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1.2 · Resolve planar forces into Cartesian components

Learn to resolve planar forces into cartesian components through clear examples and targeted practice.

University of Alberta ENGG 130: Engineering Mechanics: Statics

Force Vectors and Mechanics Foundations

ENGG 130 Engineering Mechanics: Statics study topic 1.2

A force can point in any direction in a plane, but many statics calculations are easier when the force is represented by a horizontal part and a vertical part. These parts are called Cartesian components. They are not additional forces: together they represent the same force. This lesson focuses on determining those components, with signs and units handled carefully.

What you will learn

  • Resolve a force in a plane into horizontal and vertical Cartesian components.
  • Choose component signs from the force direction and the selected coordinate axes.
  • Use trigonometry to calculate components when an angle is measured from either axis.
  • Check that the components reproduce the original force magnitude and direction.

1. Represent one force with two components

Consider a force acting at a point on a body. For resolving that force, isolate the force vector and use a coordinate system whose axes are horizontal and vertical. The positive horizontal direction is usually to the right, and the positive vertical direction is usually upward. State your choices before assigning signs.
The horizontal component is the projection of the force onto the horizontal axis, and the vertical component is its projection onto the vertical axis. The signed components are written as FxF_x and FyF_y. The vector is represented by F=Fxi+Fyj\mathbf{F}=F_x\mathbf{i}+F_y\mathbf{j}, where i\mathbf{i} and j\mathbf{j} indicate the positive horizontal and vertical directions.
For a force of magnitude FF whose angle θ\theta is measured counterclockwise from the positive horizontal axis, the component magnitudes follow from right-triangle trigonometry: the horizontal side is adjacent to the angle, and the vertical side is opposite it. Signs come from the direction of the force, not from the words sine or cosine.
Fx=Fcos⁡θ,Fy=Fsin⁡θF_x=F\cos\theta,\quad F_y=F\sin\theta
  • A force and its two components have the same units, such as newtons or kilonewtons.
  • With the angle measured from the positive horizontal axis, cosine gives the horizontal component and sine gives the vertical component.
  • A component pointing left or down is negative for the usual right-and-up positive axes.

2. Set signs from the geometry

Angles are meaningful only when their reference axis and direction are clear. In this lesson, a positive angle is measured counterclockwise from the positive horizontal axis. Thus, an angle in the first quadrant gives two positive components; an angle in the second gives a negative horizontal component and a positive vertical component.
A force described using an angle below the positive horizontal axis may be assigned a negative angle. For example, a force directed down and to the right has a positive horizontal component and a negative vertical component. Alternatively, calculate the magnitudes using the acute angle shown and assign signs by inspection. Either method works if used consistently.
If the angle is measured from the vertical rather than the horizontal, the roles of sine and cosine switch. The component next to the angle uses cosine; the component opposite the angle uses sine. Always identify the reference axis before writing a formula.
adjacent=Fcos⁡α,opposite=Fsin⁡α\text{adjacent}=F\cos\alpha,\quad \text{opposite}=F\sin\alpha
  • Sketch the force direction and axes before calculating.
  • Use the angle's stated reference axis; do not assume every angle is measured from horizontal.
  • Assign component signs from the arrow's direction relative to the positive axes.

3. Check a resolved force

Resolving a force changes its representation, not its magnitude or direction. If the calculated components are correct, their vector combination reconstructs the original force. A useful magnitude check is the Pythagorean relation between the two perpendicular components and the force magnitude.
You can also check direction by comparing the component signs with the sketch. For example, a force pointing up and left must have a negative horizontal component and a positive vertical component under the usual axes. A correct numerical magnitude with an incorrect sign still describes the wrong direction.
These component calculations are often used before writing force-equilibrium equations for a body. This lesson's central task is the component resolution itself: it does not require a separate equilibrium calculation or a moment balance.
F=Fx2+Fy2F=\sqrt{F_x^2+F_y^2}
  • Check that Fx2+Fy2=F2F_x^2+F_y^2=F^2 within rounding.
  • Check that the signs match the drawn direction.
  • Do not confuse the component magnitudes with their signed values.

4. A reliable calculation routine

First identify the force magnitude and how its angle is measured. Draw a short labelled force arrow with the coordinate axes. Next decide whether the angle is measured from horizontal or vertical and whether the force lies above or below, to the right or left of, that reference direction.
Write the component expressions before substituting numbers. Include the signs and units in the result. Finally, use the component-magnitude check and compare the signs with the sketch. Keeping these steps in order makes it easier to find a swapped sine and cosine or a missed negative sign.
F=Fxi+Fyj\mathbf{F}=F_x\mathbf{i}+F_y\mathbf{j}
  • Define axes, interpret the angle, resolve, then check.
  • Keep force units consistent throughout the calculation.
  • Round only after calculating the components.

Worked example

Force in the first quadrant

A cable exerts a 240 N240\,\mathrm{N} force at 35∘35^\circ counterclockwise from the positive horizontal axis. Find its Cartesian components.
  1. Choose axes
    Take right as positive xx and up as positive yy. The force lies in the first quadrant, so both components must be positive.
  2. Resolve the force
    The angle is measured from the positive horizontal axis, so cosine gives the horizontal component and sine gives the vertical component.
    Fx=240cos⁡35∘=196.6 N,Fy=240sin⁡35∘=137.7 NF_x=240\cos35^\circ=196.6\,\mathrm{N},\quad F_y=240\sin35^\circ=137.7\,\mathrm{N}
  3. Check the magnitude
    The components should reconstruct the stated force magnitude. Their squared values sum to the square of the original magnitude, apart from rounding.
    (196.6 N)2+(137.7 N)2≈240.0 N\sqrt{(196.6\,\mathrm{N})^2+(137.7\,\mathrm{N})^2}\approx240.0\,\mathrm{N}
Answer: Fx=196.6 NF_x=196.6\,\mathrm{N} to the right and Fy=137.7 NF_y=137.7\,\mathrm{N} upward.
Check: Both components are positive, as the force points right and up.

Worked example

Force in the second quadrant

A 500 N500\,\mathrm{N} force points up and left, making an angle of 28∘28^\circ above the negative horizontal axis. Find its signed Cartesian components.
  1. Read the direction
    Use right and up as positive. Since the force points left and up, its horizontal component is negative and its vertical component is positive. Relative to the positive horizontal axis, its angle is 152∘152^\circ.
    152∘=180∘−28∘152^\circ=180^\circ-28^\circ
  2. Calculate components
    Using the angle from the positive horizontal axis gives the signed components directly. Equivalently, the acute reference angle gives a leftward component with magnitude 500cos⁡28∘500\cos28^\circ and an upward component with magnitude 500sin⁡28∘500\sin28^\circ.
    Fx=500cos⁡152∘=−441.5 N,Fy=500sin⁡152∘=234.7 NF_x=500\cos152^\circ=-441.5\,\mathrm{N},\quad F_y=500\sin152^\circ=234.7\,\mathrm{N}
  3. Check
    The negative horizontal and positive vertical signs agree with the force arrow. The component magnitudes also reproduce the original force magnitude.
    (−441.5 N)2+(234.7 N)2≈500.0 N\sqrt{(-441.5\,\mathrm{N})^2+(234.7\,\mathrm{N})^2}\approx500.0\,\mathrm{N}
Answer: Fx=−441.5 NF_x=-441.5\,\mathrm{N} and Fy=234.7 NF_y=234.7\,\mathrm{N}.
Check: The force is up and left, matching the positive vertical and negative horizontal components.

Worked example

Angle measured from vertical

A 1.20 kN1.20\,\mathrm{kN} force points down and to the right. It is 40∘40^\circ from the downward vertical. Find its Cartesian components using right and up as positive.
  1. Choose the reference
    Because the stated angle is measured from vertical, the vertical component is adjacent to the angle and the horizontal component is opposite. The force points right and down, so the horizontal component is positive and the vertical component is negative.
  2. Resolve using the vertical reference
    Use cosine for the adjacent vertical magnitude and sine for the opposite horizontal magnitude. Apply the signs indicated by the force direction.
    Fx=1.20sin⁡40∘=0.771 kN,Fy=−1.20cos⁡40∘=−0.919 kNF_x=1.20\sin40^\circ=0.771\,\mathrm{kN},\quad F_y=-1.20\cos40^\circ=-0.919\,\mathrm{kN}
  3. Confirm the result
    The component signs describe a force directed right and down. Their combined magnitude returns approximately 1.20 kN1.20\,\mathrm{kN}.
    (0.771 kN)2+(−0.919 kN)2≈1.20 kN\sqrt{(0.771\,\mathrm{kN})^2+(-0.919\,\mathrm{kN})^2}\approx1.20\,\mathrm{kN}
Answer: Fx=0.771 kNF_x=0.771\,\mathrm{kN} and Fy=−0.919 kNF_y=-0.919\,\mathrm{kN}.
Check: The horizontal component is positive and the vertical component is negative, as required.

Common mistakes and how to avoid them

Using sine for the horizontal component even though the angle is measured from horizontal.
Correction: Identify the triangle side adjacent to the stated angle. For an angle from horizontal, the horizontal component uses cosine.
Reporting both component values as positive because component magnitudes cannot be negative.
Correction: Magnitudes are nonnegative, but signed Cartesian components can be negative. Use the axes and force direction to assign signs.
Treating an angle measured from vertical as though it were measured from horizontal.
Correction: Use the stated reference axis. The component adjacent to the angle uses cosine; the perpendicular component uses sine.
Checking only the component magnitudes and overlooking a direction error.
Correction: Check both the magnitude relation and whether each component sign agrees with the force sketch.

Lesson summary

  • A planar force can be represented by signed horizontal and vertical components.
  • For an angle from positive horizontal, use Fx=Fcos⁡θF_x=F\cos\theta and Fy=Fsin⁡θF_y=F\sin\theta.
  • For an angle from vertical, cosine gives the adjacent vertical magnitude and sine gives the opposite horizontal magnitude.
  • Choose component signs from the force direction and the selected positive axes.
  • Check that the components reconstruct the original force magnitude and direction.

Check your understanding

Question 1

A 100 N100\,\mathrm{N} force is directed down and to the right at 30∘30^\circ below the positive horizontal axis. What are its signed components?
  1. Fx=86.6 N, Fy=−50.0 NF_x=86.6\,\mathrm{N},\ F_y=-50.0\,\mathrm{N}
  2. Fx=50.0 N, Fy=−86.6 NF_x=50.0\,\mathrm{N},\ F_y=-86.6\,\mathrm{N}
  3. Fx=−86.6 N, Fy=50.0 NF_x=-86.6\,\mathrm{N},\ F_y=50.0\,\mathrm{N}
  4. Fx=86.6 N, Fy=50.0 NF_x=86.6\,\mathrm{N},\ F_y=50.0\,\mathrm{N}
Show answer and explanation
Fx=86.6 N, Fy=−50.0 NF_x=86.6\,\mathrm{N},\ F_y=-50.0\,\mathrm{N}
The angle is measured from horizontal, so the horizontal component is 100cos⁡30∘=86.6 N100\cos30^\circ=86.6\,\mathrm{N}. The downward direction makes the vertical component negative: −100sin⁡30∘=−50.0 N-100\sin30^\circ=-50.0\,\mathrm{N}.

Question 2

For a force of magnitude FF at angle α\alpha from the positive vertical axis toward the right, which expressions give the components?
  1. Fx=Fsin⁡α, Fy=Fcos⁡αF_x=F\sin\alpha,\ F_y=F\cos\alpha
  2. Fx=Fcos⁡α, Fy=Fsin⁡αF_x=F\cos\alpha,\ F_y=F\sin\alpha
  3. Fx=−Fsin⁡α, Fy=Fcos⁡αF_x=-F\sin\alpha,\ F_y=F\cos\alpha
  4. Fx=Fsin⁡α, Fy=−Fcos⁡αF_x=F\sin\alpha,\ F_y=-F\cos\alpha
Show answer and explanation
Fx=Fsin⁡α, Fy=Fcos⁡αF_x=F\sin\alpha,\ F_y=F\cos\alpha
The angle is measured from vertical, so the vertical component is adjacent and uses cosine; the rightward horizontal component is opposite and uses sine. Both are positive for this direction.

Key terms

Cartesian components
The signed projections of a vector onto perpendicular coordinate axes, such as horizontal and vertical axes.
Component
One axis-aligned part of a vector; the perpendicular components together represent the original vector.
Reference axis
The axis from which an angle is measured.
Magnitude
The nonnegative size of a vector, such as a force measured in newtons.

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