Consider two identical metersticks tied together with a loose piece of string at

Question

Consider two identical metersticks tied together with a loose piece of string attached to their ends. If you hold one stick horizontal, the other dangles vertically. You place your finger at an appropriate location beneath the horizontal stick so the two sticks will balance.
-Where along the horizontal stick should you place your finger so the metersticks will balance?
- If you did the same with one horizontal meterstick and two vertical metersticks dangling from its one end, where would you place your finger to balance the system?



I started the first question somewhat but I'm not sure if it's on the right track or not.
- X(center of mass) = (m1x1+m2x2+m3x3)/ (m1+m2+m3) = 1/4m

Explanation / Answer

Treat each object like its center of mass The top center of mass would fall at point .5m, as would the bottom one. suppose a mass of c for each stick, any mass would work, since they are the same for both sticks. formula: (m1*x1+m2*x2)/(m1+m2) where m is a mass and x is the distance from the end of the stick to the center of mass (cx1+cx2)/(2c) c(x1+x2)/(2c) (x1+x2)/2 (.5+.5)/2=.5 The center will be at .5m

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