Find perpendicular force and velocity resultants
Construct a right-angled vector triangle and state both magnitude and direction.
Cross-referenced from the related Cambridge IGCSE Chemistry 0620 or Physics 0625 teaching to the 0652 syllabus. Exact 0652 Core/Supplement objective coverage and this final-series course’s examiner fit still need specialist review. Written practice is self-checked and does not predict a Cambridge grade or demonstrate supervised laboratory competence.
Before you begin
- Recognise perpendicular directions.
- Use Pythagoras and a tangent ratio.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which triangle relation applies to perpendicular components?
What you will learn
- Find a resultant by a scale drawing or Pythagoras.
- State its direction relative to a named axis.
Add directions as well as magnitudes
For forces 3 N east and 4 N north, draw the 3 N arrow east, then the 4 N arrow north from its tip. The resultant runs from the first tail to the final tip. The right-angled triangle gives resultant magnitude √(3² + 4²) = 5 N. Adding 3 + 4 would only be valid if the forces pointed along the same line and direction.
The direction is tan⁻¹(4/3) ≈ 53.1° north of east. Name the reference direction; 36.9° east of north expresses the same resultant with another reference. A magnitude alone does not fully specify the vector. Pythagoras applies here because the two components are perpendicular.
A drawing is another calculation route
Choose a scale such as 1 cm representing 1 N. Draw the two component arrows head-to-tail with a ruler and a right angle, join the overall start and finish, then measure its length and direction. The drawing's measured result has a precision limit set by the scale and construction; it need not reproduce a calculator decimal exactly.
The same method applies to velocity components at right angles. A boat moving 6 m/s east relative to water with water flowing 8 m/s north relative to the bank has velocity 10 m/s at about 53.1° north of east relative to the bank in this model. Keep the reference frame and component units consistent. The section limits these resultant exercises to forces or velocities.
Sources for this lesson
Pause and explain
Which fully describes the resultant of 3 N east and 4 N north?
Worked example
Find the resultant of 3 N east and 4 N north.
Show the worked solution
- Magnitude = √(3² + 4²) = 5 N.
- Angle from east = tan⁻¹(4/3) ≈ 53.1°.
- State 5 N at 53.1° north of east.
Answer 5 N at approximately 53.1° north of east.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
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.
Find the resultant of 3 N east and 4 N north.
Give me a hint
Use the component triangle for magnitude and angle.
Compare my reasoning
- √(9 + 16) = 5 N.
- tan θ = 4/3 from east.
- θ ≈ 53.1° north of east.
5 N at 53.1° north of east.
Look for these in your work
- Use Pythagoras.
- Name the direction reference.
A boat has perpendicular velocity components 5 m/s east and 12 m/s north. Find its resultant relative to the bank.
Give me a hint
Use √(5² + 12²) and tan⁻¹(12/5).
Compare my reasoning
- Magnitude = √169 = 13 m/s.
- Angle from east ≈ 67.4°.
- Velocity is 13 m/s at 67.4° north of east relative to the bank.
13 m/s at 67.4° north of east.
Look for these in your work
- Keep the reference frame.
- Give magnitude and direction.
For 6 N west and 8 N south, describe a scale drawing using 1 cm per 2 N and find the resultant.
Give me a hint
The signs set the quadrant; lengths represent component magnitudes.
Compare my reasoning
- Draw 3 cm west then 4 cm south head-to-tail.
- The resultant measures 5 cm, representing 10 N.
- It is approximately 53.1° south of west.
10 N at 53.1° south of west.
Look for these in your work
- Apply the scale correctly.
- Preserve west and south directions.
Common mistakes
- Do not add perpendicular magnitudes directly.
- A direction needs a reference axis.
- Do not use this right-angle formula for an arbitrary angle.
What can you explain now?
Perpendicular velocities of 5 m/s east and 12 m/s north give what magnitude?
Compare with the explanation
13 m/s.
Pythagoras gives √(25 + 144) = 13; a full vector answer also needs direction.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Draw a new perpendicular force triangle and state its magnitude, angle and reference direction.
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.