00:01
Find the magnitude and direction of the resultant force given the vectors below where f of f sub 1 equals 600 newtons and f sub 2 equals 900 newtons using the parallelogram method and the cartesian vector notation method given that the direction of the resultant force is measured counterclockwise from the positive x axis.
00:18
Okay so we're going to use the cartesian vector notation first to solve this problem so we have force 1 at 600 newton so that's a 600 newton force in that direction and then force 2 is a 900 newton force in that direction so we're giving this 50 degree angle so we know we have a right angle here with the x and y axis so if that's 50 degrees that means this angle here would have to be 40 degrees to sum to 90 degrees so both angles both forces make a 40 degree angle with the x axis one above one below so the resultant force if we're going to find that using cartesian vector notation we can split force 1 into its vector components so that would be a, i plus b, j, where i and j are just the unit vectors and a is the horizontal component b is the vertical component so f1 would be if we split that up the horizontal component would go that way and here is the vertical components and and we use trigonometry to find that, those lengths.
01:32
So this would be the horizontal is the magnitude of the force, which is 600 times the cosine of the angle it makes with the x -axis, so cosine of 40 degrees.
01:43
And the vertical component would be the magnitude or 600 times the sign of that angle, the sign of 40 degrees.
01:52
And this angle below the x -axis, we're actually going to call that negative 40 degrees, since it's a single.
01:58
It's rotating clockwise, so that's going to come into play later when we use that angle.
02:03
So we can now say the force one, split in the vector notation or component form, is the a value would be the horizontal component, so 600 cosine of 40 degrees i, and then plus the vertical component is 600 sign of 40 degrees j.
02:27
Okay, let's do the same thing with force.
02:28
Force two.
02:29
Force two could also be split into component form.
02:34
So force two would be if we split it up here, here's the horizontal component and here's the vertical components.
02:45
The horizontal would be the magnitude of 900 times the cosine of that angle of negative 40 degrees and the vertical component is the magnitude of 900 times the sign of negative 40 degrees.
03:00
Degrees.
03:01
So if we split that up into a component form this would be 900 cosine of negative 40 degrees i plus 900 times the sign of negative 40 degrees j.
03:22
Okay so to find the resultant force we would add the horizontal and component forms together the i to the i's and the j's the jays just like they're like terms.
03:33
So the resultant force would be add the horizontal part so that would be 600 cosine of 40 degrees plus 900 cosine of negative 40 degrees, i plus the sum of the vertical components so that would be 600 sign of 40 degrees plus 900 sign of negative 40 degrees plus 900 sign of negative 40 degrees j so if we simplify that that gives us r to b let's pull up a calculator so 600 cosine of 40 plus 900 cosine of negative 40 that gives us around to three decimal places 1 ,149 .067 1 ,149 .067 i plus 149 .067 i plus let me find the other piece.
04:46
600 times sign of 40 plus 900 times sign of negative 40.
04:56
That gives us negative 192 .836.
05:01
So negative 192 .836j.
05:09
Okay, so to find the magnitude of the resultant force in this form a i plus b j the magnitude is just the square root of a square plus b squared so the magnitude of r would be the square root of we take the a part we'll just square it plus we square the b part there add those together and square root and we get the resultant to be 1 ,165 .14 1 ,000, 1 ,000 ,000 165 .14 newton's.
05:54
So it is worth drawing this vector out probably if we actually show that the horizontal component is positive so that means it's going to the right in that direction 1 ,149 .067 that's the horizontal component and the vertical component runs downward since it's negative so downward negative negative 192 .8 six so that would be associated with the point there so the resultant is moving in that direction so we need to find the angle that that makes with the x -axis so to do that to find the angle angle is always equal to the inverse tangent of the b component over the a component so we plug that in the calculator the inverse tangents of the b component was a negative 192 .836 divided by the a component was 1 ,149 .067 and that gives us negative 9 .5 degrees.
07:04
The angle is negative 9 .5 degrees.
07:07
That's this angle here.
07:11
But it asks to give it, first to give the angle as a measure from the counterclockwise from the positive x -axis.
07:18
So if we rotate counterclockwise, what would that angle be? that would just be 360 minus 9 .5 degrees.
07:27
So the angle would be at 360 minus 9 .5 is 350 .5 degrees.
07:36
So there's the answer.
07:37
There's the magnitude and direction of that force using the cartesian vector method.
07:44
So if we do this using the parallelogram rule, which is a lot quicker.
07:49
We're going to use the parallelogram method now to do the same thing.
07:54
Thing...