Two Forces Acting at a Point

Part A = -7.96N

Part B = 2.34N

Part C = 8.29N

Part D = 16.4 Degrees

**Solutions Below:**

_{1}and F

_{2}, act at a point, as shown in the picture. F

_{1}has a magnitude of 8.40N and is directed at an angle of α = 60.0° above the negative x axis in the second quadrant. F

_{2}has a magnitude of 6.20 N and is directed at an angle of β = 52.7° below the negative x axis in the third quadrant.

Part A

_{x}of the resultant force?

Since both forces are in the same overall, x-direction (negative), just solve by (remember to include the negative signs):

F_{x} = -cos(60)F_{1} + -cos(52.7)F_{2}

F_{x} = -0.5(8.40N) + -0.606(6.20N)

F_{x} = -7.96N

Part B

_{y}of the resultant force?

Same concept as Part A. Except that the first force is going up, the second is going down. Solve by:

F_{y} = sin(60)F_{1} – sin(52.7)F_{2}

F_{y} = 7.27 – 4.93

F_{y} = 2.34N

Part C

Since we already found the resultant x and y components, we can just use these to solve for the resultant magnitude:

F = sqrt(F^{2}_{x} + F^{2}_{y})

F = sqrt(-7.96^{2} + 2.34^{2})

F = sqrt(68.84)

F = 8.29N

Part D

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(γ) = F_{y} / F_{x}

tan(γ) = 2.34 / -7.96

tan(γ) = -0.294

γ = tan^{-1}(-0.294)

γ = 16.4°