The two vectors provided are F1 and F2 which have a
magnitude of 34 N and 65 N resp. and which form an angle of 240 degrees and 25 degrees
resp. with the positive x- axis.
Now we split the two
vectors into their x and y components.
We see that F1x =
-34* sin 30 and F1y = -34 * cos 30.
For the vector F2, F2x
= 65 *cos 25 and F2y = 65*sin 25.
Let the resultant vector
be denoted as F.
Fx = 65*cos 25 - 34*sin 30 =
41.91
Fy = 65*sin 25 - 34*cos 30 =
-1.97
The magnitude of the resultant vector is sqrt
(41.91^2 + 1.97^2)
=> 41.95
N
The direction of the resultant vector is arc tan (
41.91/(-1.97)) = -87.3 degrees.
Therefore the required
resultant vector of the given vectors has a magnitude of 41.95 N and an angle of -87.3
degrees to the positive x-axis.
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