Mechanical advantage....?

Question

How can a teeter-totter be adjusted so a large child and a small child can play on it together?

Please include yr calculations THX!

Answer 1

Sum of the moments must equal 0.
F x d = Moment (Torque)
let f = Weight of small child
let F= Weight of large child
let d= distance of small child from the pivot point.
let D= distance of large child from pivot point

sum of moments is (FxD)-(fxd)=0
FxD=fxd

So imagine you have two children, one weighs twice the amount as the other. The equation would look like:
F=2f
substitution:
2fxD=fxd
D=(1/2)d
In other words the heavier child would need to sit only half the distance of the smaller childs distance to the pivot point.

Answer 2

It is very simple, initialy in idle position add some equal weights on both sides right in center. Ask large child to sit and start moving the weight of other side towards the empty seat of small child the
other side will start comming up as u move further n fuerther.now
ask small child to sit. The large child side will go up now adj. the weight towards the large child till the balance is equal.Thus both can play
together,this exercise should not take more than 15 minutes.

Answer 3

I don't think that you want to do this.

The pivot point has to be moved closer to the heavy child, which will cause the smaller child to have to go much higher in the air on his end of the lever.
When he reaches the top of his travel the smaller child will have a tendency to be launched, and when he comes down, the smaller child will have a tendency to hit the ground harder.

If the heavy child weighs twice as much as the lighter child, the distance from the seesaw pivot point to the lighter child must be twice as long as the distance to the larger child.

Answer 4

HI.

I guess you need to learn about torque.

http://library.thinkquest.org/10170/torque/index.html