Would I do two normal forces for each leg?
     
  7. Jun 17, 2010 #6
    Well, in the x direction, there are only tensions and horizontal components of forces at C. However the forces at C are a bit complicated, because they also have vertical components.

    All you have to do is to write down the equilibrium equations as inky pointed out:
    - For the left leg, total F(x) = 0, total F(y) = 0, and total torque about any axis = 0.
    - Similar things for the right leg.
    So: What are the forces on the left leg? On the right leg?
     
  8. Jun 17, 2010 #7
    (1)Firstly draw right leg. How many forces acting on right leg? If we neglect friction ,check do you get 4 forces exerted on let leg.

    (2) Find sin (theta) and cos (theta) from right leg.

    (3) Use summation F(y)=0 and summation torque(C)=0

    (4) Draw left leg. How many forces acting on left leg? If we neglect friction ,check do you get 5 forces exerted on let leg.

    (5) Use summation F(y)=0 and summation torque(C)=0

    (6)Now you got 4 equations. Find normal force at A and E.

    (7) Find tension.
     
  9. Jun 17, 2010 #8
    sorry . Typing error for no.(1) and no.(4)

    (1)Firstly draw right leg. How many forces acting on right leg? If we neglect friction ,check do you get 4 forces exerted on right leg or not.

    (4) Draw left leg. How many forces acting on left leg? If we neglect friction ,check do you get 5 forces exerted on left leg or not.
     
  10. Jun 17, 2010 #9
    inky, what are those 4 forces exerted on right leg??? I got 3
    1 is on C from the left leg, which has negative Y component and positive X component
    2 is on E from the ground, which only has a positive Y component
    3 would be the tension

    the same for the left leg, I got 4 forces
    1 is on C from the right leg, with a positive Y component and negative X component
    2 is where the guy is standing, with only a negative Y component
    3 would be the tension, the same as before
    4 is the normal force from the ground to the left leg with only a positive Y component
     
  11. Jun 17, 2010 #10
    Hi santi_h87,
    You are right as well. I count two forces at point C.
     
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