Good morning. We have already defined the magnitude of torque as equal to
r F sine theta where F is the magnitude of the force causing the torque,
r is the magnitude of the position vector from the axis of rotation to where the force acts on
the object, theta is the angle between the direction of the force and the direction of
the position vector r. And we can find the direction of the torque using the right-hand
rule. Today we learn about the equation for torque which uses the cross product.
Cross product? Yeah, we learned the cross product last year?
I think it was two years ago. Whenever it was, I never thought we’d actually
use the cross product for anything. ♪ Flipping Physics ♪
Does the cross product use matrices? Yep, that’s the one.
Cool! That means the cross product is actually useful!
Yep. The cross product is very useful. Torque, as a vector, equals the cross product of r,
the position vector from the axis of rotation to the point where the force acts on the object,
and F, the force causing the torque. Please realize the cross product is not commutative,
in other words, torque equals the cross product of the position vector and the force, however,
torque does not equal the cross product of the force and the position vector.
But A cross B equals the negative of B cross A. Right mr.p?
Yes Billy, that is correct. Now, typically it has been at least one orbit of the Earth around
the Sun since you last did the cross product, so let’s do a general example to review how the
cross product works. Bobby, please read the problem and Billy, please translate the problem.
There are no words to read, just letters and numbers.
(Right. But, you can still read it.) Okay I guess I’ll say, When vector A equals
negative i plus j minus 2 k and vector B equals 2 i minus 3 j plus 4 k. Determine A cross B.
I don’t really think there is anything to translate. Is there mr.p?
(I can agree with that Billy. Please start solving the problem?)
Sure. Okay, I know the solution involves a 3 by 3 matrix, however,
I do not remember how to set that up. It’s the determinant form of the matrix.
The first row has the unit vectors i, j, and k. And the next two rows have the coefficients of
the position and force vectors. So, negative 1, 1, negative 2, and 2, negative 3, and 4.
And that equals 3 smaller determinant matrices. We start with a 2 by 2 determinant matrix multiplied
by unit vector i. And in order to figure out what goes in that matrix we cross out the row
and column that the unit vector i is in in the original 3 by 3 determinant matrix. So, the first
2 by 2 determinant matrix equals 1, negative 2, and negative 3, 4. And then we add the next 2 by
2 determinant matrix times unit vector j. Crossing out the row and column with unit vector j in it in
the original 3 by 3 determinant matrix leaves us with negative 1, negative 2, and 2, 4. And then
Actually, remember we need to subtract every other 2 by 2 matrix, so it is a negative in front of the
2 by 2 matrix that goes with the j unit vector. Thanks Bobby. And then we have plus
a 2 by 2 determinant matrix times unit vector k. Crossing out the row and column with unit vector k
leaves us with negative 1, 1, and 2, negative 3. (Well done so far. Bo, please continue from here.)
From here we have a quantity multiplied by unit vector i. And that is 1 times 4
minus negative 2 times negative 3. Minus the quantity negative 1 times 4
minus negative 2 times 2 all times unit vector j. Plus the quantity negative 1 times negative 3
minus 1 times 2 all times unit vector k. 1 times 4 minus negative 2 times negative 3
equals negative 2. Negative 1 times 4 minus negative 2 times 2 equals zero. And negative
1 times negative 3 minus 1 times 2 equals 1. So, A cross B equals negative 2 i plus k.
Do we really need to do the step with all the 2 by 2 matrices? That step seems a bit excessive.
Actually Bo, I agree. I do think it was useful to show the 2 by 2 determinant matrices step
this time while we were reviewing how to use the cross product, however, in the future,
I do not think it is necessary to show that particular step. Okay, now let’s do
a simple cross product torque example to see how this works. Bobby, please work on this one.
Sure. Well, for example 2, the position vector equals unit vector
i meters. That means the position vector is 1 meter in the positive x-direction.
And the force vector equals unit vector j newtons. So, there is an object with an axis of rotation
and a force of 1 newton is being applied in the positive y-direction 1 meter in the positive
x-direction from that axis of rotation. And we are solving for the torque caused by that force
on the object which equals the cross product of the position vector and the force vector. Again,
we set up the 3 by 3 determinant matrix with unit vectors i, j, and k in the first row.
And then, well the position vector is 1 meter in the x-direction only, so 1, zero, zero. And then
the next row, for the force is zero, 1, zero because the force is 1 newton in the positive
y-direction only. Luckily, we do not have to write down all of the 2 by 2 matrices. So, we have,
in front of the i unit vector, zero times zero minus zero times 1. And then subtracting the j
unit vector quantity which equals 1 times zero minus zero times zero all times unit vector j.
And then plus 1 times 1 minus zero times zero all times unit vector k.
That works out to be well, just unit vector k. (I forgot units.)
Yep. The units for the
1 times the 1 in front of the k unit vector are meters and newtons. So, meter newtons. Actually,
we usually call them newton meters. Correct. Now, remember that the cross
product represents the area of the parallelogram created by the position and force vectors.
But that is a square not a parallelogram. A square is a special case of a parallelogram
where all four sides have equal length, and all four corners are 90 degrees. So,
every square is a parallelogram. Right.
And if the force were at an angle that was not 90 degrees, the area would be a parallelogram
but not a square. Yeah. Got it.
But, torque is a vector and this torque is in the k direction. Can area have direction?
Billy, yes, area can have direction. The direction of an area is always normal to the plane of the
area. Which means, for the area of this cross product torque, the area is either into the
screen or out of the screen. (Uh, screen?)
Because the positive i direction is the positive x-direction, and the positive j direction is the
positive y-direction, the positive k direction, the direction of this area, is in the positive
z-direction which is out of the screen. (What screen?)
To confirm that, let’s use the right-hand rule. Step one of the right-hand rule is class?
Don’t be too cool for the right-hand rule. And step two?
Limber up! Bo,
please use the right-hand rule for this example. Starting with my right hand at the axis of
rotation, point my fingers along the position vector towards the force,
curl my fingers, in this case 90 degrees, in the direction of the force, and my thumb
points out of the screen which is positive. I will point out, that if you would like to
do this with your paper flat on your desk, the fingers of your right hand point to the
right in the direction of the r position vector, your fingers curls 90 degrees in
the direction of the force, and your thumb points up or out of the page.
Up is positive, therefore, out of the screen is also positive and up is the positive k direction.
That is exactly what I just did. Correct, that was actually my point.
Now, we could also have used the magnitude torque equation. Magnitude of the position
vector times the magnitude of the force vector times the sine of the angle between the two.
That equals 1 times 1 times the sine of 90 degrees which equals 1 newton meter. And then,
using the right-hand rule, as we just did, we get the direction out of the screen, or the positive k
direction. Both the cross product torque equation, and the magnitude torque equation combined with
the right-hand rule give us the same value for torque. Billy, please do the next example.
Absolutely. Example number 3. The r position vector is 1 meter in the i direction.
The force vector is 1 newton in the negative j direction. This is the same as example number
2 only the direction of the force is changed by 180 degrees, right mr.p?
(Correct Billy.) Then we just add a negative in front of the 1
for the j direction in the force. And that cascades all the way through the problem.
And in the end we just get the torque to be 1 newton meter in the negative k direction.
(And also please use the magnitude torque equation and right-hand rule.)
Sure. The magnitude of the torque equals the magnitude of the position vector times
the magnitude of the force times the sine of the angle between those two directions. So that equals
1 times negative 1 times, actually, no, sorry it’s positive 1 because we use the magnitudes of
the position and force vectors in this equation. times the sine of 90 degrees or 1 newton meter.
With (the paper flat on my desk, starting with the fingers of my right hand at the axis of rotation,
my fingers point to the right in the direction of the position vector, curl my fingers
90 degrees in the direction of the force, and my thumb points down or into the page
which is the negative k direction. So, we get the same answer, this torque has an area) of 1
newton meter and is directed in the negative k direction which is down or into the page.
Very nice Billy. Bo, please do one last example. And notice that, in the illustration,
I have offset the force arrow down slightly so that it does not overlap the position arrow.
Example 4. The position vector is 1 meter in the positive x-direction. The force vector is 1 newton
in the negative x-direction. We are determining the torque caused by that force on the object.
Again, this is similar to the 2 examples we did before, only the direction of the force has been
changed. We have the same 3 by 3 determinant matrix only the force row is now negative 1,
zero, zero. That means torque equals the quantity zero times zero minus zero times
zero all times i minus the quantity 1 times zero minus zero times negative 1 all times j
plus the quantity 1 times zero minus zero times negative 1 all times k. Uh
that equals zero. That makes sense. The force is pointed directly at the axis of rotation
and therefore the force will not be able to cause any rotation of the object, so torque is zero.
(And using the magnitude torque equation?) That would equal r F sine theta or well,
actually the magnitudes of the position vector and the force vector are irrelevant because the
angle between those two vectors is 180 degrees and the sine of 180 degrees equals zero. We get
the same result. The torque equals zero. Thank you Bo. Now, I want to take a moment
to go back to the first example. Let’s now call it example 1 b. Let’s replace vector
A with a position vector in meters. And let’s replace vector B with a force vector in newtons.
Where both of those vectors have the same values as when we originally did the example.
You can see we can find the torque in this situation exactly as we had before. However,
this is a three-dimensional problem where the magnitudes and directions are not straightforward.
In other words, if we were to use the magnitude torque equation instead of the cross product
torque equation in this example, we would need to use the Pythagorean theorem in three dimensions to
determine the magnitudes of both vectors, and then we would have to determine the angle between those
two three-dimensional vectors and I do not know how to do that.
But then we would also have to somehow use the right-hand rule to approximate
the direction of the torque. All that sounds really complicated. However, because we used
the cross-product torque equation, the solution is rather simple, and we get an exact direction
for the direction of the torque. So, please, get used to using the cross-product equation
for torque. Thank you very much for learning with me today, I enjoy learning with you.