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## Measurement

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# 2D equilibrium -- balancing games

## Video transcript

I got this to balance
by carefully measuring equal distances, and
hanging equal weights at those distances. This balance is
called equilibrium. Equilibrium is when the state
of the system isn't changing. In this case, the 2D equilibrium
case, the state of the system is the position of
these objects, which still isn't changing,
unless I throw a coin in to knock it out of equilibrium. In other types of equilibrium,
the state of a system could be its
temperature, or pressure, or something else entirely. But now, I'll focus
on 2D equilibrium, the balanced state where
the positions don't change. Some things balance
in the middle, and these things tend
to be symmetrical. Other things balance off to one
side, always the heavier side. What if I want to move this
cup a different distance away from the middle and still
keep it all balanced? One way of doing this
is moving the other cup to the new distance as well. Sure enough, it balances again. But there's another
way of doing this. I could've left the right
hand cup at the old position and just taken out
some of its weight. This also balances. So we found two different ways
of balancing the same thing. We can either change the
weight or change the distance. In equilibrium, where we have
forces F1 and F2 balancing each other at
distances D1 and D2, the counterclockwise
force times distance must equal the clockwise
force times distance. Force times distance has
a special name, torque, from the Latin word to twist. In 2D equilibrium, clockwise
and counterclockwise torques are balanced. Before I calculate
some torques, I need to check some
of the masses. The mass of five
coins is 13 grams. And the mass of the cup plus
the string is also 13 grams. And since we're on
Earth, this means that they both have
the same weight, which I'll call W. Start it off with
weights 2W one cup plus 5 coins hung on each side, distance
I'll call D from the middle. And I got it balanced, and
the torques 2 times W times D, are equal on both sides. And it balances. But when I moved
the distance to 1/2D and kept the weight at
2W, five coins plus a cup, there were two ways
that I could balance it. I could either move the
other cup closer to 1/2D, or I could keep
the other cup a D and just empty it
of the five coins. So now its weight
is only W. This leaves the counterclockwise
torque, 1/2D times 2W, equal to the clockwise
torque, D times W. But what if we
move the left hand cup farther from the middle to
2D, and leave its weight at 2W. One way of balancing it
is to move the right hand cup farther out as well. Or we can move the right
hand cup back to D, and increase its weight to 4W. Let's check that this would
make the torques work out. On the left hand side, we have
2D times 2W, which is 40W. And on the right
hand side, we have D times 4W, which is also 40W. How many coins will we
need in the cup for that? One cup is 1W, and so
we need three more W, So that's 15 coins. So four W is equal to
one cup plus three more W, which we get with 15 coins. I'll move the cup to a
distance D from the middle and add a total of 15 coins,
for a total weight of 4W. Sure enough, it balances,
because in equilibrium, the counterclockwise torque has
to equal the clockwise torque.