
#1
Oct2807, 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? 



#2
Oct2807, 10:24 PM

Sci Advisor
HW Helper
PF Gold
P: 5,966

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 .




#3
Oct2807, 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.) 



#4
Oct2807, 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?




#5
Oct2807, 11:08 PM

P: 169

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 T_{1} and T_{2}, then T_{1} + T_{2} = 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 T_{1} and/or T_{2}. Then, solve for T_{1} and T_{2} from the two equations obtained. It would be quicker if you take torque about the point where the cable meets scaffolding (either of them). 



#6
Oct2807, 11:24 PM

P: 20

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



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