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So, that should be integrated from 0 to h,
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that'll give me h^3 / 3.
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So, I get 2·M·b·h^3 / 3A·h.
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Now, you know there is one more thing we have to do.
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We must replace the area of the triangle
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by one half base times altitude, which is b·h.
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I also forgot to divide everything by the mass,
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because the center of mass is this weighted average
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divided by the total mass,
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so I've got to divide by M.
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Well, I claim, if you do this and cancel everything
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you will get the answer of (2 / 3)·h.