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# Physics question?

Consider three force vectors F~1, F~2, and F~3

with magnitude 68 N and direction 250◦

; magnitude 57 N and direction 313◦

; and magnitude 71 N and direction 113◦

, respectively. All direction angles 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 kF~ k, where F~ = F~1 + F~2 + F~3 ?Draw the vectors to scale on a graph to determine the

answer.

Answer in units of N. Your answer must be within ± 5.0%

### 2 Answers

- oldschoolLv 79 months ago
Sum the horizontal components:

68*cos(250) + 57*cos(313) + 71*cos(113) = -12.1

Sum the vertical components:

68*sin(250) + 57*sin(313) + 71*sin(113) = -40.2

12.1²+40.2² = 42²

42N at arctan(-40.2/-12.1) = 253.2°

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- oubaasLv 79 months ago
Rx = -71*sin 23-68*sin 20+57*sin 43 = -12.13 N

Ry = 71*cos 23-68*cos20-57*cos 43 = -40.23

R = √12.13^2+40.23^2 = 42.0 N

heading = 270-arctan Rx/Ry = 270-16.8 = 253.2°

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