Engineering Mechanics | System of Forces | Lecture 3 | Resultant of two forces| Parallelogram Law
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Engineering Mechanics | System of Forces | Lecture 3 | Resultant of two forces| Parallelogram Law
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94 просмотра · 3 нед. назад
Engineering Mechanics – Chapter 1: System of Forces | Lecture 3 – Resultant of Two Forces
In this lecture, we learn how to find the resultant of two forces using the Parallelogram Law of Forces and Triangle Law of Forces, followed by step-by-step solved numerical problems.
This is an important topic in Engineering Mechanics and forms the foundation for studying resolution and composition of forces, equilibrium, moments and other force-system problems.
📘 Topics covered in Lecture 3:
Resultant of Two Forces
Composition of Forces
Parallelogram Law of Forces
Triangle Law of Forces
Magnitude of Resultant Force
Direction of Resultant Force
Angle made by the resultant
Vector representation of forces
Graphical representation
Step-by-step solved numericals
Exam-oriented problems
For two forces F₁ and F₂ acting at an angle θ:
R = √(F₁² + F₂² + 2F₁F₂ cosθ)
Direction of the resultant:
tan α = (F₂ sinθ)/(F₁ + F₂ cosθ)
🎯 Useful for: First Year B.E./B.Tech students, Diploma Engineering students and candidates preparing for GATE, ESE, SSC JE, RRB JE, State AE/JE and other engineering competitive examinations.
📚 Engineering Mechanics – Chapter 1: System of Forces Series
Lecture 1: Force, Characteristics of Force & Graphical Representation
Lecture 2: Types of Systems of Forces – Coplanar & Non-Coplanar Forces
Lecture 3: Resultant of Two Forces – Parallelogram Law & Triangle Law
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