How Does Throwing a Medicine Ball Affect Beam Displacement?

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SUMMARY

The discussion focuses on calculating the new x-position of a person at the left end of a uniform beam after a medicine ball is thrown from the right end. The initial setup involves a beam with a mass of 104 kg and a length of 2.7 m, with two individuals of masses 61 kg and 68 kg positioned at either end, along with an 8 kg medicine ball. The correct center of mass was determined to be 1.43, leading to the conclusion that the beam moves approximately 0.186 m when the ball is thrown, although initial calculations suggested an incorrect center of mass value of 1.53.

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1. Homework Statement
A person with mass m1 = 61 kg stands at the left end of a uniform beam with mass m2 = 104 kg and a length L = 2.7 m. Another person with mass m3 = 68 kg stands on the far right end of the beam and holds a medicine ball with mass m4 = 8 kg (assume that the medicine ball is at the far right end of the beam as well). Let the origin of our coordinate system be the left end of the original position of the beam as shown in the drawing. Assume there is no friction between the beam and floor.

3)After the ball is thrown to the person on the left, what is the new x-position of the person at the left end of the beam? (How far did the beam move when the ball was throw from person to person?)


2. Homework Equations
Center of mass equation (can't figure out how to type it on here)

3. Attempt at a solution
I kept the original center of mass of 1.53 since it doesn't move and created an equation to solve for the new position.

1.53 = \frac{(61+8) (0-x) + 104 (1.35-x) + 68 (2.7-x)}{241}

And I get x = .186m, which is not correct.
 
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I would check your center of mass value to start -- I got 1.434.
 
The system accepted 1.53 as the answer. I could try 1.434 though.
 
Wow. I don't know why it accepted something so far off for the first question. Using 1.43 worked, thanks.
 
No problem -- those darn online things.
 

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