Mr.p: Good morning. Bobby, please read the problem, and Bo, please translate.
"Flipping Physics" Intro
Bobby: A uniform 0.093 kg meter stick is supported at the 15 cm and 92 cm marks.
When a 0.250 kg object is placed at the 6.0 cm mark,
what are the magnitudes of the forces supporting the meter stick?
The mass of the meter stick is 0.093 kg and the mass of the object is 0.250 kg.
I suppose we should draw a picture of the meter stick supported at the 15 and 92 cm marks with the
object on the meter stick at the 6.0 cm mark. I'm not sure what to do with those distances right now.
We're solving for the forces caused by the supports, which are both normal forces.
So force normal one and two each equal question mark. That's all I got right now.
Billy: We can draw a free body diagram of all the forces acting on the meter stick.
Bo: Yeah. The force of gravity on the meterstick acts down at ... the meterstick is "uniform"
so the force of gravity acts at the center of mass of the stick, so right in the middle at the 50 cm mark.
The force of gravity on the object pushes down on the meterstick at the 6 cm mark.
Let's say force normal one acts up on the support at the 15 cm mark and
force normal 2 acts up on the support at the 92 cm mark.
Mr.p: This is a good start. Billy, what should we do from here?
Billy: Well, the net force in the y-direction on the meter stick equals the force of gravity acting on the object,
which is negative because it is down, plus force normal number 1, which is positive because it's up,
minus the force of gravity acting on the meterstick, minus because the force of gravity is down,
plus force normal number 2 which is also positive because it is up.
The net force in the y-direction acting on the meterstick also equals
the mass of the meterstick times the acceleration in the y-direction of the meterstick.
However, the meterstick is at rest, so its acceleration in the y-direction equals zero.
Oh, it is in translation equilibrium
and we can substitute mass times acceleration due to gravity for both forces of gravity.
But we don't know either force normal, so this is one equation with two unknowns,
which we cannot currently solve.
Bo: Equation holster. Billy & Bobby: Equation holster!
Mr.p: That is correct.
We cannot currently do anything more with the net force equation,
so we put it in our equation holster for when we need it later.
Now, notice the meter stick is not moving, in fact, it is also not rotating.
This means it has a zero angular velocity and zero angular acceleration.
Billy: Oh, it is in rotational equilibrium.
The meterstick is not only in translational equilibrium, it is also in rotational equilibrium!
Mr.p: That is correct, Billy. Remember,
this is a special case where the object is at rest, is not rotating, and is in what we call static equilibrium.
When an object is in static equilibrium,
the net torque acting on the object equals zero about not just one axis of rotation, but any axis of rotation.
Therefore, in this problem we can set the net torque equal to zero about any axis of rotation.
Bobby, could you please do that?
Booby: You want me to find the net torque about an axis of rotation?
But the meterstick is not rotating?
Mr.p: Correct. Therefore, you can pick any axis of rotation and the net torque will always equal zero.
Please give it a try.
Bobby: Okay. The net torque acting on a meterstick with an axis of rotation at ... one end?
I don't know does it matter where we put the axis of rotation?
Bo: Let's put the axis of rotation at the location of one of the normal forces.
Billy: Oh right, because then the torque caused by that force normal will be zero!
Bobby: Okay, let's pick force normal 1, or the 15 cm mark, as our axis of rotation.
The net torque then equals the torque caused by the force of gravity on the object
plus the torque caused by force normal 1.
plus the torque caused by the force of gravity on the meterstick plus the torque caused by force normal 2.
Billy: I thought the torque caused by force normal number 1 was zero because torque equals the r value
times force times the sine of the angle between the direction of "r" and the force.
Because the axis of rotation is where the force normal number 1 acts on the meterstick,
the "r" value for force normal number 1 is zero
which makes the torque caused by force normal number 1 equal to zero.
Bo: And shouldn't the torque caused by the force of gravity on the meterstick be negative
because it would cause the meterstick to rotate clock -wise or into the board, which is negative?
Bobby: Right.
But the torque caused by the force of gravity on the object and the torque caused by force normal 2
are both positive because they would cause the meter stick to rotate counterclockwise
or out of the board which is positive.
All that equals rotational inertia times angular acceleration,
but the angular acceleration of the meter stick is zero,
so the net torque acting on the meterstick equals zero.
Mr.p: Bo, please keep going from there.
Bo: For each torque, we can substitute in the equation for torque Billy just mentioned.
The angle in each of those torque equations is 90 degrees because each "r",
the distance from the axis of rotation to where the force is applied, is horizontal and each force is vertical.
The sine of 90 degrees equals one.
So now we have r for the object times the mass of the object times acceleration due to gravity
minus r for the stick times the mass of stick times the acceleration due to gravity
plus r for normal force 2 times normal force 2 all equal to zero.
Move everything but r 2 times normal force 2 over to the other side
and divide by r 2 to solve for normal force 2.
Now we need the distances or "r" values for the stick, the object, and normal force 2.
Mr.p: Billy, please determine those three torque "r" values.
Billy: Well, 15 cm equals 6 centimeters plus r for the object,
therefore r for the object equals 15 minus 6 or 9 cm.
15 cm plus r for the meter stick equals 50 cm, therefore r for the meterstick equals 50 minus 15 or 35 cm.
15 centimeters plus r for the force normal number 2 equals 92 cm,
therefore r for force normal number 2 equals 92 minus 15 or 77 cm.
We can now substitute all our values into the equation for which Bo just solved.
Force normal number 2 equals the quantity 35 times 0.093 times 9.81 minus 9 times 0.25 times 9.81
all divided by 77 which equals 0.12804 or 0.13 with two significant digits.
And the units are newtons.
We need to convert all the r values from cm to meters because newtons are kg meters per second squared.
So multiply all r values by one meter over 100 cm and substitute all those new r values back into the equation
and we get ... we get the exact same answer we got before. That's weird.
Bobby: That is because if you substitute all the units into the equation,
you can see the cm cancel out and we end up with kg meters per second squared, which are newtons.
So we do not actually have to convert to meters. That's weird.
Mr.p: It is a fine habit to convert your values to base SI units,
however you are correct that we actually did not need to convert the r values
to meters in this particular problem.
Okay, we have solved for force normal number 2,
anybody know how we can solve for force normal number 1?
Billy: Oh, we use the equation holster! Mr.p: That is correct Billy.
We can use the equation we previously placed in our equation holster.
Billy: May I do it? Mr.p: Sure, go ahead.
Billy: We can solve the equation for force normal number 1 and substitute in known values to get,
force normal number 1 equals 0.25 times 9.81 minus 0.12804 plus 0.093 times 9.81
which equals 3.23679 or 3.2 newtons with 2 significant digits!
Mr.p: Correct.
Now realize we also could have solved for force normal number 1
by summing the torques and picking a different axis of rotation.
This is because the meterstick is at rest and therefore in static equilibrium,
which means the net torque acting on the meterstick about any axis of rotation equals zero.
But, you know we have to check our predictions, right?
I've replaced the dominoes which were supporting the meter stick with force sensors
and you can see right now those force sensors are reading zero.
And you can see when I add the meterstick the forces go up.
Then when I add the 250 gram mass at 6 cm you can see... There it is!
I think you would agree with me that it matches our predictions and we can say: 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.