Question

Consider three force vectors F~ 1 with magnitude 43 N and direction 38◦ , F~ 2 with magnitude 26 N and direction −140◦ , and F~ 3 with magnitude 27 N and direction 110◦ . 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 F of the net force vector F~ = F~ 1 + F~ 2 + F~ 3? Answer in units of N.

Answer 1

34.70 N

Step-by-step explanation:

Given :

F~ 1 = 43 N in direction 38◦

F~ 2 = 26 N in direction −140◦

F~ 3 = 27 N in direction 110◦

Therefore,

F~x = 43 cos (38) + 26 cos (-140) + 27 cos (110)

= 43 (0.7) + 26 (-0.7) + 27 (-0.3)

= 3.8

F~y = 43 sin (38) + 26 sin (-140) + 27 sin (110)

= 43 (0.6) + 26 (-0.6) + 27 (0.9)

= 34.5

so, F~ =
`$ √(3.8^2 + 34.5^2)$`

= 34.70 N