Question

Consider three force vectors F~1 with magnitude 55 N and direction 110◦, F~2 with magnitude 32 N and direction −150◦, and F~3 with magnitude 24 N and direction 28◦. All direction angles θ are measured from the positive
x-axis: counter-clockwise for θ > 0 and clockwise for θ < 0.
What is the magnitude of the resultant vector F~ = F~1 + F~2 + F~3 ?
Answer in units of N.

Answer 1

F1 20 degrees west of north
F2 30 degrees south of west
F3 28 degrees north of east
x direction (East) = -55 sin 20 - 32 cos 30 + 24 cos 28
= - 55 * .342 - 32* .866 + 24 * .883
= -18.8 -27.7+21.2 = -25.3
y direction (North) = 55 cos 20 -32 sin 30 + 24 sin 28
= 51.7 - 16 + 11.3 = 47
so
| F | = sqrt ( 25.3^2 + 47^2)