## minumum rope length

The rope connected to the triangle is pulling away at 3000N
i need to determine the minimum length of ac so that the tension in either ab or ac does not exceed 4000N. AC and BC are the same length.

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 can someone please just tell me the equation that relates rope length to tension.
 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor Hi Ry122! I don't understand … your diagram says BC = 5m, and angles BCA and CBA are 30º. So AC is fixed, isn't it? What am I missing?

## minumum rope length

oops, I made a mistake. angle CBA is unknown and length BA is different from AC

 Is the tension in the rope pulling away (3000N) equal to the tension in BC? if so ill know how to answer the question

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 Quote by Ry122 Is the tension in the rope pulling away (3000N) equal to the tension in BC?
No, certainly not.

You have to use Newton's second law: sum of the forces in any particular direction is zero.

But I still don't understand what stops the triangle from rotating.

 a large mass is attached to the triangle at BC.
 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor Sorry … this is no good. Will you please type out the whole original question?
 sure The tension in the tow rope pulling the car in Newtons is 3000N. Determine the minumum length of the rope l, between A and B, so that the tension in either AB or AC does not exceed 4000N
 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor So where does the 30º come into it? Or is that gone too? Is angle CAB unknown as well as angle CBA?
 angle CBA = theta angle BCA = 30 degrees BA = l BC = 1.2m
 can anyone help me with this?
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 Quote by gozita73 can anyone help me with this?
You need to show your attempt first.

HINT: Start by breaking the forces down into x and y components, then solve the equations for the desired unknown. Initially, the desired unknown will be theta. Once you have theta, you can use a little trig to find the length, L, which is what you truly desire.

CS

 I get: Sum of Fx 0 = Fab cos (theta) + Fac cos (30) Sum of Fy 0 = Fab sin (theta) + Fac cos (30) - 3000N Would Fab and Fac equal max. tension (4000N)? otherwise, I'm completely stuck

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