0:00
Hi there.
00:01
So for this problem, we are told that three astronauts propelled by jet backpads push and guide a 137 kilograms asteroid.
00:11
So the mass of the asteroid is given 137 kilograms.
00:16
And toward a processing dock, asserting the forces that is shown in this figure.
00:22
So the magnitude of the force f1 is given and that magnitude is equal to 31 newtoms.
00:28
The magnitude of the force f2 is also given, and that force is equal to 56 newtoms.
00:35
And the magnitude of the force f3 is also given, and that is equal to 45 newtoms.
00:42
The angle theta 1 is also given, and that angle is equal to 30 degrees.
00:48
And the angle theta 3 is also given, and that is equal to 60 degrees.
00:55
So with that said, for part a of this problem, we are asked about what is the magnitude of the asteroid's acceleration.
01:06
So to obtain that, the first thing that we need to do is to add together these three forces by using newton's second law.
01:15
So we note that the sum of all of these forces will give us the net force.
01:22
And so that will be the force f1, the force f2, and the force f3.
01:28
And then this should be equal to the mass times the acceleration.
01:33
So solving for the acceleration in a vector form, so that will be one divided by the mass times the sum of these three forces.
01:41
So what we need to do first is to write these three forces in a vector form, so we can add them together, and then we can obtain the magnitude of that acceleration vector.
01:59
So let's start by writing the force f1 in a pectoral form.
02:03
So that will be just simply 31 newtons.
02:09
And as you can see, this force is in the first quadrant.
02:13
So both components should be positive.
02:18
So the x company, because the angle is given with respect to the horizontal, is the cosine of the angle t -t -1, which is 30 degrees.
02:27
This in the x direction and the sign of 30 degrees in the y direction.
02:38
So with that said, we just simply now simplify this term in here.
02:45
So using our calculator for the x component, so that will be 31 times cosine of 30 degrees.
02:52
So that will be 206 .85 in the x component.
03:00
And the y company is 31 times sign of 30 degrees.
03:05
So that will be 15 .5 in the y direction.
03:12
Now for the force f2, we do something similar, but in this case, the force f2 is just along the x direction, as you can see from the figure.
03:23
So it's just simply its magnitude that we're given that this is newton in the x direction.
03:30
And for the force f3, as you can see from the figure, this force is in the fourth quadrant.
03:37
So the x component is positive, but the y component is negative because it is pointing down more.
03:45
So we will have the magnitude of this force f3, which is 45 newtons.
03:52
This times the cosine of the angle, the angle that we are given.
03:56
So the x component is positive, as i said.
03:59
And that is 60 degrees, and this in the edge eruption, minus the sign of 60 degrees in the y direction.
04:11
So now we simplify this term in here, so that will be 45 times the cosine of 60 degrees, so that will give us the value of 22 .5 in the x direction, and for the y direction, we will have 45 times the sign of 60 degrees, so that will give us a value of 38 .97 in the y direction.
04:38
So now we just add together these three forces...