>> Welcome to a video on vector applications involving force and work.
Let's go and take a look at our first example.
If a 4000 pound car comes to rest on a hill with a grade of 5 degrees, how much force is needed
to be applied so it doesn't roll back down the hill?
So the grade is 5 degrees, so let's go ahead and label this.
And since the car weighs 4000 pounds, this red vector here would be our force vector.
Or we could let X component equal zero and the Y component equal negative 4000.
Now, in relation to the ramp, this force vector has 2 components: one component that's parallel
to the ramp, another component that is perpendicular to the ramp.
So the amount of force that needs to be applied so it doesn't roll back down the hill would have
to be the force that counteracts this force here.
Well, if we take a look at this, if we project the force vector onto the ramp
that would give us the amount of force that's pushing it down the ramp,
and therefore the amount of force we need in order to stop it from rolling backwards.
So we want to project the force vector onto the ramp, so we need to come
up with a vector to represent the ramp.
Remember, if we want to project the force vector onto let's call the ramp vector V,
the magnitude of vector V does not affect the projection of F onto V. So we'll go ahead
and let the ramp be a unit vector, so we can call vector V,
where the X component would be cosine theta or cosine 5 degrees,
and the Y component would be sine 5 degrees.
Now we have all the information we need to project F onto V. And down here in the corner,
I've already taken our projection formula and replaced U with our force vector,
so now we can just go ahead and apply this formula here.
So our projection of F onto V is going to be equal to F dotted with V divided
by the magnitude of V squared times vector V. So if we dot F and V,
we're going to have zero times cosine 5 degrees plus negative 4000 times sine 5 degrees divided
by the magnitude of vector V squared.
Vector V is a unit vector and therefore the magnitude is equal to 1, 1 squared is still 1.
We now need to multiply this by vector V, which is cosine 5 degrees sine 5 degrees.
So this would be zero and negative 4000 times sine 5 degrees is going
to give us approximately negative 348.6 times our unit vector
of cosine 5 degrees sine 5 degrees.
Well, the magnitude of this projection will give us the amount of force needed
to prevent this car from rolling back down the hill.
So here we have a scalar times a unit vector, so the magnitude of our projection is going
to equal positive 348.6 and this would be pounds.
This would be the amount of force needed to keep this car from rolling backwards
down a hill with a grade of 5 degrees.
Now let's talk about work.
If F is a constant along vector V as we just saw, then the work done is given
by the magnitude of the projection of F onto V times the magnitude of V, or vector F dotted
with vector V. Now let's take a look at a word problem.
Here a warehouse cart is being pulled by exerting a force of 50 pounds on a handle,
as we see here in black, that makes a 3 degree angle.
So this angle here would be 3 degrees.
Determine the work done in pulling the cart 20 feet.
Okay, so on this problem we'll go ahead and let this represent the force vector.
And since the cart is being pulled 20 feet horizontally,
we could let this represent vector V. Let's talk about the force vector for a moment.
We know what the forms a 3 degree angle
with vector V. We also know the magnitude would have to be 50.
So we can let the force vector equal 50 times a unit vector in the same direction,
so we'd have cosine 3 degrees, sine 3 degrees.
So this would be our force vector.
And then for vector V, since it's being pulled 20 feet horizontally,
we could represent that using the vector 20 zero.
Now that we have vector F and vector V, we can go ahead and use our second formula,
and the dot product of these 2 vectors will give us the amount
of work needed to perform this task.
So the work is going to be equal to F dotted with V. For vector F, let's go ahead
and find the values of cosine 3 degrees and sine 3 degrees.
If we use a reference triangle in the first quadrant, well,
cosine 3 degrees would be square root 3 over 2.
So 50 times square root 3 over 2 would be 25 square root 3.
The sine of 3 degrees is 1/2, 50 times 1/2 would be 25.
And we're going to dot this with the vector 20 zero.
So we have 25 square root 3 times 20, that would be 500 square root 3 plus zero.
So this is the amount of work done in pulling the cart 20 feet.
Let's go ahead and get a decimal approximation for this.
I've already done that, it's 866.03.
And the units would be foot pounds.
Because the force is in pounds and the distance is in feet.
Okay, that's going to do it for this video.
I hope you found these examples helpful. ------------------------------512803428f26--