Help on an old physics test problem?

Question

I missed this question on a test I had and I just can't figure out how to do it :( Please help me figure it out, thanks!

A water molecule consists of an oxygen atom with two hydrogen atoms bound to it as shown in the figure in the link below the problem. The bonds are 0.100 nm in length and the angle between the two bonds is 106°. Use the coordinate axis shown and determine the location of the center of gravity of the molecule. Take the mass of an oxygen atom to be 16 times the mass of hydrogen.

x = _____ nm
y = _____ nm

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Answer 1

OK. To do this, you have 3 masses at 3 points.

At 0,0, you have a 16x mass.
at (.100 nm cos 53, .100 nm sin 53) you have a mass of x (the coordinates taken from the length of the bond times the sin or cosine of the angle).
at (.100 nm cos 53, -.100 nm sin 53) you have another mass of x.

Step 1.) Determine the center of gravity along the y-axis.
Since the Oxygen is at 0,0, and the 2 hydrogen atoms are equidistant from the x-axis, y must equal 0 nm, because the y components cancel.

y = 0 nm.

Step 2.) Determine the center of gravity along the x-axis.
Oxygen = 16 at 0
The 2 hydrogen molecules are at the same x coordinate, so you can say that 2 Hydrogen are at .100 nm cos 53.
Center of Gravity(x) = (m1 * x1 + m2 * x2) / (m1 + m2)
Center of Gravity(x) = (2 * .100 nm cos 53 + 16 * 0) / 18
Center of Gravity(x) = 0.00669 nm

So the final coordinates for the center of gravity are (0.00669, 0) nm.

Answer 2

it is obvious that the middle of mass ought to lie on the x-axis, by way of fact the molecule is symmetric approximately it. First, the middle of mass of the two Hydrogen atoms: it is going to be the mid element of the line starting to be a member of them, which would be on the x-axis at one hundred cos fifty 3 (nm). So think of them the two at that element. we've got: Oxygen (mass sixteen) at x = 0 and imaginary double Hydrogen (mass 2) at one hundred cos fifty 3. the middle of mass is at x = (sixteen x 0 + 2 x 100cos fifty 3)/(sixteen + 2) = (one hundred cos fifty 3)/9 ~ 6.687 nm of direction, y = 0 nm.