Resolve forces and calculate a resultant with its direction
Calculate signed perpendicular components, add them and recover a resultant’s magnitude and direction in the correct quadrant.
Exam scope and assessment notes
Mechanics (Paper 4, AS & A Level) · Mathematics 9709, 2026–2027, section 4.1. Paper 1 algebra and trigonometry are assumed. Bodies are modelled as particles: forces act at one point and rotational effects are not considered. Resolve numerically; a scale drawing is not a substitute for calculation. Force diagrams are qualitative unless stated otherwise. Where mass is supplied in this sequence, use the stated g = 10 m s⁻². Check non-negative tension and contact reaction and use F ≤ μR except at limiting friction. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Use sine, cosine and Pythagoras.
- Locate signed components in Cartesian quadrants.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For P at θ above the horizontal, what is its horizontal component?
What you will learn
- Resolve a force relative to explicitly chosen axes.
- Calculate the magnitude and directed angle of a resultant from its components.
Components depend on where the angle is measured
A force P at angle θ above the positive horizontal has rightward component Pcos θ and upward component Psin θ. If θ is measured from the vertical instead, the roles of sine and cosine swap. Components can be negative when a force points against a chosen positive direction. State the axes and angle reference before calculating; the component adjacent to the given angle uses cosine.
For two perpendicular forces 8 N right and 6 N up, the resultant components are 8 and 6. Its magnitude is √(8² + 6²) = 10 N, and its angle above the rightward horizontal is arctan(6/8) ≈ 36.9°. Its magnitude is not the sum 14 N. A force triangle can illustrate the geometry, but the numerical calculation is required rather than an approximate scale measurement.
Add signed components before finding the resultant
For several forces, add all components along each axis to get H and V. Then resultant magnitude is √(H² + V²), with direction set by their signs. For 10 N east, 6 N west and 8 N south, H = 4 and V = −8; the resultant is 4√5 N, directed 63.4° below east. An unsigned inverse tangent alone would not establish the quadrant.
If H = V = 0, the resultant is zero and has no direction angle. Forces can cancel without each being zero. A resultant acts as the combined force in the particle model; it is not an extra force to add to the original force diagram. To balance a non-zero resultant, apply an equal-magnitude opposite-direction force, called an equilibrant.
Pause and explain
What is the resultant magnitude of perpendicular 8 N and 6 N forces?
Worked example
Find the resultant of 8 N right and 6 N up, giving magnitude and direction.
Show the worked solution
- The resultant’s perpendicular components are H = 8 N and V = 6 N.
- Its magnitude is √(64 + 36) = 10 N.
- Its quadrant I angle is arctan(6/8) ≈ 36.9° above the rightward horizontal.
Answer 10 N, 36.9° above the rightward horizontal.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Resolve a 12 N force at 60° above the rightward horizontal into horizontal and vertical components.
Give me a hint
Use the components adjacent and opposite to the specified horizontal angle.
Compare my reasoning
- The rightward component is 12cos60° = 6 N.
- The upward component is 12sin60° = 6√3 N.
- Their squared magnitudes total 36 + 108 = 144, recovering the force magnitude 12 N.
6 N right and 6√3 N up.
Look for these in your work
- Use the stated angle reference.
- Check the recovered magnitude from both components.
Find the resultant of forces 10 N east, 6 N west and 8 N south, including its direction.
Give me a hint
Take east and north as positive and add signed components first.
Compare my reasoning
- H = 10 − 6 = 4 N and V = −8 N.
- The magnitude is √(16 + 64) = 4√5 N ≈ 8.94 N.
- It points southeast, at arctan(8/4) ≈ 63.4° below east.
4√5 N ≈ 8.94 N, 63.4° below east.
Look for these in your work
- Keep the west and south signs when summing.
- State a direction consistent with both component signs.
A 30 N force points at an acute angle above the rightward horizontal and has upward component 18 N. Find its horizontal component and the resultant after adding 16 N right and 18 N down.
Give me a hint
The acute-angle condition fixes the positive horizontal component.
Compare my reasoning
- The horizontal component is √(30² − 18²) = 24 N right.
- The added horizontal force gives H = 24 + 16 = 40 N.
- The vertical forces cancel: V = 18 − 18 = 0, giving a 40 N rightward resultant.
Original horizontal component 24 N; combined resultant 40 N right.
Look for these in your work
- Use the given acute direction to select the component sign.
- Combine all signed components before finding the resultant.
Common mistakes
- Identify the axis from which the angle is measured before choosing sine or cosine.
- Add signed components, not force magnitudes.
- A resultant replaces the original forces for balance; it is not another external force.
What can you explain now?
What is the equilibrant of a 10 N resultant directed rightwards?
Compare with the explanation
10 N leftwards.
It has equal magnitude and opposite direction, making the total resultant zero.
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Make it stick
Resolve relative to labelled axes, add signed components and calculate the resultant’s magnitude, quadrant and equilibrant.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.