# Need help with Center of Mass problem

1. Mar 15, 2006

### frankfjf

Alright, here's the problem:

In Fig. 9-74, two identical containers of sugar are connected by a cord that passes over a frictionless pulley. The cord and pulley have negligible mass, each container and its sugar together have a mass of 540 g, the centers of the containers are separated by 52 mm, and the containers are held fixed at the same height. What is the horizontal distance between the center of container 1 and the center of mass of the two-container system (a) initially and (b) after 15 g of sugar is transferred from container 1 to container 2? After the transfer and after the containers are released, (c) what is the acceleration of the center of mass, taking up to be the positive direction?

Part a I get, it's 26mm, but for some reason I keep getting part b and c wrong.

I attempted to use the formula and ended up with:

Xcom = ((525g)(-26mm) + (555g)(26mm)) / (525g + 555g) = 0.72

But the quiz I'm taking returns that as a wrong answer.

For part c I used:

Acom = ((9.8)(555g - 525g)^2) / (525g + 555g)^2 = .008

But part c also returns as wrong.

What am I not seeing? Thanks!

2. Mar 15, 2006

### Staff: Mentor

You are measuring position from the midpoint between the two containers. Instead, measure from the center of container 1.

3. Mar 15, 2006

### frankfjf

I don't understand, how would I need to change my equation to change the place I'm measuring from?

4. Mar 15, 2006

### frankfjf

Oh wait, nevermind. I understand now. If I add the 0.72 to the X-coordinate of the lighter container I obtain the true answer.

That solves b, but then, what am I doing wrong in c?

5. Mar 15, 2006

### Staff: Mentor

Be sure you are using the correct sign and the right units.