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Statics: Tension in a cable

  1. Jun 17, 2013 #1
    1. The problem statement, all variables and given/known data
    A force P of magnitude 90 lb is applied to member ACE, which is supported by a frictionless pin at D and by the cable ABE. Since the cable passes over a small pulley at B, the tension may be assumed to be the same in portions AB and BE of the cable. For the case when a = 3 in., determine (a) the tension in the cable, (b) the reaction at D.
    tCGrJb4.png


    2. Relevant equations
    ƩM=0
    ƩF_x=0
    ƩF_y=0

    3. The attempt at a solution

    I am stuck on part a). Here is my equilibrium equation:
    ƩM_D=90*9 + T*3 - (5/13)*(9)*T - (12/13)*(7)*T = 0

    Solving for T I get the wrong answer.
     
  2. jcsd
  3. Jun 18, 2013 #2
    Perhaps you could explain each term in the moments equation.
     
  4. Jun 18, 2013 #3
    90*9: 90 is the P force, and 9 is the perpendicular distance from C to D (12-3)

    T*3: This is for the portion of the cable running from B to E. Perpendicular distance is a=3.

    (5/13)*(9)*T: horizontal component of the AB portion of the cable. I got 5/13 from the ratio of the triangle (hypotenuse is 13 by pythagorean theorem). 9 is the perpendicular distance.

    (12/13)*(7)*T: vertical component of the AB portion of the cable. Same triangle ratio, with 7 as the perpendicular distance.
     
  5. Jun 18, 2013 #4
    Let's look at the force applied at A. What is the direction and the magnitude of the lever arm relative to D?
     
  6. Jun 18, 2013 #5
    I thought it was the AB tension from the last two terms in my equation. I don't see any other forces acting on A.

    edit: Its direction would be 67.4° from the horizontal, and the magnitude would be whatever T is.
     
    Last edited: Jun 18, 2013
  7. Jun 18, 2013 #6
    You are talking about the magnitude and the direction of force. I am talking about the lever arm: this is the distance from the point about which moments are calculated to the point to which the force is applied.

    I notice that you did in fact decomposed the lever arm into horizontal and vertical components, which is almost correct. Almost, because I think you neglected the signs. Which is why it is important to know what the direction of the lever arm is.
     
  8. Jun 18, 2013 #7
    Wouldn't it be different for both components? For the vertical T component, it goes to the right from A to C (+7in). For the horizontal component, it goes up from C to D (+12in).
     
  9. Jun 18, 2013 #8
    Actually, the directions are exactly the opposite of what you said. Recall that the vector of the displacement must be taken from D, not toward D. Despite this, I now think your original equation is correct. What is value of tension that you get from it? Why do you think it is not correct?
     
  10. Jun 18, 2013 #9
    Whoops, it looks like I did get it right the first time, I just did the calculations wrong. I did it again and got the right answer. Sorry about that!
     
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