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Torque Ladder Question

  1. Aug 31, 2006 #1
    Two Ladders 4m and 3m long, respectively,are hinged at their tops (point A) and tied together by a horizontal rope .6m above the floor. The ladders weigh 400N and 300N respectively and the center of gravity of each is at its center. If the floor is frictionless, find
    (a) the upward force at the bottom of each ladder
    (b) the tension in the rope and
    (c) the force that one ladder exerts on the other at point A
    (d) If a load of 1000N is now suspended from point A, find the tension in the rope.
    Draw the 4m ladder on your left (so your drawing is the same as mine) at an angle of 36.87 degrees to the floor, at the r.h.s. is the 3m ladder at an angle of 53.13 degrees to the floor, this means that the angle at point A is 90 degrees. I can do (a) the answers are 326N and 374N respectively, pointing vertically up ward. (b) Let x be the point where the rope is attached to the 4m ladder , taking moments about x:
    -326*.8*1 -400*.8 -300*3.3 + 3.75*F3y + 374*4.2 =0
    I get 0=3.75*F3y => F3y =0 => F3x=0 i.e.,no tension in the rope at where it is attached to the 3m ladder.
    (c)At point A I'm also at a loss to identify the forces.
    Thanks for your time and effort.
     
  2. jcsd
  3. Aug 31, 2006 #2

    Doc Al

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    Staff: Mentor

    If you wish to find the tension in the rope, you cannot take moments about the point of attachment of the rope, since the tension exerts no torque about that point. Use the top of the ladder as your pivot point for taking moments.
     
  4. Sep 4, 2006 #3
    (b)Taking moments about the top of the ladder:
    = -326*3.2 + 1.8*F4 + 400*1.6 -300*.9 -1.8F3 + 374*1.8
    = 0 + 1.8(F4 - F3); where F4 is the tension in the rope attached to the 4m ladder & -F3 is the tension in the rope attached to the 3m long ladder. I end up F4=F3, and still not knowing the tension in the rope?
    (c)Let F5 be the force on the 4m ladder due to the 3m ladder and let F6 be the force on the 3m ladder due to the 4m ladder; then we have:
    Sum of the Fx components = F4 -F3 - F5x + F6x =0
    Sum of the Fy components = 326N -400N + F5y -F6y -300N +374N = 0,
    implying F5x = F6x, F5y=F6y, leaving me still not knowing the force that one ladder exerts on the other?Thanks.
     
  5. Sep 4, 2006 #4

    Doc Al

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    Staff: Mentor

    Only analyze one ladder at a time! For example, analyze the forces acting on the 4 m ladder, taking moments about the top of the ladder. (Of course F4 = F3. It's a single rope--the tension is the same throughout.)
     
  6. Sep 5, 2006 #5
    I at last got the answer to part (b) it is 224N. In part (c) I get the wrong answer when I take moments about where the rope is attached to the 4m ladder i.e., : -326*.8 -400*.8 + F5*3 = 0;
    it implies that F5 = 193.6N, F5 being the force exerted by the 3m ladder on the 4m ladder.
    At the top of the 4m ladder:
    Sum of Fx = F4 - F3 + F5x -F6x = 0
    Sum of Fy = 326N -400N -300n + 374N + F5y + F6y = 0, therefore
    F5y = -F6y; F5x = F6x. Where F6 is the force exerted by the 4m ladder on the 3m one. If I get help on part (c), I should be able to do part (d) myself. Thanks again for the help.It would help if PF had a drawing tool in this case.
     
  7. Sep 6, 2006 #6

    Doc Al

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    Staff: Mentor

    Realize that the force exerted at the top of the 4m ladder (due to the 3m ladder) has both vertical and horizontal components. Rather than taking moments, just set the vertical and horizontal components of the net force on the 4m ladder equal to 0.
     
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