- #1

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.