You have 8 coins. 1 coin weighs less than the others. You can do 2 weighings on scales like in the gif...

You have 8 coins. 1 coin weighs less than the others. You can do 2 weighings on scales like in the gif. How do you find the coin which weighs less?

Wouldn't you need three weighings?

4 coins in each
Get the lighter pile
Put 2 in each
Get the lighter pile
Put 1 in each
The lighter pIle is the coin

Each weighing can give three results (left heavier, right heavier, equal weight) so you have to put a third of the coins on each side.

With my hands.

That's just a statement user

Weigh 3 coins on each side, if they balance,t he two missing coins contain the lighter coin.
If not, the side with less has it, which means you can weigh one on each side. The one if even, the one taken from the pile is the lighter. If uneven, the lighter coin is the lighter coin.

it was stated that you can only do 3 weighings

Here's an interesting one. You have 8 coins, and you know that one of the coins has a defect that makes it either lighter or heavier, but you do not know which. Can you come up with a scheme that lets you identify the defect coin, as well as if it is lighter or heavier with 3 weighings?

set two aside and weigh 3v3, if it's balanced then you know the lighter coin is in your removed pile and if it's not balanced then you can do the 1v1 with the 3 coins on the scale to reveal the lighter coin