# Cable Tension problem

by galuda
Tags: cable, tension
 P: 20 An 8.9kg watermelon is placed at one end of a 6.1m, 238 N scaffolding supported by two cables. One supporting cable is at the opposite end of the scaffolding, and the other is 0.76 m from the watermelon. The acceleration of gravity is 9.8 m/s^2. How much tension is in the cable at the end of the scaffolding? How much tension is in the cable closest to the watermelon? I really have no clue how to go about this. Any relevant formulas I should know?
 Sci Advisor HW Helper PF Gold P: 6,049 The system is in equilibrium. Sum of forces in y direction and sum of torques about any point must equal zero. Don't forget the torque from the scaffolding .
 P: 169 For equilibrium of a body: 1. ΣF = 0 2. ΣT = 0 where, F is the force and T is the torque. (bold ones are vector quantities.)
 P: 20 Cable Tension problem ok well with the watermelon converted over to N i've got a total of 325.28N * 9.8m/s^2 = 3187.744 for my downward force. If that is correct then how do I apply the distance from the melon to figure the tension on the individual cables?
P: 169
 Quote by galuda ok well with the watermelon converted over to N i've got a total of 325.28N * 9.8m/s^2 = 3187.744 for my downward force. If that is correct then how do I apply the distance from the melon to figure the tension on the individual cables?
Remember: Always try to be atleast dimensionally correct!
How come, "325.28N * 9.8m/s^2 = 3187.744 for my downward force"?? You must be knowing, unit of force is N.. then how, N * m/s^2 = N??
In fact, if tensions in the strings are T1 and T2, then
T1 + T2 = 238N + 8.9kg*9.8m/s^2 = 325.22N.

Now, take torque about any point and equate to zero to get one more equation involving T1 and/or T2. Then, solve for T1 and T2 from the two equations obtained. It would be quicker if you take torque about the point where the cable meets scaffolding (either of them).
 P: 20 ah ok sorry about that. Thank yall very much for the assistance.

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