Frequency problem (static equilibrium)

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toesockshoe
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Homework Statement


The lower end of a uniform beam is attached to a vertical wall by a frictionless pivot. The beam extends away from the wall and upward, making a 62° angle with the wall, and it is held in place by a horizontal wire attached from its upper end to the wall. The wire's length and mass are 4.98-m, 0.737-kg and the beam's weight is 349-N. The speed of sound is 344 m/s. When the wind blows, the wire vibrates in its 4th overtone

Homework Equations


T=Ialpha
flambda = v

The Attempt at a Solution


First I wanted to find the tension force so I can find v...

[itex]v = \sqrt {\frac{F_{t}}{\mu}}[/itex]
Because the system is in static equilibrium I set [itex]\tau_{F_{g}} = \tau_{T}[/itex] ... [itex]\frac{m_{b}gL}{2} * sin(\theta) = TLcos(\theta)[/itex] ... [itex]T = \frac{m_{b}g}{2}tan(\theta)[/itex]

Thus, [itex]v= \sqrt{\frac{\frac{m_{b}g}{2}tan(\theta)}{\frac{0.737}{4.98}}} = 9.456[/itex]
[itex]\lambda_{5} = \frac{2L}{5}[/itex] ... solving for [itex]f* \frac{2L}{5} = 9.456[/itex], I get the frequency is 4.747... which isn't the correct answer... can someone tell me where I'm going wrong? Thanks.
 
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toesockshoe said:

Homework Statement


The lower end of a uniform beam is attached to a vertical wall by a frictionless pivot. The beam extends away from the wall and upward, making a 62° angle with the wall, and it is held in place by a horizontal wire attached from its upper end to the wall. The wire's length and mass are 4.98-m, 0.737-kg and the beam's weight is 349-N. The speed of sound is 344 m/s. When the wind blows, the wire vibrates in its 4th overtone

Homework Equations


T=Ialpha
flambda = v

The Attempt at a Solution


First I wanted to find the tension force so I can find v...

[itex]v = \sqrt {\frac{F_{t}}{\mu}}[/itex]
Because the system is in static equilibrium I set [itex]\tau_{F_{g}} = \tau_{T}[/itex] ... [itex]\frac{m_{b}gL}{2} * sin(\theta) = TLcos(\theta)[/itex] ... [itex]T = \frac{m_{b}g}{2}tan(\theta)[/itex]

Thus, [itex]v= \sqrt{\frac{\frac{m_{b}g}{2}tan(\theta)}{\frac{0.737}{4.98}}} = 9.456[/itex]
Check the value of v.
 
ehild said:
Check the value of v.

bloody hell.