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## Homework Statement

1.68 - ((51(0) + 109(1.5) + 66(3) + 13(0))/(51 + 66 + 109 + 13)) [/B]A person with mass m1 = 51 kg stands at the left end of a uniform beam with mass m2 = 109 kg and a length L = 3 m. Another person with mass m3 = 66 kg stands on the far right end of the beam and holds a medicine ball with mass m4 = 13 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.

1. What is the location of the center of mass of the system?

2. The medicine ball is throw to the left end of the beam (and caught). What is the location of the center of mass now?

3.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?)

4.To return the medicine ball to the other person, both people walk to the center of the beam. At what x-position do they end up?

## Homework Equations

xcm = 1/MtotΣ mi*ri

## The Attempt at a Solution

I solved number one by using the formula, I did (51(0) + 109(1.5) + 66(3) + 13(3))/(51 + 109 + 66 + 13) = 1.68.

For the second part there is no external force so the center of mass stays the same at 1.68. It is only at #3 where I seem to go wrong.

I thought that you had to calculate the shift by using the same equation by adjusting it to fit the new situation. So I did 1.68 - ((51(0) + 109(1.5) + 66(3) + 13(0))/(51 + 66 + 109 + 13)) = 1.68 - 1.512 = 0.167 however this answer is wrong.

Number 4 requires number 3, so I need to figure this question out first.

Please help

Thank you

Nobody wants to help ? this is due very soon. Than you for your time though.

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