Static equilibrium of diving board

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SUMMARY

The discussion focuses on solving a static equilibrium problem involving a uniform diving board with a mass of 27 kg and a length of 4.3 m, supported by two beams at points A (0 m) and B (1.1 m). A diver weighing 68 kg stands at the end of the board. The net force equation is established as ForceNet = 0 = md*g + mb*g - FA - FB, while the torque equation is TorqueNet = md*g*l + mb*g*(l/2) - FA*(0) - FB*1.1. The correct approach to find the forces on supports A and B involves summing torques about different pivot points.

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barz1
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Homework Statement


A uniform diving board of mass=27kg and length x=4.3m is supported by two beams. beam A at x=0m and beam B at x=1.1m. A 68kg diver stands at the other side of the board.
A. Find force on support A.
B. Find force on support B.


Homework Equations


Net force = 0
Net torque = 0
md=68kg -mass of diver
mb=27kg - mass of board
l = 4.3m - length of board


The Attempt at a Solution


I am just having a lot of trouble figuring out the forces and toques for these static equilibrium problems. So I set my ForceNet = 0 = md*g + mb*g - FA - FB. (Not sure if this is correct)
TorqueNet = md*g*l + mb*g*(l/2) - FA*(0)? - FB*1.1
So I guess I don't understand what to use as my pivot point because this torque equation gets rid of my FA all together.

Thanks!
 
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barz1 said:

The Attempt at a Solution


I am just having a lot of trouble figuring out the forces and toques for these static equilibrium problems. So I set my ForceNet = 0 = md*g + mb*g - FA - FB. (Not sure if this is correct)

Yep that is correct.

barz1 said:
TorqueNet = md*g*l + mb*g*(l/2) - FA*(0)? - FB*1.1
So I guess I don't understand what to use as my pivot point because this torque equation gets rid of my FA all together.

What you did was take the sum of the torques about the point A, so this will give you FB

Remember, the sum of the torques about any point is zero.

So if you want FA, you can sum the torques about point B.
 

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