Static equilibirum, elasticity, fracture problem

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SUMMARY

The discussion focuses on calculating the tensions in two ropes supporting a 19.0 kg ball, with rope A at a 22-degree angle to the vertical and rope B at a 53-degree angle. The weight of the ball is calculated as 186.39 N (19 kg * 9.81 m/s²). The tension in rope A is derived using the formula TensionA = (Weight of the ball) / cos(22 degrees), while a similar approach is applied to find the tension in rope B. The key forces present include the weight of the ball and the tensions in both ropes.

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1. A 19.0 kg ball is supported from the ceiling by rope A. Rope B pulls downward and to the side on the ball. If the angle of A to the vertical is 22 degrees and if B makes an angle of 53 degrees to the vertical find the tension in rope A and then find the tension in rope B
 

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Can you list the forces present?
Please show your attempt.
 
yeah the forces present are the weight of the ball (19.0 kg) and the tensions in both ropes i tried to work out the tension in rope A by looking at its tension as a vector with both an X and Y components and used that to make a triangle out of the tension in. I figured the y would be equal to the weight of the ball which would be (19kg)(9.81) and the hypotenuse would work out to be cos(22 degrees)= (19kg)(9.81)/tension or TensionA= (19kg)(9.81)/cos(22degrees) and the same thing for Rope B and I am not sure where i went wrong
 

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