00:01
Hi there, so for this problem, we are given this information that is shown in the figure.
00:07
Now let's label the magnitude, the force f1 as just 312, and the force f2 in magnitude is 147 pounds.
00:20
Now with that said, we need to obtain the resultant force, this for part 8.
00:25
Now the resultant force in vector form is the sum of these two four.
00:31
Forces but in a vector form.
00:35
We know that the force f2 is only directed and the x ereption so in component that will be just simply 312 .0.
00:50
Now the force f1 is given in components.
00:59
So the first component is the magnitude that we're given for that that is 147, this times the cosine of the angle that is 20 .6 degrees.
01:09
Okay, and the other component is 147, and this sign of 20 .6 degrees.
01:21
So with that, if we add this to, the net force is just simply 312 plus 142 ,000, 25 times the cosine of 20 .6 degrees.
01:42
And the other component is just simply 145 times the sign of 20 .6 degrees.
01:51
So that's the solution for part a of this problem.
01:56
Now for part b, the question is about its magnitude.
02:03
Now the magnitude, well, let me just first simplify the expression that we obtained from before, so that will be 312 plus 147 times the cosine of 20 .6...