Answer to Question #214331 in Physics for frisha

Question #214331

Masses of 50g, 100g, and 200g are placed at the 30 cm, 60 cm, and 80 cm mark, respectively, of a meterstick. If the mass of the meterstick is 300g, locate the center of gravity of the system.

Expert's answer

"50(50-30)+300(50-x)\\\\=100(60-50)+200(80-50)\\\\x=30\\ cm"

Learn more about our help with Assignments: Physics

Comments

No comments. Be the first!

Leave a comment

Related Questions

1. The center of gravity of a non-uniform bamboo pole of length 3m is 1m from its larger end and its we
2. A uniform ladder weighing 100N leans against a smooth vertical wall with its lower end resting on a
3. A uniform beam 3m long weighs 100N. Loads of 50 N and 150 N are placed on the beam at points which a
4. A uniform bar of weight 40 N is 4 m long. Weights of 60N and 100N are placed on the bar at points wh
5. A non-uniform bar 2 meters long is supported by means of a string tied at its midpoint. To keep the
6. The center of gravity of a non-uniform bamboo pole of length 3m is 1m from its larger end and its we
7. A uniform bar weighing 60 N has a length of 4 m and is supported at its end by strings. A load W=200

Who Can Help Me with My Assignment

There are three certainties in this world: Death, Taxes and Homework Assignments. No matter where you study, and no matter…
How to Finish Assignments When You Can’t

Crunch time is coming, deadlines need to be met, essays need to be submitted, and tests should be studied for.…
How to Effectively Study for a Math Test

Numbers and figures are an essential part of our world, necessary for almost everything we do every day. As important…