Frequency problem (static equilibrium)

[toesockshoe]

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...

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

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

 

[ehild]

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...

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

Thus, v= \sqrt{\frac{\frac{m_{b}g}{2}tan(\theta)}{\frac{0.737}{4.98}}} = 9.456

Check the value of v.

 

[toesockshoe]

Check the value of v.

bloody hell.