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23 January, 04:55

A meter stick is supported by a pivot at its center of mass. Assume that the meter stick is uniform and that the center of mass is at the 50 cm mark. a) If a massm1 = 80 g is suspended at the 30 cm mark, at which cm mark would a mass m2 = 110 g need to be suspended for the system to be in equilibrium?

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  1. 23 January, 05:51
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    64.5 cm

    Explanation:

    30 * 80 + x * 110 = 50 * 190 = > x = 64.5
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