Mr.p: Good morning.
A typical static equilibrium problem involves a painter on a scaffold supported at two points.
Sometimes the scaffold supports are two vertical ropes from above
and sometimes the scaffold supports are two structures holding the scaffold from below.
Regardless, the way to solve the physics problem is the same.
We are going to model this using a meter stick as the scaffold,
two dominos which will support the scaffold from below,
and a 200 gram mass which will represent the painter.
"Flipping Physics" Intro
Billy, please read the problem, and Bobby, please translate.
Billy: What is the closest to the end of a 93 gram uniform meter stick you can place a 200.0 gram object
and have the system stay balanced?
The meterstick is supported at the 20.0 cm and 80.0 cm marks.
Bobby: The mass of the meterstick is 93 grams. The mass of the object is 200.0 grams.
Supports are at the 20.0 and 80.0 cm marks and we are solving for x, the location of the object.
Billy: What is a scaffold?
Bo: According to Meriam Webster,
a scaffold is "a temporary or movable platform for workers (such as bricklayers, painters, or miners)
to stand or sit on when working at a height above the floor or ground."
So it's the wooden platform in mr.p's illustrations.
Mr.p: Bo, please solve the problem.
Bo: Well, let's start with a free body diagram.
The force of gravity on the object acts down at the location of the object.
The force of gravity on the stick acts down at the center of mass of the meterstick, or the 50 centimeter mark.
There are two normal forces caused by the dominos,
both act up at their respective locations, the 20 and 80 centimeter marks.
Now that we have the free body diagram, let's sum the forces.
The net force in the y-direction acting on the meterstick equals...
Bobby: Hold up, Bo. Are you sure you want to do that?
Mr.p said last time that we don't always need to sum the forces.
Bo: Okay. Let's start by summing the torques instead.
Let's sum the torques acting on the meterstick
with the axis of rotation at the 20 cm mark, or where force normal 1 is,
because then the torque caused by force normal 1 will be zero.
Like we usually do, let's make counterclockwise or out of the board positive.
The net torque then equals the torque caused by the force of gravity acting on the object,
which is positive because it would cause the meter stick to rotate counterclockwise, or out of the board.
There is no torque caused by force normal one because that force acts directly at the axis of rotation.
The torque caused by the force of gravity of the stick
is negative because it would cause the meterstick to rotate clockwise or into the board.
The torque caused by force normal 2 is
positive because it would cause the meter stick to rotate counterclockwise or out of the board.
That all equals rotational inertia times angular acceleration. The angular acceleration is zero.
So, the meterstick is in rotational equilibrium because the net torque equals zero.
But I don't know what to do now. I mean, we don't know the value of force normal 2 or where the object is.
So, we cannot solve that one equation with two unknowns.
Mr.p: Yeah, so let's talk about the normal forces. I have replaced the 2 dominos with 2 force sensors.
This is so we can measure and understand what happens to the normal forces
as the painter moves closer to the end of the scaffold.
Notice when the painter is directly in the middle of the scaffold,
the magnitude of both normal forces is the same.
And as the painter moves toward the left end,
the normal force on the left increases and the normal force on the right decreases.
And eventually, we will get to the point where the painter is so far to the left,
that the normal force on the right is decreased down to zero.
And if the painter moves any farther to the left, the scaffold will be unbalanced and the painter will fall.
Bo: Ah. I get it. At the point we are talking about normal force 2 is reduced to zero.
So the torque caused by normal force 2 is reduced to zero. However, the meterstick is still balanced,
so its angular acceleration is still zero and it is still in rotational equilibrium.
Bobby: Right, so the torque caused by the object equals the torque caused by the meterstick.
Billy: And we can substitute r times force of gravity times sine of theta for both torques.
We know both angles are 90 degrees
because both forces of gravity are down and both r values are horizontal.
The sine of 90 degrees is 1.
We can substitute mass times acceleration due to gravity for both forces of gravity and
BBB: Everybody brought the acceleration due to gravity to the party!
Mr.p: Therefore we can solve for the r for the object, it equals 30 times 93 divided by 200 or 13.95 cm.
Notice r for the meterstick equals 30 cm because 50 cm equals 20 centimeters plus r for the meterstick.
But what is the answer to the question?
What is the closest to the end of the meterstick the painter can go
without the system becoming unbalanced?
Billy: Oh right, we solved for r for the object not x.
But 20 equals x plus r for the object, therefore x equals 20 minus r for the object or,
20 minus 13.95, or 6.05 which is 6.0 with two significant digits.
So, 6.0 cm from the end of the meterstick is how close to the end of the meter stick the object can go
before the meterstick leaves rotational equilibrium and starts to angularly accelerate.
Mr.p: And there you have it. 6.0 cm from the left end is our answer, and of course we need to test our answer.
So we are going to bring the painter so the painter is 6.0 cm
from the left end and you can see it is now balanced.
In fact, I can remove this domino because, this domino was not providing any support.
The force normal from that domino was zero. And as you can see....
the physics works, the physics works, uh huh uh huh, the physics works,
the physics works, uh huh uh huh, the physics works.
Thank you very much for learning with me today, I enjoyed learning with you.