Balancing Act: Forces on See-Saws in a Square Arrangement

[cmcd]

Homework Statement

A team of Mexican acrobats are developing a new act. Four see-saws are arranged in a square so that the ends of the see-saws overlap as shown in the plan view below. Each member of the team has a mass of exactly 50kg.

(a) Calculate the contact forces between the see-saws when the acrobats, under the sombreros are positioned as shown in the digram - {linked}

(b) What will happen if one member of the team over-eats and becomes heavier than the three others?
Can the other three acrobats position themselves such that the system is balanced?

Homework Equations

[tex]\sum{V_{forces}}=0[/tex]

[tex]\sum{M_{pivotPoint}}=0[/tex]

The Attempt at a Solution

[/B]
[tex]R_A=\text{ Force exerted on see-saw in question on the left}[/tex]
[tex] R_B=\text{ Force exerted on see-saw in question on the right}[/tex]
[tex] N=\text{ The reaction force due to the fulcrum in the middle of the see-saw}[/tex]
[tex] L =\text{ The length of the see-saw}[/tex]

[tex]\sum{M_{Fulcrum}} =+R_A*L/2 - 50*g*L/3 +R_B*L/2[/tex]

This is just for one of the four see-saws. For part (a) I suppose each see-saw should be the same. Is this right? Or do I have to include the forces the see-saw in question exerts on the other see-saws?

 

[haruspex]

Yes, by symmetry you can deduce the reactions between see-saws are all the same.

 

[cmcd]

Can the system be seen as a set of two x two see-saw balancing acts, where the reaction force = 0 where they meet? N must be equal to 50g Newtons right? Can the reaction forces = 0?

 

[haruspex]

Can the system be seen as a set of two x two see-saw balancing acts, where the reaction force = 0 where they meet? N must be equal to 50g Newtons right? Can the reaction forces = 0?

Sure, but by the symmetry you can reduce it to a single see-saw with equal and opposite forces applied at the ends.

 

[cmcd]

That's great thanks!
I got the reaction force = [tex]50g/3[/tex]
for part (a)

 

[haruspex]

That's great thanks!
I got the reaction force = [tex]50g/3[/tex]
for part (a)

Looks right.

 

[cmcd]

Yeah. Can you take a look at this for me, I think it has been abandoned.
https://www.physicsforums.com/threads/gravity-of-new-planet.774667/#post-4878577