# 4 see-saws in a balancing act

1. Oct 11, 2014

### cmcd

1. The problem statement, all variables and given/known data
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?
2. Relevant equations

$$\sum{V_{forces}}=0$$

$$\sum{M_{pivotPoint}}=0$$

3. The attempt at a solution

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

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

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?

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Last edited: Oct 11, 2014
2. Oct 11, 2014

### haruspex

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

3. Oct 11, 2014

### 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?

Last edited: Oct 11, 2014
4. Oct 11, 2014

### haruspex

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

5. Oct 12, 2014

### cmcd

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

Last edited: Oct 12, 2014
6. Oct 12, 2014

### haruspex

Looks right.

7. Oct 12, 2014