Answer to Question #257604 in Electric Circuits for Jayson

Question

Two masses of 50.0 grams and 100 grams are placed on a 200-gram uniform meter stick. The 50 g mass is placed on the 10.0 cm mark while the 100 g is at 80.0 cm mark. Locate the center of gravity of the system of weights including the meterstick.

Answer 1

We know that

$$50\times(70-10)+200(70-x)=100(80-70)$$

$$50\times60+200\times(70-x)=100\times10$$

$3000-1000=-200(70-x)$

$x=60cm$