Let the tension on the left wire be T_{L} and tension in the right wire be T_{R}

Balancing the forces in the horizontal direction gives:

$\begin{array}{rcl}\begin{array}{rc}{\mathbf{T}}_{\mathbf{R}}\mathbf{c}\mathbf{o}\mathbf{s}{\mathbf{\varphi}}_{\mathbf{2}}& \mathbf{-}{\mathbf{T}}_{\mathbf{L}}\mathbf{c}\mathbf{o}\mathbf{s}{\mathbf{\varphi}}_{\mathbf{1}}\end{array}& \mathbf{=}& \mathbf{0}\\ {\mathbf{T}}_{\mathbf{R}}& \mathbf{=}& \frac{{\mathbf{T}}_{\mathbf{L}}\mathbf{c}\mathbf{o}\mathbf{s}{\mathbf{\varphi}}_{\mathbf{1}}}{\mathbf{c}\mathbf{o}\mathbf{s}{\mathbf{\varphi}}_{\mathbf{2}}}\end{array}$

A nonuniform, horizontal bar of mass m is supported by two massless wires against gravity. The left wire makes an Φ_{1 }with the horizontal, and the right wire makes an Φ_{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, Φ_{1}, and Φ_{2}.

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