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Applied Mathematics — Second Secondary
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Applied Mathematics — Second Secondary
Unit 1: Statics
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Forces
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Forces — Lesson quiz
6 questions
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Resolution of Forces
Text lesson
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Meaning of resolving a force
Exact values of the trigonometric ratios
From the components back to the force
Worked examples
Common mistakes
Lesson summary
Resolution of Forces — Lesson quiz
6 questions
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Resultant of Coplanar Concurrent Forces
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Resultant of Coplanar Concurrent Forces — Lesson quiz
6 questions
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Equilibrium of a Rigid Body under Coplanar Forces
Text lesson
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Equilibrium of a Rigid Body under Coplanar Forces — Lesson quiz
6 questions
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Equilibrium on a Rough Horizontal Plane
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Equilibrium on a Rough Horizontal Plane — Lesson quiz
6 questions
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Unit 2: Solid and Analytic Geometry
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Lines and Planes in Space
Text lesson
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Lines and Planes in Space — Lesson quiz
6 questions
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The Pyramid and the Cone
Text lesson
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The Pyramid and the Cone — Lesson quiz
6 questions
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Lateral and Total Area of Pyramid and Cone
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Lateral and Total Area of Pyramid and Cone — Lesson quiz
6 questions
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Volume of the Right Pyramid and Cone
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Volume of the Right Pyramid and Cone — Lesson quiz
6 questions
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Equation of a Circle
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Equation of a Circle — Lesson quiz
6 questions
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Text lesson
Resolution of Forces
Applied Mathematics — Second Secondary
Resolution of Forces
Explanation
Worked examples
Worked examples
Example (1): Resolving a force of 10 N at 30°
Find the horizontal and vertical components of a force of 10 N that makes an angle of 30° with the positive x-axis.
Solution:
F
x
= 10 × cos 30° = 10 × √3 ÷ 2 = 5√3 ≈ 8.66 N.
F
y
= 10 × sin 30° = 10 × 1 ÷ 2 = 5 N.
8.66² + 5² = 75 + 25 = 100 = 10², which is correct.
Example (2): Resolving a force of 20 N at 60°
A force of 20 N makes an angle of 60° with the x-axis. Find its components.
Solution:
F
x
= 20 × cos 60° = 20 × 1 ÷ 2 = 10 N.
F
y
= 20 × sin 60° = 20 × √3 ÷ 2 = 10√3 ≈ 17.32 N.
Notice that the vertical component is larger than the horizontal one because the angle is greater than 45°.
Example (3): A force making an angle of 45°
A person pulls a box with a rope whose tension is 12 N.
The rope makes an angle of 45° with the horizontal.
Find the horizontal component of the tension.
Solution:
Horizontal component = 12 × cos 45° = 12 × √2 ÷ 2 = 6√2.
6√2 ≈ 8.49 N.
In this case the vertical component is also equal to 6√2 N, because sin 45° = cos 45°.
Example (4): Finding a force from its components
A force has a horizontal component of 6 N to the right and a vertical component of 8 N upwards.
Find the magnitude and the direction of the force.
Solution:
F = √(6² + 8²) = √100 = 10 N.
tan θ = 8 ÷ 6 = 4 ÷ 3, so θ ≈ 53.13°.
The force of 10 N makes an angle of about 53.13° with the positive x-axis.
Resolution of Forces · 4 / 6
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