i suck

A nonuniform, horizontal bar of mass m is supported by two massless wires against gravity. The left wire makes an angle phi_1 with the horizontal, and the right wire makes an angle phi_2. The bar has length L.

Find the position of the center of mass of the bar, x, measured from the bar's left end.

Express the center of mass in terms of L, phi_1, and phi_2.



i haet physics...how the fuck does this work and what's the answer?

halp me

Draw a picture and I'll do it. I'm too hungover to understand what you're trying to say.

Not the OP but wow I suck. I am a second year physics major and I can't tell the answer. Forgot all about center of mass stuffs.

Sounds like some trigonometry problem. You've got 2 angles and an adjacent side and you need to find a portion of that side(length to center of gravity/total length). I failed physics in high school, so I'm surprised I could even offer that much towards your solution. Sorry!

>>1
Was a diagram included or is it purely a word problem?  If it is the latter, then I don't believe there is enough information to solve the problem.

Let's call the forces from the left and right wires onto the bar A and B, and let's rename the angles a and b.

The bar is not moving or accelerating, so the forces must cancel. For the horizontal components, this means: A cos a = B cos b, or A/B = cos b / cos a.

Since there's also no rotational acceleration, moments must cancel (I don't really remember this part very well from high school, so I may be wrong): x A sin a = (L-x) B sin b, or (A/B) sin a / sin b = (L-x)/x = L/x - 1.

Combining those results, we get:
(cos b / cos a) * (sin a / sin b) = L/x - 1
tan a / tan b + 1 = L/x
x = L / (tan a / tan b + 1)

I get x = L / [ 1 + (sin(phi_1)*cos(phi_2))/(sin(phi_2)*cos(phi_1))].  Seems to work, annoying to explain.  My only extra assumption is that the wires attach to the ends of the bar.

x= L*tan(phi_2)/(tan(phi_1)+tan(phi_2))

>>1
...ball-sack