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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
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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
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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
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Lines and Planes in Space — Lesson quiz
6 questions
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The Pyramid and the Cone
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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
Lesson summary
Lesson summary
Resolving a force means finding two perpendicular components that have the same effect as the original force.
Horizontal component F
x
= F cos θ, and vertical component F
y
= F sin θ.
From the components
F = √(Fx² + Fy²), and tan θ = Fy ÷ Fx.
Exact values
sin 30° = cos 60° = 1 ÷ 2, sin 60° = cos 30° = √3 ÷ 2, and sin 45° = cos 45° = √2 ÷ 2.
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