F1 = 8.2N[125o]
F2 = 6.40N[233.4o]
A. Fx = 8.2*cos125 + 6.40*cos233.4 =
-8.519 N.
B. Fy = 8.2*sin125 + 6.40*sin233.4 =
1.579 N.
C. F^2 = -8.519^2 + 1.579^2 = 75.07
F = 8.66 N.
D. Tan Ar=Fy/Fx = 1.579/-8.519=-0.18535
Ar = -10.5o = Reference angle.
A = -10.5 + 180 = 169.5o = 10.5o Above
the negative x axis.
Two forces, F⃗ 1 and F⃗ 2, act at a point, as shown in the picture. (Figure 1) F⃗ 1 has a magnitude of 8.20N and is directed at an angle of α = 55.0∘ above the negative x axis in the second quadrant. F⃗ 2 has a magnitude of 6.40N and is directed at an angle of β = 53.4∘ below the negative x axis in the third quadrant.
a)
What is the x component Fx of the resultant force?
Express your answer in newtons.
B)
What is the y component Fy of the resultant force?
Express your answer in newtons.
C)
What is the magnitude F of the resultant force?
Express your answer in newtons.
D)
What is the angle γ that the resultant force forms with the negative x axis? In this problem, assume that positive angles are measured clockwise from the negative x axis.
Express your answer in degrees.
1 answer
1 answer