A meterstick is found to balance at the 49.7cm mark when placed on a fulcrum. Wh

Question

A meterstick is found to balance at the 49.7cm mark when placed on a fulcrum. When the 50 gram mass is attached at the 10cm mark, the fulcrum must be moved to 39.2cm mark for balance. What is the mass of the meterstick? I drew a diagram and then tried to use the center of gravity equation because it was the only equation that I had that used masses in it. What can I do from here?
A meterstick is found to balance at the 49.7cm mark when placed on a fulcrum. When the 50 gram mass is attached at the 10cm mark, the fulcrum must be moved to 39.2cm mark for balance. What is the mass of the meterstick? I drew a diagram and then tried to use the center of gravity equation because it was the only equation that I had that used masses in it. What can I do from here?

Explanation / Answer

From the first part of the question we know that the mass of therod acts not at its center as normally assumed, but at the 49.7 cmmark.

In the next part the rod is made to balance at the 39.2 cm when amass of 50 g is on the 10 cm.

So we have 50 grams acting 39.2 - 10 = 29.2 cm away from thefulcrum. Also the mass of the rod (M) is acting on the other sideat 49.7 - 39.2 = 10.5 cm away from the folcrum.

So 50 * 29.2 = M * 10.5 ===> M = 50*29.2/10.5

Mass of rod = M = 139.048 grams Please rate the answer

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