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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
Practice
From the components back to the force
From the components back to the force
If the horizontal and vertical components of a force are known
its magnitude and direction are found from:
Magnitude
F = √(Fx² + Fy²)
Angle with the positive x-axis
tan θ = Fy ÷ Fx
When finding the angle we must check the quadrant in which the force lies.
If both components are positive, the angle lies in the first quadrant, that is between 0° and 90°.
Important note
The calculator gives an angle in the first quadrant.
If one of the components is negative, the angle must be adjusted to the quadrant in which the force lies.
Resolution of Forces · 3 / 6
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