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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.

Respuesta :

Answer:

34.70 N

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~ = [tex]$ \sqrt{3.8^2 + 34.5^2}$[/tex]

          = 34.70 N

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