"A uniform 1-meter stick that is supported by a fulcrum at the 25-cm mark is in equilibrium when a 1-kg rock is suspended at the 0 cm end. Is the mass of the meter stick greater than, equal to, or less than the mass of the rock?"

I thought the answer would be that the mass of the whole meter stick must be greater than the rock, since 3/4 of the meter stick is balancing the rock plus the remaining 1/4 of the meter stick. (rock + 1/4 meter stick = 3/4 meter stick)

But my professor's answer is: "Equal. The center of
mass of the meter stick is also 25 cm from the fulcrum (as is the rock). For these torques to balance the weights must be equal."

Can someone please explain this to me? Maybe plugging in some actual numbers would help.

Thank you!

Solution Preview

See the attached figure...
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<br>a) The center of mass of a uniform meter stick is at 0.5 m from one end. That is we can assume the mass of the rod to be concentrated at the 0.5 m point (a point ...

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