Tension in Cables Supporting a Watermelon on Scaffolding

[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]

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]

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