A7.6 kg watermelon is placed at one end of a 5.9 m, 245 n scaffolding supported by two cables. one supporting cable is at the opposite end of the scaffolding, and the other is 0.63 m from the watermelon. how much tension is in the cable at the end of the scaffolding? the acceleration of gravity is 9.8 m/s 2 .

98.952≅ 99 N

Explanation:

Since you don't know the tension in the cable 0.63 m from the watermelon, use this as your axis.

Clockwise moments:

(245)(2.32)

This is the weight of the scaffolding, acting in the middle of the board.Since we took the cable as the axis, instead of 3.25 m as the distance, it's 2.15 m, the distance from the cable to the middle of the board.  

Counterclockwise moments:

(T)(5.27) + (7.6)(9.8)(0.63)

The other cable and the weight of the watermelon.  

These need to be equal to each other.  

(T)(5.27) + (7.6)(9.8)(0.63) =  (245)(2.32)

Solving for T, and you should get 98.952≅ 99 N

measure their densities same density, same sub

unbalanced force

explanation: