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The seesaw is divided equally into $6$ parts and is already tilted to the left side with the first $2$ blocks. Assuming all the green blocks weigh the same, which side would the seesaw tilt if the $3^{\text{rd}}$ block is placed on the right edge of the seesaw, or would the seesaw not tilt at all?

Is this question even solvable? What is the proof?

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closed as off topic by Austin Mohr, Cameron Buie, Erick Wong, Arkamis, EuYu Nov 4 '12 at 2:31

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This question would be better served at Physics.SE: physics.stackexchange.com –  Austin Mohr Nov 3 '12 at 21:58
    
It seems to be identical to this question at Yahoo! answers. –  Douglas S. Stones Nov 3 '12 at 23:10
    
i posted it on yahoo also –  Truth Nov 4 '12 at 1:08

1 Answer 1

up vote 3 down vote accepted

You need to find the moment about the center hinge to decide, which side it would tilt. Let the weight of each of the green squares be $w$ and the distance between the blue pointers be $a$.

The two green squares are placed at a distance of $\dfrac{3a}2$ to the left of the hinge while one green square is placed at a distance of $3a$ to the right of the hinge.

enter image description here

The clockwise moment about the center hinge due to two green squares on the left is given by $2w \times \dfrac{3a}2 = 3aw$.

The counter-clockwise moment about the center hinge due to one green square on the right extreme is given by $w \times 3a = 3aw$.

Hence, the net clockwise moment about the hinge is $3aw - 3aw = 0$. Hence, the seesaw won't tilt.

The plots in the question and the solution were made using TikZ.

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thanks. the picture was helpful. –  Truth Nov 4 '12 at 1:08
    
But are we supposed to treat the blocks as points? This is a much more interesting problem if we don't. –  Phira Nov 4 '12 at 10:26

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