00:01
For this problem, we are told that a 1 ,500 pound car is stopped in traffic on a hill that is at an incline of 10 degrees.
00:09
We're as to determine the force that is required to keep the car from rolling down the hill.
00:14
And we are to round our answer to the nearest 10th.
00:17
So if we think about the situation here, let's pretend our car is just a dot here.
00:24
So we know that the incline is 10%.
00:28
So, the angle that is being made by the road and the horizontal is 10 degrees.
00:37
We know that the force going on the car is pointing straight down, and that is going with a force of 1 ,500 degrees downwards, and the force that is going to be required to keep the car from rolling down the hill would be equal to, effectively, the perpendicular component, or pardon me, the parallel, it would be the negative of the parallel component of that force, since we have that there would be, or we can split this apart into being, that original vector as being a parallel and perpendicular component.
01:21
If we think about the geometry of the situation, we know that we would end up having a right angle triangle being made with the, or if we think about this, i'll draw it here, actually i should switch colors just to make it a little bit more clear.
01:40
So we can figure out what the angles here are by recognizing that we should have a right angle triangle being formed as shown there, or we'd have 10 degrees here, 90 degrees here.
01:59
So if everything, you'd have everything.
02:01
Has to add up to 180 degrees, that means that this angle, shown here, would be an angle of 80 degrees, which then in turn means that for that vector, for the, for the triangle that i have drawn there, the component -wise breakdown of the force vector, that this internal angle is going to be equal to 80 degrees, which in turn means, well, let's see here...