How Do You Calculate the Center of Mass of a Nonuniform Bar Suspended by Wires?

[PSEYE]

Homework Statement

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=?

Homework Equations

-F₁ · sin(φ₁) + F₂ · sin(φ₂) = 0
___________________________________
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)

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.

 

[tiny-tim]

take moments …

x = L / ( (tan(φ₂)/tan(φ₁) + 1)

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]!

 

[PhanthomJay]

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

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