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1.3 · Add planar force vectors and find a resultant

Learn to add planar force vectors and find a resultant through clear examples and targeted practice.

University of Alberta ENGG 130: Engineering Mechanics: Statics

Force Vectors and Mechanics Foundations

ENGG 130 study topic 1.3

A force resultant is a single force vector that has the same combined force effect as a given set of forces. To find it, represent every force using the same perpendicular axes, add the components, and convert the resulting components back to a magnitude and direction. If forces act at different points on a rigid body, their positions also matter: the force resultant alone may not represent the same turning effect. In that case, compare moments about a chosen point as well. This lesson uses planar statics only. A force is described by its magnitude, direction, and point of application; units and sign conventions are stated throughout.

What you will learn

  • Resolve a force into horizontal and vertical components using its magnitude and direction.
  • Add planar force vectors by summing their components.
  • Find a resultant's magnitude and direction, and check that it represents the original forces.
  • For forces applied at different points, compare the resultant's moment with the moments of the original forces.

1. Represent each force with components

Start by identifying the system: the particle or rigid body to which the listed forces act. Draw and label the forces when their directions or application points could be confusing. Choose horizontal xx and vertical yy axes, and use the same axes for every force. A common sign convention is positive to the right and upward.
An angled force can be replaced in calculations by two perpendicular components. If its magnitude is FF and its angle θ\theta is measured counterclockwise from the positive xx-axis, the components are Fx=Fcos⁡θF_x=F\cos\theta and Fy=Fsin⁡θF_y=F\sin\theta. The signs follow from the direction: a leftward or downward component is negative. For example, an angle in the second quadrant has a negative horizontal component and a positive vertical component.
This component representation follows from right-triangle trigonometry. It does not change the force; it expresses the same vector in a form that can be added coordinate by coordinate. Check the stated angle reference before using sine and cosine, since an angle measured from the vertical axis requires a different setup.
Fx=Fcos⁡θ,Fy=Fsin⁡θF_x=F\cos\theta,\quad F_y=F\sin\theta
  • Use one shared coordinate system for all forces.
  • Include component signs; do not treat every component as positive.
  • Angles in this lesson are measured counterclockwise from positive horizontal unless stated otherwise.

2. Add components and recover the resultant

The resultant R\mathbf R is the vector sum of the forces. Add all horizontal components to obtain RxR_x, and add all vertical components to obtain RyR_y. Then use the Pythagorean theorem to find the magnitude. The direction is obtained from the component ratio, with the quadrant checked against the signs of RxR_x and RyR_y.
A calculator's inverse tangent can return an angle in the wrong quadrant if it is used without checking signs. For instance, a vector with negative horizontal and positive vertical components points into the second quadrant. State the direction relative to a clear reference, such as counterclockwise from positive xx.
The units of a force resultant are force units, such as newtons or kilonewtons. Do not add magnitudes directly unless the vectors point along the same line and direction. For perpendicular or angled forces, component addition preserves both direction and size.
Rx=∑iFix,Ry=∑iFiy,R=Rx2+Ry2R_x=\sum_i F_{ix},\quad R_y=\sum_i F_{iy},\quad R=\sqrt{R_x^2+R_y^2}
  • Add horizontal components together and vertical components together.
  • Use component signs to identify the resultant's quadrant.
  • The resultant magnitude is not generally the sum of the force magnitudes.

3. Check what the resultant represents

For forces concurrent at one point, the resultant acts through that point. The component sums directly show that the original forces and resultant have equal total force. Their moments about the common point are all zero, so the moment comparison is also satisfied.
For forces applied at different points on a rigid body, force addition still gives the net force, but the points of application affect the turning effect. Choose a reference point OO and calculate each force's moment about it. In a plane, a counterclockwise moment is commonly taken as positive and a clockwise moment as negative. A force perpendicular to a position vector has moment magnitude equal to force times perpendicular distance.
A single force can represent a nonconcurrent force system only if its line of action gives both the same net force and the same moment about the reference point. Some systems also have a remaining couple; this lesson's worked nonconcurrent example has a nonzero force resultant whose line of action can be located. Checking force and moment prevents mistaking a matching net force for a fully equivalent replacement.
MO=∑i(ri×Fi)zM_O=\sum_i (\mathbf r_i\times\mathbf F_i)_z
  • Force-resultant components must equal the sums of the original components.
  • Forces applied at different points require a moment comparison as well.
  • Keep the chosen moment sign convention consistent.

Worked example

Two perpendicular forces

A small ring is acted on by a 30 N force to the right and a 40 N force upward, both through the ring. Find the resultant's magnitude and direction.
  1. Set axes and components
    Take right and upward as positive. The two forces already lie along the axes, so their components are F1x=30 NF_{1x}=30\,\text{N}, F1y=0F_{1y}=0, F2x=0F_{2x}=0, and F2y=40 NF_{2y}=40\,\text{N}.
    Rx=30 N,Ry=40 NR_x=30\,\text{N},\quad R_y=40\,\text{N}
  2. Find magnitude and direction
    Use the component vector as a right triangle. Both components are positive, so the resultant points into the first quadrant.
    R=302+402=50 N,θ=tan⁡−1(40/30)=53.1∘R=\sqrt{30^2+40^2}=50\,\text{N},\quad \theta=\tan^{-1}(40/30)=53.1^\circ
  3. Verify the force and moment
    The resultant has the same horizontal and vertical components as the two original forces. Because all forces act through the ring, their moments about that point and the resultant's moment there are zero.
    ∑Fx=30 N=Rx,∑Fy=40 N=Ry,∑MO=0\sum F_x=30\,\text{N}=R_x,\quad \sum F_y=40\,\text{N}=R_y,\quad \sum M_O=0
Answer: The resultant is 50 N50\,\text{N} at 53.1∘53.1^\circ counterclockwise from the positive xx-axis.
Check: Resolving the answer gives 50cos⁡53.1∘≈30 N50\cos 53.1^\circ\approx30\,\text{N} and 50sin⁡53.1∘≈40 N50\sin 53.1^\circ\approx40\,\text{N}.

Worked example

Angled forces at one point

A particle is acted on by a 80 N force at 30∘30^\circ, a 60 N force at 150∘150^\circ, and a 40 N force downward. Find the resultant.
  1. Resolve each force
    Use the stated angle convention. The 60 N force is in quadrant two, so its horizontal component is negative. The downward force has a negative vertical component.
    F1=(69.28,40.00) N,F2=(−51.96,30.00) N,F3=(0,−40) N\mathbf F_1=(69.28,40.00)\,\text{N},\quad \mathbf F_2=(-51.96,30.00)\,\text{N},\quad \mathbf F_3=(0,-40)\,\text{N}
  2. Add components
    Sum the horizontal and vertical components separately. The remaining vertical component is positive, so the resultant is in the first quadrant.
    Rx=17.32 N,Ry=30.00 NR_x=17.32\,\text{N},\quad R_y=30.00\,\text{N}
  3. Find and verify the resultant
    Calculate the magnitude and direction from the summed components. The force check reproduces both component sums; all forces pass through the same point, so the net moment about that point is zero.
    R=17.322+302=34.64 N,θ=60.0∘R=\sqrt{17.32^2+30^2}=34.64\,\text{N},\quad \theta=60.0^\circ
Answer: The resultant is approximately 34.64 N34.64\,\text{N} at 60.0∘60.0^\circ counterclockwise from positive xx.
Check: The resultant components are approximately (17.32,30.00) N(17.32,30.00)\,\text{N}, matching the sums. The moment about the common point is zero for both the original forces and resultant.

Worked example

Resultant for forces at different points

A horizontal beam is 4.0 m long. A 100 N upward force acts 1.0 m from its left end, and a 150 N downward force acts 3.0 m from its left end. Find the force resultant and its line of action.
  1. Add the force components
    Take right and upward as positive. There are no horizontal forces. The net vertical force is the upward force minus the downward force.
    Rx=0,Ry=100−150=−50 NR_x=0,\quad R_y=100-150=-50\,\text{N}
  2. Sum moments about the left end
    Take counterclockwise as positive. The upward force at 1.0 m produces a positive moment; the downward force at 3.0 m produces a negative moment.
    MO=(100)(1.0)−(150)(3.0)=−350 N⋅mM_O=(100)(1.0)-(150)(3.0)=-350\,\text{N}\cdot\text{m}
  3. Locate the resultant
    The resultant is downward. Let its distance from the left end be xRx_R. Its moment must match the original moment, so a downward force at that location produces the same clockwise moment.
    (−50)xR=−350 N⋅m,xR=7.0 m(-50)x_R=-350\,\text{N}\cdot\text{m},\quad x_R=7.0\,\text{m}
  4. Verify force and moment
    The resultant's vertical component equals the total vertical force, and its moment about the left end equals the sum of the original moments. Its line of action lies beyond the beam's right end; this is valid for the equivalent resultant.
    ∑Fy=−50 N=Ry,MO=(−50)(7.0)=−350 N⋅m\sum F_y=-50\,\text{N}=R_y,\quad M_O=(-50)(7.0)=-350\,\text{N}\cdot\text{m}
Answer: The force resultant is 50 N50\,\text{N} downward, acting on a vertical line 7.0 m7.0\,\text{m} from the left end.
Check: The original forces give a net force of −50 N-50\,\text{N} and a moment of −350 N⋅m-350\,\text{N}\cdot\text{m} about the left end. The located resultant gives the same two quantities.

Common mistakes and how to avoid them

Adding force magnitudes directly even when forces point in different directions.
Correction: Resolve forces into signed components, add those components, and then calculate the resultant magnitude.
Giving every component a positive sign.
Correction: Use the force direction and chosen axes to assign signs before summing.
Reporting an inverse-tangent angle without checking the quadrant.
Correction: Check the signs of both resultant components and state the direction with a clear reference axis.
Assuming the net force alone makes a single force equivalent when the forces act at different points.
Correction: Also compare moments about a reference point and locate the resultant's line of action when possible.

Lesson summary

  • Resolve each planar force into components using the same axes and angle convention.
  • Add components to obtain the resultant vector, then find its magnitude and direction.
  • Verify the resultant by comparing force components; for forces at different points, compare moments too.
  • A negative component or moment indicates a direction opposite to the chosen positive direction.

Check your understanding

Question 1

A force has components Rx=−6 NR_x=-6\,\text{N} and Ry=8 NR_y=8\,\text{N}. What is its magnitude and quadrant?
  1. 10 N10\,\text{N}, quadrant II
  2. 14 N14\,\text{N}, quadrant II
  3. 10 N10\,\text{N}, quadrant IV
  4. 2 N2\,\text{N}, quadrant I
Show answer and explanation
10 N10\,\text{N}, quadrant II
The magnitude is (−6)2+82=10 N\sqrt{(-6)^2+8^2}=10\,\text{N}. A negative horizontal and positive vertical component place the vector in quadrant II.

Question 2

Two forces act at one point: 20 N20\,\text{N} right and 20 N20\,\text{N} left. What is their resultant?
  1. 40 N40\,\text{N} right
  2. 20 N20\,\text{N} left
  3. 0 N0\,\text{N}
  4. 40 N40\,\text{N} upward
Show answer and explanation
0 N0\,\text{N}
The horizontal components sum to 20−20=0 N20-20=0\,\text{N}, and both vertical components are zero.

Key terms

Component
The signed part of a vector along one chosen coordinate axis.
Resultant
A single vector equal to the vector sum of a specified set of vectors.
Moment
A measure of a force's turning effect about a chosen point; in planar statics its unit is force times distance.
Line of action
The straight line along which a force acts.

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Published by DoAssignment. This AI-assisted lesson follows University of Alberta ENGG 130: Engineering Mechanics: Statics, study topic 1.3. It is a study resource, not an official curriculum publication.

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