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Laws of Motion - Equilibrium of a Particle

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Equilibrium of a particle in mechanics refers to the state where the net external force acting on the particle is zero: ∑F⃗=0\sum \vec{F} = 0.

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According to Newton's First Law, if the net force is zero, a particle at rest will remain at rest (Static Equilibrium), and a particle in motion will continue to move with a constant velocity (Dynamic Equilibrium).

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For a particle to be in equilibrium under the influence of concurrent forces (forces acting at the same point), the vector sum of all forces must be the null vector.

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In terms of rectangular components, equilibrium implies that the sum of components along each axis must independently be zero: ∑Fx=0\sum F_x = 0, ∑Fy=0\sum F_y = 0, and ∑Fz=0\sum F_z = 0.

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Lami's Theorem is a useful tool for equilibrium involving exactly three concurrent forces. It states that each force is proportional to the sine of the angle between the other two forces.

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A Free Body Diagram (FBD) is essential for solving equilibrium problems, where all forces acting on the particle are represented as vectors originating from a single point.

📐Formulae

∑F⃗ext=0\sum \vec{F}_{ext} = 0

F⃗1+F⃗2+F⃗3+...+F⃗n=0\vec{F}_1 + \vec{F}_2 + \vec{F}_3 + ... + \vec{F}_n = 0

∑Fx=0;∑Fy=0;∑Fz=0\sum F_x = 0; \quad \sum F_y = 0; \quad \sum F_z = 0

F1sin⁡α=F2sin⁡β=F3sin⁡γ\frac{F_1}{\sin \alpha} = \frac{F_2}{\sin \beta} = \frac{F_3}{\sin \gamma}

💡Examples

Problem 1:

A mass of 6 kg6 \text{ kg} is suspended by a rope of length 2 m2 \text{ m} from the ceiling. A force of 50 N50 \text{ N} in the horizontal direction is applied at the midpoint PP of the rope. What is the angle the rope makes with the vertical in equilibrium? (Take g=10 m/s2g = 10 \text{ m/s}^2)

Solution:

At the midpoint PP, three forces act: the tension T1T_1 in the upper part of the rope, the tension T2T_2 in the lower part (which equals the weight W=mg=6×10=60 NW = mg = 6 \times 10 = 60 \text{ N}), and the horizontal force F=50 NF = 50 \text{ N}. Let the angle with the vertical be θ\theta. For horizontal equilibrium: T1sin⁡θ=F=50T_1 \sin \theta = F = 50. For vertical equilibrium: T1cos⁡θ=W=60T_1 \cos \theta = W = 60. Dividing the two equations: tan⁡θ=5060=56\tan \theta = \frac{50}{60} = \frac{5}{6}. Therefore, θ=tan⁡−1(56)\theta = \tan^{-1}(\frac{5}{6}).

Explanation:

The problem is solved by resolving the tension force into horizontal and vertical components and setting the net force in each direction to zero.

Problem 2:

Three forces F⃗1\vec{F}_1, F⃗2\vec{F}_2, and F⃗3\vec{F}_3 keep a particle in equilibrium. If F⃗1=3i^−4j^\vec{F}_1 = 3\hat{i} - 4\hat{j} and F⃗2=−2i^+5j^\vec{F}_2 = -2\hat{i} + 5\hat{j}, find the magnitude of F⃗3\vec{F}_3.

Solution:

For equilibrium, F⃗1+F⃗2+F⃗3=0\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = 0. This implies F⃗3=−(F⃗1+F⃗2)\vec{F}_3 = -(\vec{F}_1 + \vec{F}_2). Calculating the sum: F⃗1+F⃗2=(3−2)i^+(−4+5)j^=1i^+1j^\vec{F}_1 + \vec{F}_2 = (3-2)\hat{i} + (-4+5)\hat{j} = 1\hat{i} + 1\hat{j}. Thus, F⃗3=−i^−j^\vec{F}_3 = -\hat{i} -\hat{j}. The magnitude is ∣F⃗3∣=(−1)2+(−1)2=2 N|\vec{F}_3| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2} \text{ N}.

Explanation:

Since the particle is in equilibrium, the vector sum of all forces must be zero. We find the third force by taking the negative of the resultant of the first two.

Equilibrium of a Particle Class 11 Notes & Examples