CivilSolve

Statics & Mechanics of Materials

Resultant of Concurrent Forces

Adds concurrent coplanar forces (each as magnitude + direction angle CCW from +x, or as signed x/y components) into a single resultant: magnitude, direction, components — plus the equilibrant (the force that would balance the system). Use for: 'find the resultant of the forces acting at point O', ring/hook/particle equilibrium setups, 'what force balances the system'. Not for: rigid-body problems with moments (beam-reactions) or trusses.

Direct solving is free and needs no account. Have a word problem instead? Submit it as text.

Inputs

Load a sample problem:
forcesThe concurrent forces (all acting through one point)
forces 1

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. 1.magnitude 30 Nangle deg 0
  2. 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

  1. 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

  2. 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. 1.magnitude 100 Nangle deg 30
  2. 2.magnitude 100 Nangle deg 150
  3. 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

  1. 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

  2. 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. 1.fx 300 lbf
  2. 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

  1. 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

  2. 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°
verified by
hand-recomputed
derivation
R = √(300² + 400²) = 500 lbf; θ = atan2(−400, 300) = −53.130°.