Add concurrent 2D forces to find resultant magnitude, direction, components, and the equilibrant. Follow the vector calculations step by step.
Direct solving is free and needs no account. Have a word problem instead? Submit it as text.
Worked examples
These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.
Example 1
3-4-5: 30 N along +x and 40 N along +y
Given
forces
1.magnitude 30 Nangle deg 0
2.magnitude 40 Nangle deg 90
Assumptions
All forces are concurrent (act through one point) and coplanar; angles are measured counterclockwise from the positive x-axis.
Solution steps
Sum the components
Each force splits into x and y parts; the resultant's components are the sums.
Rx = ΣF·cosθ; Ry = ΣF·sinθ
Rx = 0.03 kN; Ry = 0.04 kN
Combine into the resultant
Magnitude by Pythagoras; direction from the two-argument arctangent (quadrant-correct).
R = √(Rx² + Ry²); θ = atan2(Ry, Rx)
R = 0.05 kN at θ = 53.13° from +x
= 0.05 kN
Results
Resultant magnitude R
0.05kN
Resultant direction (deg CCW from +x)
53.13
Resultant x-component Rx
0.03kN
Resultant y-component Ry
0.04kN
Equilibrant direction (deg CCW from +x)
-126.9
Where this answer was checked
source
Vector addition, exact Pythagorean triple — R = 50 N at 53.13°
verified by
hand-recomputed
derivation
R = √(30² + 40²) = 50; θ = atan(40/30) = 53.130°.
Example 2
symmetric three-force system cancels to zero
Given
forces
1.magnitude 100 Nangle deg 30
2.magnitude 100 Nangle deg 150
3.magnitude 100 Nangle deg 270
Assumptions
All forces are concurrent (act through one point) and coplanar; angles are measured counterclockwise from the positive x-axis.
Solution steps
Sum the components
Each force splits into x and y parts; the resultant's components are the sums.
Rx = ΣF·cosθ; Ry = ΣF·sinθ
Rx = -1.837e-17 kN; Ry = -1.421e-17 kN
Combine into the resultant
Magnitude by Pythagoras; direction from the two-argument arctangent (quadrant-correct).
R = √(Rx² + Ry²); θ = atan2(Ry, Rx)
R = 2.322e-17 kN at θ = 0° from +x
= 2.322e-17 kN
Results
Resultant magnitude R
2.322e-17kN
Resultant direction (deg CCW from +x)
0
Resultant x-component Rx
-1.837e-17kN
Resultant y-component Ry
-1.421e-17kN
Equilibrant direction (deg CCW from +x)
-180
Notes
The resultant is essentially zero — the system is already in equilibrium.
Where this answer was checked
source
120°-spaced equal forces sum to zero — equilibrium detection
verified by
hand-recomputed
derivation
100 N at 30°, 150°, 270°: Rx = 100(0.866 − 0.866 + 0) = 0; Ry = 100(0.5 + 0.5 − 1) = 0.
Example 3
US components: 300 lbf right, 400 lbf down
Given
forces
1.fx 300 lbf
2.fy -400 lbf
Assumptions
All forces are concurrent (act through one point) and coplanar; angles are measured counterclockwise from the positive x-axis.
Solution steps
Sum the components
Each force splits into x and y parts; the resultant's components are the sums.
Rx = ΣF·cosθ; Ry = ΣF·sinθ
Rx = 300 lbf; Ry = -400 lbf
Combine into the resultant
Magnitude by Pythagoras; direction from the two-argument arctangent (quadrant-correct).
R = √(Rx² + Ry²); θ = atan2(Ry, Rx)
R = 500 lbf at θ = -53.13° from +x
= 500 lbf
Results
Resultant magnitude R
0.5kip
Resultant direction (deg CCW from +x)
-53.13
Resultant x-component Rx
0.3kip
Resultant y-component Ry
-0.4kip
Equilibrant direction (deg CCW from +x)
126.9
Where this answer was checked
source
Component input path, 3-4-5 — R = 500 lbf at −53.13°