A barbell is loaded with two 20 kg plates on its right side and two 20 kg plates on its left side. The barbell is 2.2 m long, and its unloaded mass is 20 kg. The two 20 kg plates on the right side are locked in place at 35 cm and 40 cm from the right end of the bar. The two 20 kg plates on the left side have slipped. One is 30 cm from the left end of the bar, and the other is 20 cm from the left end of the bar. Where is the center of gravity of this barbell with the four plates?

[tex]x_g=\frac{(20\ kg)(20\ cm)+(20\ kg)(30\ cm)+(20\ kg)(110\ cm)+(20\ kg)(180\ cm)+(20\ kg)(185\ cm)}{20\ kg\ +\ 20\ kg\ +\ 20\ kg\ +\ 20\ kg\ +\ 20\ kg} \\\\x_g=\frac{(20\ kg) (525\ cm)}{100\ kg} \\\\x_g=105\ cm[/tex]

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sqdancefan sqdancefan · Answer 2 · 2024-02-29T18:36:47+01:00

Answer:

105 cm from the left, or 5 cm left of center

Explanation:

You want the center of gravity of a barbell system with a 2.2 m bar and 4 plates arranged on the bar, each element having a mass of 20 kg. Two plates are 20 and 30 cm from the left end, and two plates are 35 and 40 cm from the right end.

Center of mass

The center of mass is the weighted average of the positions of each of the masses. Since each mass is the same as the others, it will simply be the average of their locations.

The reference location from which positions are measured can be anywhere. For the purpose here, we can choose the left end of the bar.

Average

The two masses on the right end are 220 -{40, 35) = {180, 185} cm from the left end. So the average position of the 5 masses is ...

(20 +30 +110 +180 +185)/5 = 105 . . . . cm from the left end

The center of gravity of the barbell and four plates is 105 cm from the left end, or 5 cm left of center.