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Cable Tension problem

by galuda
Tags: cable, tension
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galuda
#1
Oct28-07, 10:13 PM
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?
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PhanthomJay
#2
Oct28-07, 10:24 PM
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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 .
saket
#3
Oct28-07, 10:29 PM
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.)

galuda
#4
Oct28-07, 10:33 PM
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?
saket
#5
Oct28-07, 11:08 PM
P: 169
Quote Quote by galuda View Post
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).
galuda
#6
Oct28-07, 11:24 PM
P: 20
ah ok sorry about that. Thank yall very much for the assistance.


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