The magnitude and direction of two forces acting on an object are 80 pounds, S63°E, and 50 pounds, N67°E, respectively.

Question

Find the magnitude, to the nearest hundredth of a pound, and the direction angle, to the nearest tenth of a degree, of the resultant force. And find the direction angle in degrees. (Please solve and show the steps thank you!)

Answer 1

You can do this by breaking down each of the vectors into their component x and y parts and then add them together which will give you the x and y components of the resultant vector. The magnitude of this vector will be the hypotenuse of the triangle formed.

S63E means you rotate 63 degrees east from due south which would put you at 27 degrees south of horizontal which is -27degrees

The x and y components are x=80cosine(-27degrees) =-71.25082

y=80sin(27degrees)=36.31923

N67E means you rotate 63 degree east from due north which puts you at 23 degrees from the horizontal

The x and y components are x=50cosine(23degrees)=46.02524

y=50sin(23degrees) = 19.53655

Adding these together gives will give you the cx and y components of your resultant vector which would be

y =-16.78267 x= 117.30575 Draw a right triangle from the origin to calculate the hypotenuse which the resultant vector using pythagorean theorem and you get 118.50

To get the angle use tan(x) = opposite(-16.78267) / adjacent(117.30575)

tan x = .14306

To find x take inverse tan of both sides and you get x=-8.14 degrees. which would be S 81.86 E. Rounding to nearest tenth gives you S 81.9 E