- #1

farleyknight

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## Homework Statement

A nonuniform bar is suspended at rest in a horizontal position by two massless cords as shown in Fig. 12-45. One cord makes the angle θ = 15.5° with the vertical; the other makes the angle φ = 58.5° with the vertical. If the length L of the bar is 6.08 m, compute the distance x from the left end of the bar to its center of mass.

http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c12/fig12_39.gif

## Homework Equations

[tex]\tau_{net} = 0[/tex]

## The Attempt at a Solution

Okay, so this chapter is mainly on static equilibrium, and so we're going to use the rotational equilibrium to determine the value of x in this problem.

There are exactly three forces acting on the bar: The force from the left cord, the force from the right cord, (both equal in magnitude), and the force of gravity. If we make the center of coordinate system the same as the center of gravity (unknown, I realize, but we can still use x), then the y-coordinate of the force of gravity falls out of our equation, and we can take the torques as:

[tex]\tau_{net} = T \sin( \bar{\theta} )(-x) + T \sin( \bar{\phi} )(L - x)[/tex]

where I've taken [tex]\bar{\theta} = 90 - \theta[/tex], the right triangle compliment of [tex]\theta[/tex]

Solving for x, we have

[tex]T \sin( \bar{\theta} ) x + T \sin( \bar{\phi} ) x = T \sin( \bar{\phi} ) L [/tex]

Pulling out T and moving our sines to the right

[tex]x = \frac{ \sin( \bar{\phi} ) L }{ \sin( \bar{\phi} ) + \sin( \bar{\theta} ) }[/tex]

Then plugging in values

[tex]x = \frac{ \sin(31.5) (6.08) }{ \sin(31.5) + \sin(74.5) } = 2.13 m[/tex]

However, this isn't correct. I imagine my mistake is probably a simple one.. since I managed to do about half the homework problems in this chapter without any real need for help. But that was about a week ago or so, and it seems the ideas flew out of my head :) So any help would be grateful.