Welcome to a video on torque
and application of cross products.
You've probably heard of a torque wrench,
which is a wrench that tightens bolts or screws
to a certain specification.
Well, in physics, the cross product is used
to measure torque, or twisting effect.
Torque is also called the moment of a force.
If a force F is applied to a terminal point
on a vector R, the torque is given by
the cross product of vector R and vector F.
So this tells us the torque vector
would be perpendicular, or orthogonal,
to both vector R and vector F.
The magnitude of the torque measures the tendency
of vector R to rotate, where theta is the angle
between vector R and vector F.
So there's two ways to determine
the magnitude of the torque.
We can determine the magnitude
of the cross product of R and F,
or we can find the product of
the magnitude of R, magnitude of F, and sine theta,
again where theta is the angle between the two vectors.
So we're going to take a look at one problem
and solve it using two methods,
one using this formula here,
and a second method using this formula here.
If you apply 40 pounds of force at the end
of a one foot wrench at an angle of 110 degrees,
what is the torque produced?
So we want to find the magnitude of the torque,
and for this method we're going to find
the magnitude of R crossed with F.
Let's go ahead and make a diagram
to represent this situation.
Now, even though we know we have to be in R three
to cross two vectors, we're going to represent this in two D
and just let the Z component equal zero.
The terminal point of vector R is going to be
the terminal point of this one foot wrench,
so we'll let R equal the unit vector
along the negative Y axis.
So we're going to let R equal zero, negative one,
and zero for the Z component.
Next, the 40 pounds of pressure is applied
at an angle of 110 degrees from this terminal point.
So if we start to measure the angle from here,
we go over to 90 plus 20 more.
This is where we would apply the force of 40 pounds.
This is 110 degrees.
But if we want to cross the two vectors,
we do have to have it in component form,
so we're going to move this vector so
the initial point is at the origin.
Looks something like this.
So if this is 110 degrees, so is this.
Since this is 90, this would be 20.
That's helpful because that means that vector F
would be equal to 40, since that's its magnitude,
times the unit vector in the same direction,
which would be cosine 20 degrees, sine 20 degrees,
and again we're going to let the Z component equal zero.
Let's go ahead and perform this scalar multiplication here.
So vector F is going to be equal to 40 times cosine 20 degrees,
that's approximately 37.6,
40 times sine 20 degrees is
approximately 13.7, and the Z component is zero.
Okay, so now what we have to do is cross R and F,
and then find the magnitude.
Well, R crossed with F is going to be equal to
the three by three determinant,
where the first row is going to be I, J, K.
The second row will be the components from vector R,
zero, negative one, zero.
The third row will be the components from vector F.
And now we need to evaluate this determinant.
I'm going to go ahead and use the diagonal method
for this one, so we'll have I, J, K
and then I and J again.
The second row is going to be zero, negative one,
zero, zero, negative one.
And the third row is going to be 37.6,
13.7, zero,
37.6, and 13.7.
So now we'll find the product of these diagonals,
one, two, three,
then find the product of these diagonals.
We'll take the sum of the blue products
and subtract the sum of the green products.
Let's go ahead and do that.
There's going to be a lot of zeroes here.
This product is zero, this product is zero,
and so is this one.
Zero minus, and now we'll take a look at the green products.
This will be negative 37.6K.
And then we have a product of zero here,
and a product of zero here.
So the torque vector is equal to 37.6K.
Well, since K is a unit vector, the magnitude
of this vector is going to be 37.6.
So we'd have 37.6 pound feet of torque.
And this would be counterclockwise based upon
the right hand rule.
Now let's go ahead and solve the same problem
using a different method.
Again, we already know that vector R
has components zero, negative one, zero,
and the force vector was 40 times
cosine 20 degrees, sine 20 degrees.
This problem is actually a lot easier to solve
using this formula here.
Let's see if we can do that.
Let's try it.
The magnitude of the torque is going to be equal to
the magnitude of vector R.
Well, vector R is a unit vector, so this would be
one times the magnitude of the force vector,
well this is a unit vector here, so the magnitude
would be 40, and then remember theta is the angle
between the two vectors, it's given as 110 degrees.
So one times 40 times the sine of 110 degrees
should give us the same value that we obtained before.
Let's go and check on our calculator.
Let's make sure our calculator's in degree mode, it is.
So we have 40 times sine 110 degrees.
As you can see, the result is the same.
It's approximately 37.6.
And again the units here would be pound feet.
So again, this is the amount of torque
being applied in this situation.
So as you can see, setting up and solving it this way
was much easier and much faster,
but either way the result is obviously the same.
So even though this is an example of an application
of cross products, we can see it's much easier
to solve using this second formula here,
even though the result is obviously the same.
I hope you found this example helpful.