Answer to Question #345719 in Physics for claire

Question #345719

A uniform meterstick is supported at the 50 cm mark. A 25N weight is placed at the 10 cm mark. Where should a 200N weight be placed for the meterstick to be in equilibrium?

1
Expert's answer
2022-05-28T08:22:32-0400

The equilibrium condition

"M_1=M_2"

"F_1d_1=F_2d_2"

"d_2=\\frac{F_1}{F_2}d_1=\\rm\\frac{25\\: N}{200\\: N}*40\\: cm=5\\: cm"

Answer: the second weight should be placed at 55 cm mark.


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