Two forces, F 1 and F 2, act at a point, as shown in the picture. (Figure 1) F 1
ID: 1586523 • Letter: T
Question
Two forces, F 1 and F 2, act at a point, as shown in the picture. (Figure 1) F 1 has a magnitude of 8.80 N and is directed at an angle of = 65.0 above the negative x axis in the second quadrant. F 2 has a magnitude of 6.80 N and is directed at an angle of = 53.8 below the negative x axis in the third quadrant.
a- What is the x component Fx of the resultant force?
b- What is the y component Fy of the resultant force?
c- What is the magnitude F of the resultant force?
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.
Explanation / Answer
What is the x component Fx of the resultant force?
Since both forces are in the same overall, x-direction (negative), just solve by (remember to include the negative signs):
Fx = -cos(65.0)F1 + -cos(53.8)F2
Fx = -0.422(8.80N) + -0.590(6.80N)
Fx = -7.72N
What is the y component Fy of the resultant force?
Same concept as Part A. Except that the first force is going up, the second is going down. Solve by:
Fy = sin(65)F1 – sin(53.8)F2
Fy = 7.27 – 4.93
Fy = 2.48N
What is the magnitude F of the resultant force?
Since we already found the resultant x and y components, we can just use these to solve for the resultant magnitude:
F = sqrt(F^2x + F^2y)
F = sqrt(7.72^2+2.48^2)
F = 8.10N
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.
Since we have all 3 sides to a triangle (x and y components, plus the magnitude/hypotenuse), we can easily solve using trigonometric ratios. For example, using tangent (opposite / adjacent):
tan() = Fy / Fx
tan() = 2.48 / -7.72
tan() = -0.321
= tan-1(-0..321)
= 17.79°
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