Cable Tension problem

galuda

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?

PhanthomJay

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

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

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

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

ah ok sorry about that. Thank yall very much for the assistance.