Flyball governor and force on slider

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SUMMARY

The flyball governor regulates the speed of steam turbines by adjusting the steam flow based on the rotational speed of the shaft. Given parameters include a length (L) of 300 mm, an angle (gamma) of 45 degrees, a mass (m) of 1 kg for each ball, and a rotational speed (N) of 1452 RPM. The centripetal force acting on the balls can be calculated using the formula F = mω²r, where ω is derived from N. To find the force on the slider A, one must project the forces acting on the system using trigonometric functions.

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http://www.mech.uq.edu.au/courses/mech2210/yat/q/governor.jpg

A flyball governor is a device used to regulate the speed of steam turbines in steam power plants. The rotation of the shaft causes the two balls move outward. As the balls move out, they pull on the bearing A. The position of the slider A is linked to a valve admitting steam into the turbine. This way, the flow of speed is automatically restricted when the shaft starts rotating too fast.

The following data is given for the flyball governor shown in the figure:

L =300 mm
gama= 45 degrees
m = 1 kg (the mass of each ball)
N = 1452 RPM
What is the force (in N) pulling on the slider A if the shaft rotates at the speed given below? Ignore friction and the weights of the conmponents other than the two balls.

I have NO idea where to start with this one! Everything I do just seems to lead in circles and not get me anywhere...Any help would be much appreciated...
 
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In the inertial coordinate system where the flywheel is at rest, there is a centripetal force acting on the balls:

F = m\omega^2 r = m \left(\frac{2\pi}{N}\right)^2 r

which is directed radially away from the axis of rotation, and r is the radius. Use trigonometry to split this force into components directed along the supporting beams and then combine them to find the net resulting force on the bearing. Don't forget to use SI units in your calculation.
 
I'm not sure how to split the force and recombine it. Do you mind stepping me through it?

Thanks
 
Let the tension along L be T. Then, projecting

2T\sin \gamma = F\, , \quad 2T = \frac{F}{\sin \gamma}

The factor 2 is because there are two L's. Further, project the forces vertically to find the force on A:

F_A = 2T \cos \gamma = \frac{F}{\tan \gamma}
 
HI, I found a similar qns and were asked to
a)draw the free body diagram of the ball and sliding collar
b)calculate y which is 2Lcos(gama) in this case, as a function of omega, that is the angular speed.
c)the min speed of rotation for th eball to "fly" (gama > 0)

anyone has any idea/clues on this qns?
thanks.
 

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