0:00
Hi there.
00:01
So for this problem we have two forces, f1 and f2.
00:09
Now, we are given the magnitude for f1, which is equal to 9 neutens.
00:22
And the magnitude for f2 is also given, and that is 6 neutrons.
00:33
So we are also given the information about the direction of these peters.
00:39
So let me first draw an axis.
00:45
This is the y -adxes and this in here is the x -adxes.
00:52
So the problem states that the petro of force f1 is directing 60 degrees above the x -axis in the second quadrant.
01:10
So we know that this is the first quadrant, this is the second quadrant.
01:17
So the vector lives in here.
01:21
This is the vector f1.
01:24
And the angle is with respect to the x -adxes.
01:29
So that given angle of 60 degrees, it's this one in here.
01:35
So that is the direction for the first vector.
01:39
Now for the second vector, we are given the information that its direction, is 53 .1 below the x -atis in the third quadrant.
01:51
Now, we know that this in here is the third quadrant, and the vector now lives in here, and the angle is with respect to the x -atxis, so is this angle in here 53 .1 degrees.
02:09
So this is the vector f2.
02:12
Now what we need to obtain for the first part of this problem is the ets and y components of the resultant force.
02:25
So we need to find the x component of the net force and the y component of that force.
02:36
To obtain the x component of the vector 1 of the vector force 1 we need to see first and the angle at which this is given.
02:54
Now, we can see from the picture that the angle is with respect to the horizontal for both vectors, and they both point, and they both point to the left, and if we said that as the negative x -axis, we will obtain that both of the left, that components of these spectres are going to be negative.
03:24
So we have a negative in here.
03:26
The magnitude of this vector, which is 9 neutums, and that component is given by the cosine of 60 degrees.
03:39
That's because the angle is with respect to the horizontal.
03:43
This value gives us minus 8 .50.
03:59
7 newton.
04:01
So this is the x component of the vector f1.
04:05
Now for the vector f2, the x component of that vector, again, it is negative because it is pointed to the left.
04:14
And the angle is also given with respect to the horizontal.
04:18
So this is also given by the cosine of that angle given.
04:24
So it will be the magnitude of this betr times the cosine of 5 .5.
04:29
3 .1 degrees...