Two forces, F? 1 and F? 2, act at a point, as shown in the picture. (Figure 1) F? 1 has a magnitude of 9.20 N and is directed at an angle of ? = 65.0 ? above the negative xaxis in the second quadrant. F? 2 has a magnitudeof 5.80 N and is directed at an angle of ? = 53.9 ? below the negative x axis in the third quadrant.

Part A
What is the x component Fx of the resultant force?
Express your answer in newtons.

Part B
What is the y component Fy of the resultant force?
Express your answer in newtons.

Part C
What is the magnitude F of the resultant force?
Express your answer in newtons.

Part D
What is the angle ? 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.

Explanation:

A) x-component of forces is 7.297 N

For x-component FX;

FX = F1 cos 65 + F2 cos 53.9

FX = 9.20 cos 65 + 5.8 cos 53.9

NB both forces are positive if resolved along the x-axis.

FX = 3.88 + 3.417 = 7.297 N

B) 6-component of forces is -3.654 N

For y-component FY

FY = -F1 sin 65 + F2 sin 53. 9

FY = - 9.2 sin 65 + 5.8 sin 53.9

NB F1 is negative along y-axis

FY = - 8.34 + 4.686 = - 3.654 N

C) The magnitude of the resultant force is 8.16 N

Magnitude = (FX^2 + FY^2)^0.5

(7.297^2 + (-3.654)^2)^0.5 = 8.16N

D) Angle made with negative x-axis is 26.56°

Tan^ -1 of FY/FX gives the angle of the resultant force.

= tan^-1 of -3.654/7.297

=tan^-1 of -0.5

= -26.56°

I.e 26.55° from the negative x-axis in a clockwise direction.

nah.   i'm just sleepy.

a. spokes of a wheel.

as indicated in the question radial contains rad, also found in radius, so the fracture's would span the radius.