# Find the x coordinate of the center of mass of the system of three chocolate blocks

1. Oct 21, 2007

### Heat

1. The problem statement, all variables and given/known data

Three odd-shaped blocks of chocolate have the following masses and center-of-mass coordinates:
(1) 0.310 kg, ( 0.200 m, 0.310 m);
(2) 0.410 kg, ( 0.110 m, -0.380 m);
(3) 0.200 kg, ( -0.280 m, 0.610 m).

Find the x coordinate of the center of mass of the system of three chocolate blocks.

3. The attempt at a solution

This is how I commenced,

I took (mass1*x1+mass2*x2+mass3*x3)/(m1+m2+m3) and got .051

this should be right but I got it wrong. I then redid the calculation and this time I got .05
not that much difference, I only got one try left, and don't think this is right because they are so close together.

I did the y the same way and got it right. :S

Please and thank you.

Last edited: Oct 21, 2007
2. Oct 21, 2007

### arildno

Well, then you have put in the wrong numbers somewhere (I assume that even though you have written m3*x2, that should read m3*x3)
It is impossible for us to know how you clicked about on your calculator.

You should, at least, calculate the following:
$$\frac{0.310*0.200+0.410*0.110-0.2*0.28}{0.92}=\frac{0.062+0.0451-0.056}{0.92}=\frac{0.0457}{0.92}$$
which ought to be about what you've found already..

Last edited: Oct 21, 2007
3. Oct 21, 2007

### Heat

for some reason, either my calculator is wack, for the numerator I keep on getting

.062+.0451-.056 = .0511

for the denominator I get .92

.0511/.92 = .055543478

edit: I tried it on windows, and I get .0511 too. :O

4. Oct 21, 2007

### princessfrost

for your x i got .055543. try that and see if it works. I did it how you had it up there and that is what I got. I did this on my homework yesterday and got it right so it should be right. :)

5. Oct 21, 2007

### Heat

yep, it's right thanks. I it just as it showed in the calculator this time without any rounding. :)

thank you princessfrost and arildno for your help.

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