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Cable tension problem

  1. Oct 16, 2009 #1
    1. The problem statement, all variables and given/known data

    hello i have a problem with this question.

    the cylinder shown in fig1 has a mass of 10 kg. if theta = 45 and the system is in eqlibrium determain the tension in cables ca, cb.

    2. Relevant equations

    all i want is the equations and a guide to the i.e what each component is m=mass or g=gravity. and it would be really helpful is some provided outher useful formula for similar questions. as i have 3 or 4 and one includes a spring but i dont want to be cheeky and ask all 3 questions so if i got a list of formula i think i could apply them ok providedi know each component.


    3. The attempt at a solution

    10x9.81 = 98.1 and because the system is in equlibrium the total tension must equal 98.1.

    Tca= 30=w1=10x 9.81
    98.1/sin30 =196.2N

    Tcb = Fca cos 45 = 196.2 cos 45 = 138.59N
    these two answers are far to big i think and i dont know of any way to find the tension
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Oct 16, 2009 #2

    Redbelly98

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    Welcome to Physics Forums :smile:

    Actually, no. That is only true for a mass hanging by a single, vertical cable.

    This makes no sense. In part, this equation says that 30=10x9.81???

    The approach, after drawing a free-body diagram, is to set up 2 separate force equations for the horizontal and vertical force components.
     
  4. Oct 16, 2009 #3
    if i were to do

    +> Fcb cos 45 = Fca cos 30

    +^ Fca sin 30 = Fcb sin 45 = 98.1N

    then

    Fcb = Fca cos 30/ cos 45

    then sub Fcb with:

    Fa sin 30+ Fca cos(30) x sin 45 / cos 45

    Fca [sin 30 + (cos 30 x sin 45/cos 45)]

    Fca=98.1

    sub in Fca into Fcb = 9.81 x cos 30 / cos 45


    Would that be the right kind of theory??? If not i am lost thanks for you help and any futher is greatly appriciated.

    Thank You
     
  5. Oct 16, 2009 #4

    Redbelly98

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    Okay, that is the sum of the upward rope forces. If you set that whole expression equal to the downward 98.1 N force, you can then find Fca.

    No.
     
  6. Oct 17, 2009 #5
    How would i do that? Could you give me the equation please?

    Thank you
     
  7. Oct 17, 2009 #6

    rl.bhat

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    Fca [sin 30 + (cos 30 x sin 45/cos 45)]

    Fca=98.1

    You have to modify this step.
    Fca [sin 30 + (cos 30 x sin 45/cos 45)] = 98.1
    Find Fca
     
  8. Oct 17, 2009 #7
    i dont know how to do that could you show me please?
     
  9. Oct 17, 2009 #8

    Redbelly98

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    It's basic algebra. I would start by evaluating the expression,

    (cos 30 x sin 45/cos 45)​
     
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