Overview of Statics in JEE Maths

Statics is a branch of mechanics that deals with objects at rest. It focuses on the study of forces, equilibrium, moments, torque, and center of gravity. Statics is a key part of JEE Maths as it forms the foundation for solving problems related to forces and mechanics. A strong understanding allows students to solve both JEE Main and Advanced questions efficiently.

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Basic Concepts in Statics

• Force: A push or pull on an object causing a change in motion or shape.

• Equilibrium: A state where an object is at rest or moves with constant velocity, and the net force is zero.

• Torque: The rotational effect of a force about a point.

• Moment of a Force: Product of force and perpendicular distance from the pivot: M=F×dM = F \times dM=F×d

• Centre of Gravity: The point where the total weight of a body acts.

Equilibrium of Forces

An object is in equilibrium when the sum of all forces acting on it is zero.

• Formula: ΣF = 0

• Example: A block on a table with forces: 10 N to the right and 10 N to the left → Net force = 0 → Block is in equilibrium.

Resolution of Forces

Any force can be resolved into perpendicular components:

• Fx = F cos θ

• Fy = F sin θ
  This is useful when analyzing forces acting at angles in JEE problems.

Moments and Torque

• Moment measures the rotational effect of a force about a point.

• Formula: Moment = Force × Perpendicular distance from pivot

• Torque is often used interchangeably with moment.

• Example: Force of 10 N applied at 2 m from pivot → Moment = 10 × 2 = 20 Nm

Conditions for Equilibrium

For planar problems:

• ΣF = 0 → Net force is zero

• ΣM = 0 → Net moment about any point is zero

• ΣFx = 0, ΣFy = 0, ΣM = 0 for 2D systems

Centre of Gravity

• Centre of gravity is the point where the total weight of a body acts.

• For uniform objects, it is at the geometric center.

• Example: Centre of gravity of a uniform rod of length L → L/2 from one end

Simple Examples in Statics

1. Find the resultant of two forces 5 N and 12 N at 90° → R = √(5² + 12²) = 13 N

2. A uniform rod of length 4 m, weight 20 N → Centre of gravity from one end = 2 m

3. Torque of 15 N force applied at 3 m → Moment = 15 × 3 = 45 Nm

4. Block on table: Forces 8 N right, 5 N left → Net force = 3 N to the right

Applications in JEE Exams

• Solving problems on blocks on inclined planes

• Analyzing forces in beams and rods

• Calculating torque and moments in mechanical systems

• Finding centre of gravity of uniform and composite bodies

• Equilibrium problems involving multiple forces

Tips for JEE Exam Preparation

• Memorize key formulas: ΣF = 0, ΣM = 0, Torque = F × d

• Practice resolving forces into components

• Use diagrams for visual understanding of equilibrium

• Solve previous years’ JEE Statics questions

• Apply centre of gravity concepts to composite objects

Conclusion

Statics is an important part of JEE Maths that deals with forces, equilibrium, moments, torque, and centre of gravity. By understanding basic principles and practicing simple examples, students can solve problems confidently and efficiently. StudentBro.in provides step-by-step guides, examples, and practice questions to help aspirants master statics for JEE Main & Advanced exams.