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Simple Tension problem

  1. Jul 28, 2009 #1
    1. The problem statement, all variables and given/known data

    A 125 lb weight is suspended from the ceiling by means of two perpendicular flexible cables. Find the tension (in pounds) in each cable. There is a picture attached. I'm new at posting so I'll try my best to describe this simple image. There is a line that runs horizontal representing the ceiling. On the left side of the ceiling, there is a line that runs South east at a 55 degree angle ending at the 125lb weight. On the right side of the ceiling there is a line that runs south west at a 40 degree angle ending at the 125lb weight. That is about it. The ceiling and the cables form a triangle.

    2. Relevant equations
    None given


    3. The attempt at a solution
    This is a problem in the vector section of my multivariable calculus book. The solution to a similar problem is given by defining the cables (arbitrarily called T1 and T2) as T1=(T1 cos40)i +(T1sin40)j and T2=(-T2cos55)i +(T2sin55)j. From there the solution states that T1 + T2 + F(force or -125j) = 0. And they solve algebraically. I am having trouble with the concept. I do not want to plug in numbers without knowing why. I would like to know why we use cosine and sine. (I have a vague idea, but not a firm understanding). I would also like to know how to solve the problem. If anyone has an idea, it would be appreciated. I feel as though this is a very basic and simple problem and I would like to get some good insight to it. Thanks in advance.
     
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  3. Jul 28, 2009 #2

    Dick

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    You use sine and cosine to split the forces at the specified angles into i and j components. Then you add all the vectors. The result must be zero because the suspended weight is not accelerating. If the sum of the vectors is zero, then the sum of the i components and the sum of the j components must both be zero. That gives you two equations in two unknowns for the magnitudes of the tensions. Does that help?
     
  4. Jul 29, 2009 #3
    This should probably be in the introductory physics homework help subforum.
     
  5. Jul 29, 2009 #4
    Thank you Dick and AUMathTutor for responding. I believe I need more information on vectors in the general sense. Dick, your answer makes sense to me. I believe that I understand the idea of separating the i and the j components. However it occurs to me that I am missing a very rudimentary understanding of vectors because I do not fully understand your statement: "If the sum of the vectors is zero, then the sum of the i components and the sum of the j components must both be zero." It is not that this statement is overly complex or that it is worded in a way that is hard to comprehend. It is that I cannot state with complete accuracy that I understand completely what you are saying. I understand the idea that since the weight is not accelerating that the force should equal 0. However I do not understand, in depth, that the the i components and the j components should both be zero.

    I believe this to be a result of my own lack of knowledge on the subject. I am missing some key vocabulary. Therefore, I would like to change the aim of this query to pointing me in the correct direction of basic vector knowledge. Thank you again Dick and AUMathTutor for taking the time to respond. I believe that my lack of knowledge is preventing me from understanding the solution. Also, sorry for posting this in the incorrect subject.
     
  6. Jul 29, 2009 #5

    HallsofIvy

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    The force from each cable on the weight is a vector pointing in the direction of the cable. You can resolve that into [itex]\vec{i}[/itex] and [itex]\vec{j}[/itex] components (horizontal and vertical). For example, if one wire makes angle [itex]\theta[/itex] with the horizontal and has tension T, we can draw vertical horizontal and vertical lines making a right triangle with the wire as hypotenuse. In terms of force, that hypotenuse has "length" T so the horizontal and vertical "lengths" (forces) are [itex]T cos(\theta)[/itex] and [itex]T sin(\theta)[/itex] respectively. Since the weight is not moving, the total force on it must be 0 and that means the horizontal forces must sum to 0 while the vertical force due to the wires must sum to the weight of the object. That gives two equations to solve for the two tensions.
     
  7. Jul 29, 2009 #6

    Dick

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    If ai+bj=ci+dj and if i and j were ordinary numbers then you wouldn't be able to tell me much about the relation between a, b, c and d. But they aren't, i and j are vectors representing directions that are at right angles to each other. The only way that equality can be true is if a=c and b=d. The up-down components have to be equal AND the left-right components have to be equal. Can you picture why that has to be? (In technical language, i and j are linearly independent.) Similarly, if ai+bj=0, then a=0 and b=0.
     
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