physics

the problem reads this "A person with mass m1 = 62 kg stands at the left end of a uniform beam with mass m2 = 93 kg and a length L = 3.5 m. Another person with mass m3 = 65 kg stands on the far right end of the beam and holds a medicine ball with mass m4 = 10 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.
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?) "
I know we have to do something like 1.822826=[(60+100d]+93(1.75+d)+65(3.5+d)/230
(btw 1.822826 is the location of center of mass)
I keep getting d=.1271928947 but it's wrong, can someone double check my work please

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  1. the error could be related to significant figures
    ... just because the calculator shows them, doesn't mean they are there

    all of the data in the problem is 2 sig fig
    ... except for the mass of the ball, which is one sig fig

    assuming the ball is exactly 10 kg
    ... the best the answer can be is two sig fig
    ... yours looks like 10 ... like a calculator display

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  2. I also tried cutting it down to .13 but that was wrong, the website that I use sometimes wants us to be 99%accurate so it'll want more numbers than needed
    I think this might be a calculation error on my part

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  3. total mass = 231
    original cg = [93*1.75 + 75*3.5 ]/231 = 1.841
    new cg = [93*1.75 + 65*3.5]/231 =1.689
    so the rod moved 0.152

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  4. thank you damon, I was using the wrong numbers :)

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  5. You are welcome.

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