# A Bar Suspended by Two Wires

1. Aug 26, 2008

### PSEYE

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

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

What is the position of the center of mass of the bar, measured as distance from the bar's left end?

x=?

2. Relevant equations

-F₁ · sin(φ₁) + F₂ · sin(φ₂) = 0
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F₁x = -F₁ · sin(φ₁)
F₁y = F₁ · cos(φ₁)
F₂x = F₂ · sin(φ₂)
F₂y = F₂ · cos(φ₂)

forces in terms of magnitude and angle
x = L / ( (F₁/F₂)·(cos(φ₁)/cos(φ₂)) + 1)

3. The attempt at a solution

x = L / ( (tan(φ₂)/tan(φ₁) + 1)

this seems right, but I'm repeatedly getting it wrong no matter how I input the answer.

2. Aug 26, 2008

### tiny-tim

take moments …

Hi PSEYE!

Sorry, but I've no idea how you got that result.

You have three unknown forces, F1 F2 and W, which you don't want in the final equation.

The only ways I know to eliminate unknown forces are:

i] take components perpendicular to them … which isn't going to work in this case … or

ii] take moments about a point through their line of action.

Try ii]!

3. Aug 26, 2008

### PhanthomJay

If phi_1 and phi_2 are the angles with the horizontal, you've got your sines and cosines mixed up.