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KazhiimKazhiim __BANNED USERS regular
edited September 2008 in Help / Advice Forum
Oh my god. I am dying here. Fuck statics.


I need to find the tension of each line segment (except the two holding the weights) such that the system is in equilibrium, as well as the missing angle. I know that the x- and y-axis components need to equal zero, but every equation I come up with has too many variables.

I'm completely fucking stumped. Not asking for the answer, since that's verboten, but even a little hint as to the equation I'm setting up would be nice. Thanks.

Kazhiim on


  • shadydentistshadydentist Registered User regular
    edited September 2008
    Well... break it down into component forces at each intersection of 3 ropes. Solving this depends on knowing that nothing is accelerating, so all forces must balance, and that ropes must have equal tension at both ends. Also, the force of the rope MUST be in the direction of the rope.

    The one on the left has 60 lb down, F1 in the direction of rope, and F2 right. You know that F1(y direction)=F1sin(50)=60 lb, so solve for F1. F1cos(50) = tension left, which also has to equal tension right. You now know the tension to the left of the weight on the right, as well as the weight. Take the arctan of this and you have your answer.

    Make sense?

    shadydentist on
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  • HiroconHirocon Registered User regular
    edited September 2008
    The way you've drawn the problem, it looks like the string connecting the two knots is perfectly horizontal. Do you know that it is perfectly horizontal? If so, then I can find the solution pretty easily.

    First find the tension in the 50 degree string by balancing the vertical forces in the left knot. Then find the tension in the middle string by balancing the horizontal forces in the left knot. Then set up two equations with two unknowns to find the tension and angle of the rightmost string.

    Hirocon on
  • KazhiimKazhiim __BANNED USERS regular
    edited September 2008
    oh, wow

    I am dumber than shit

    Kazhiim on
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