Consider three force vectors ~ F1, ~ F2, and ~ F3
with magnitude 43 N and direction 53�; mag-
nitude 94 N and direction 162�; and magni-
tude 59 N and direction 288�, respectively. All
direction angles are measured from the posi-
tive x axis: counter-clockwise for � > 0 and
clockwise for � < 0.
What is the magnitude of the resultant vec-
tor k~F k, where ~F = ~ F1 + ~ F2 + ~ F3 ? Draw the
vectors to scale on a graph to determine the
answer.
Answer in units of N Your answer must be
within ± 5.0%
To find the magnitude of the resultant vector k~F k, we need to add the three force vectors ~F1, ~F2, and ~F3, and then find the magnitude of the resultant vector.
First, let's find the components of each force vector using the given magnitude and direction:
For ~F1 with a magnitude of 43 N and direction 53 degrees:
Fx1 = 43 N * cos(53°)
Fy1 = 43 N * sin(53°)
For ~F2 with a magnitude of 94 N and direction 162 degrees:
Fx2 = 94 N * cos(162°)
Fy2 = 94 N * sin(162°)
For ~F3 with a magnitude of 59 N and direction 288 degrees:
Fx3 = 59 N * cos(288°)
Fy3 = 59 N * sin(288°)
Next, let's calculate the sum of the x-components and y-components of the force vectors:
Fxtotal = Fx1 + Fx2 + Fx3
Fytotal = Fy1 + Fy2 + Fy3
Now, we can find the magnitude of the resultant vector using the formula:
k~F k = sqrt(Fxtotal^2 + Fytotal^2)
Finally, let's calculate the magnitude of the resultant vector:
k~F k = sqrt((Fxtotal)^2 + (Fytotal)^2)
Now, we can plug in the values:
k~F k = sqrt((Fxtotal)^2 + (Fytotal)^2)
After performing the calculations, we can find the magnitude of the resultant vector.