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Resultants of Force Systems Example

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Find the magnitude and direction of the resultant of two concurrent forces in the plane.

Concept

The resultant of a force system is a single force that has the same effect as all forces combined. For concurrent forces, sum components: R_{x} = \sum F_{x} and R_{y} = \sum F_{y} . Then magnitude R = R_{x}^{2} + R_{y}^{2} and direction θ = arctan (R_{y} / R_{x}) (with quadrant from signs of R_{x}, R_{y}). R_{x} = \sum F_{x}, R_{y} = \sum F_{y} R = R_{x}^{2} + R_{y}^{2}, θ = arctan (\frac{R _{y}}{R _{x}}) Notation: R_{x}, R_{y} = resultant components, R = magnitude, θ = angle from positive x-axis.

Two forces act at a point as shown: F₁ = 20 kN at 30° below the horizontal (to the right), and F₂ = 15 kN vertically downward. Find: Magnitude of the resultant force. Direction of the resultant (angle from the positive x-axis).

F₁ = 20 kN at 30° below +x (down and to the right) F₂ = 15 kN downward (-y)

F_{1 x} = 20 cos (30°) = 17.32 kN, F_{1 y} = - 20 sin (30°) = - 10 kN (downward) F₂: F_{2 x} = 0, F_{2 y} = - 15 kN (downward).

R_{x} = 17.32 + 0 = 17.32 kN, R_{y} = - 10 - 15 = - 25 kN R = (17.32)^{2} + (- 25)^{2} = 925 \approx 30.4 kN θ = arctan (- 25/17.32) \approx - 55.3° (below +x)

Resultant: R \approx 30.4 kN at about 55.3° below the positive x-axis.

R_{x} = \sum F_{x}, R_{y} = \sum F_{y}; R = R_{x}^{2} + R_{y}^{2}