Find center of mass using rotational equilibrium

[farleyknight]

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

\tau_{net} = 0

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:

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

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

Solving for x, we have

T \sin( \bar{\theta} ) x + T \sin( \bar{\phi} ) x = T \sin( \bar{\phi} ) L

Pulling out T and moving our sines to the right

x = \frac{ \sin( \bar{\phi} ) L }{ \sin( \bar{\phi} ) + \sin( \bar{\theta} ) }

Then plugging in values

x = \frac{ \sin(31.5) (6.08) }{ \sin(31.5) + \sin(74.5) } = 2.13 m

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.

 

[rl.bhat]

In this problem T1 and T2 are not the same.
Since the rod is in equilibrium, there is no vertical motion, horizontal motion and the rotation of the rod.
Write the conditions for these three cases. Then solve for x.

 

[farleyknight]

In this problem T1 and T2 are not the same.
Since the rod is in equilibrium, there is no vertical motion, horizontal motion and the rotation of the rod.
Write the conditions for these three cases. Then solve for x.

Okay, so then we would have

F_{net,x} = - T_1 \cos(\bar{ \theta }) + T_2 \cos( \bar{ \theta } )

And also

F_{net,y} = -mg + T_1 \sin(\bar{ \theta }) + T_2 \sin(\bar{ \theta })

We create a relation for T_1 and T_2:

T_1 = T_2 \frac{ \cos(\bar{\phi}) }{ \cos(\bar{\theta}) }

We then use this relation to solave for T_1 in the second equation

mg = T_1 \sin( \bar{ \theta } ) + T_1 \frac{ \cos(\bar{\theta}) }{ \cos(\bar{\phi}) } \sin(\bar{\phi})

or more simply

mg = T_1 \sin(\bar{\theta}) + T_1 \cos( \bar{\theta} ) \tan( \bar{ \phi } )We can then find T_1 and T_2 as values of m and g

T_1 = \frac{ mg }{ \sin(\bar{\theta}) + \cos(\bar{\theta})\tan(\bar{\phi}) }

or T_1 = 11.425 m

and

T_2 = \frac{ mg \cos(\bar{\phi}) }{ \cos(\bar{\phi}) ( \sin(\bar{\theta}) + \cos(\bar{\theta})\tan(\bar{\phi})) }

or T_2 = 3.581 m

We then plug these into the equation for torque

\tau_{net} = 0 = T_1 \sin(\bar{\theta})(-x) + T_2 \sin(\bar{\phi})(L - x)

Solving for x we have

x = \frac{ T_2 \sin(\bar{\phi}) }{ T_1 \sin(\bar{\theta}) + T_2 \sin(\bar{\phi}) }

or

x = \frac{ 3.581 \sin(\bar{\phi}) }{ 11.425 \sin(\bar{\theta}) + 3.581 \sin(\bar{\phi}) }

But evaluating this gets me a pretty small number, namely x = 0.1453, which isn't correct.

 

[rl.bhat]

Your T1 calculation is wrong. You can skip this step.
From F(netx) you get
T1*cos(90 - θ) = T2*cos(90 - φ) ------(1)
Using Τ(net) you get
T1*sin(90 - θ)*x = T2*sin(90-φ)*( L - x )---(2)
Take the ratio of eq 1 and 2, and solve for x.