⚖️ Equilibrium of Forces
⚖️ Chapter 18: Statics | NEB Mathematics Notes Class 12
Based on Latest Syllabus 2080 | Forces, Equilibrium, Triangle of Forces, Lami’s Theorem
✅ Updated according to latest syllabus of 2080 | Complete notes with 15 solved problems
⚖️ Statics deals with the study of bodies at rest or in equilibrium under the action of forces. This chapter introduces force, moments, equilibrium, triangle of forces, and Lami’s theorem. The notes have been updated according to the latest syllabus of 2080.
📚 Exercise
📖 Introduction to Statics
Statics is the branch of mechanics that deals with the study of bodies at rest or in equilibrium under the action of forces. It focuses on the analysis of forces and their effects on objects that are not in motion.
Key Concepts:
Key Concepts:
- Force: A push or pull that can cause motion or deformation
- Equilibrium: State where net force and net moment are zero
- Concurrent Forces: Forces whose lines of action intersect at a common point
- Coplanar Forces: Forces acting in the same plane
📐 Triangle of Forces
Triangle of Forces Theorem: If three forces acting at a point are in equilibrium, they can be represented in magnitude and direction by the three sides of a triangle taken in order.
Converse: If three forces acting at a point are represented in magnitude and direction by the three sides of a triangle taken in order, then they are in equilibrium.
Converse: If three forces acting at a point are represented in magnitude and direction by the three sides of a triangle taken in order, then they are in equilibrium.
📐 Lami’s Theorem
Lami’s Theorem: If three coplanar, concurrent forces acting at a point are in equilibrium, then each force is proportional to the sine of the angle between the other two forces.
\[ \frac{F_1}{\sin \alpha} = \frac{F_2}{\sin \beta} = \frac{F_3}{\sin \gamma} \] where \(\alpha, \beta, \gamma\) are the angles opposite to forces \(F_1, F_2, F_3\) respectively.
\[ \frac{F_1}{\sin \alpha} = \frac{F_2}{\sin \beta} = \frac{F_3}{\sin \gamma} \] where \(\alpha, \beta, \gamma\) are the angles opposite to forces \(F_1, F_2, F_3\) respectively.
📌 Conditions for Equilibrium
ΣFx = 0
Sum of horizontal forces = 0
Sum of horizontal forces = 0
ΣFy = 0
Sum of vertical forces = 0
Sum of vertical forces = 0
ΣM = 0
Sum of moments about any point = 0
Sum of moments about any point = 0
📝 Solved Problems (15 Important Questions)
Q1 Three forces of magnitudes 3N, 4N, and 5N are in equilibrium. Find the angle between the 3N and 4N forces.
Using triangle of forces: \(5^2 = 3^2 + 4^2 + 2(3)(4)\cos\theta\)
\(25 = 9 + 16 + 24\cos\theta\) ⇒ \(25 = 25 + 24\cos\theta\) ⇒ \(24\cos\theta = 0\) ⇒ \(\cos\theta = 0\) ⇒ \(\theta = 90^\circ\)
\(25 = 9 + 16 + 24\cos\theta\) ⇒ \(25 = 25 + 24\cos\theta\) ⇒ \(24\cos\theta = 0\) ⇒ \(\cos\theta = 0\) ⇒ \(\theta = 90^\circ\)
✅ Angle = 90°
Q2 Three forces P, Q, R are in equilibrium. If P = 10N, Q = 12N, and the angle between P and Q is 60°, find R.
By triangle of forces: \(R^2 = P^2 + Q^2 + 2PQ\cos\theta\)
\(R^2 = 100 + 144 + 2(10)(12)\cos60° = 244 + 240 × 0.5 = 244 + 120 = 364\)
\(R = \sqrt{364} = 2\sqrt{91} \approx 19.08N\)
\(R^2 = 100 + 144 + 2(10)(12)\cos60° = 244 + 240 × 0.5 = 244 + 120 = 364\)
\(R = \sqrt{364} = 2\sqrt{91} \approx 19.08N\)
✅ \(R = 2\sqrt{91} N\)
Q3 Using Lami’s theorem, find the force P if forces P, 20N, and 30N are in equilibrium with angles 120°, 150°, and 90° respectively opposite to them.
\(\frac{P}{\sin 90°} = \frac{20}{\sin 150°} = \frac{30}{\sin 120°}\)
\(\frac{P}{1} = \frac{20}{0.5} = 40\) ⇒ \(P = 40N\)
\(\frac{P}{1} = \frac{20}{0.5} = 40\) ⇒ \(P = 40N\)
✅ \(P = 40N\)
Q4 A weight of 100N is suspended by two strings making angles 30° and 60° with the horizontal. Find tensions in the strings.
Let T₁ at 30°, T₂ at 60° from horizontal.
Horizontal: \(T_1\cos30° = T_2\cos60°\) ⇒ \(T_1 × 0.866 = T_2 × 0.5\) ⇒ \(T_2 = 1.732T_1\)
Vertical: \(T_1\sin30° + T_2\sin60° = 100\) ⇒ \(0.5T_1 + 0.866T_2 = 100\)
Substituting: \(0.5T_1 + 0.866(1.732T_1) = 0.5T_1 + 1.5T_1 = 2T_1 = 100\) ⇒ \(T_1 = 50N\), \(T_2 = 86.6N\)
Horizontal: \(T_1\cos30° = T_2\cos60°\) ⇒ \(T_1 × 0.866 = T_2 × 0.5\) ⇒ \(T_2 = 1.732T_1\)
Vertical: \(T_1\sin30° + T_2\sin60° = 100\) ⇒ \(0.5T_1 + 0.866T_2 = 100\)
Substituting: \(0.5T_1 + 0.866(1.732T_1) = 0.5T_1 + 1.5T_1 = 2T_1 = 100\) ⇒ \(T_1 = 50N\), \(T_2 = 86.6N\)
✅ \(T_1 = 50N, T_2 = 86.6N\)
Q5 Three forces 5N, 6N, and 7N are in equilibrium. Find the angle between the 5N and 6N forces.
\(7^2 = 5^2 + 6^2 + 2(5)(6)\cos\theta\) ⇒ \(49 = 25 + 36 + 60\cos\theta\) ⇒ \(49 = 61 + 60\cos\theta\) ⇒ \(-12 = 60\cos\theta\) ⇒ \(\cos\theta = -0.2\) ⇒ \(\theta = \cos^{-1}(-0.2) \approx 101.54°\)
✅ \(\theta \approx 101.54°\)
Q6 A body of weight 50N is suspended by two strings of lengths 3m and 4m from two points 5m apart. Find tensions.
The strings form a triangle with sides 3,4,5 which is right-angled at the point where weight hangs.
Strings make angles: sinθ₁ = 4/5, sinθ₂ = 3/5, cosθ₁ = 3/5, cosθ₂ = 4/5
Horizontal: \(T_1\cosθ₁ = T_2\cosθ₂\) ⇒ \(T_1(3/5) = T_2(4/5)\) ⇒ \(3T_1 = 4T_2\)
Vertical: \(T_1\sinθ₁ + T_2\sinθ₂ = 50\) ⇒ \(T_1(4/5) + T_2(3/5) = 50\) ⇒ \(4T_1 + 3T_2 = 250\)
Solving: \(T_1 = 40N, T_2 = 30N\)
Strings make angles: sinθ₁ = 4/5, sinθ₂ = 3/5, cosθ₁ = 3/5, cosθ₂ = 4/5
Horizontal: \(T_1\cosθ₁ = T_2\cosθ₂\) ⇒ \(T_1(3/5) = T_2(4/5)\) ⇒ \(3T_1 = 4T_2\)
Vertical: \(T_1\sinθ₁ + T_2\sinθ₂ = 50\) ⇒ \(T_1(4/5) + T_2(3/5) = 50\) ⇒ \(4T_1 + 3T_2 = 250\)
Solving: \(T_1 = 40N, T_2 = 30N\)
✅ \(T_1 = 40N, T_2 = 30N\)
Q7 Using Lami’s theorem, find the forces in two strings supporting a weight of 100N making angles 120° and 150° with each other.
Angles between strings = 120° and 150°, so forces are opposite to these angles.
Let angles opposite to forces be α, β, γ. Here γ = 90° (between the two strings). Then α = 150°, β = 120°, γ = 90°.
\(\frac{T_1}{\sin 150°} = \frac{T_2}{\sin 120°} = \frac{100}{\sin 90°}\) ⇒ \(T_1/0.5 = 100/1 = 100\) ⇒ \(T_1 = 50N\)
\(T_2/0.866 = 100\) ⇒ \(T_2 = 86.6N\)
Let angles opposite to forces be α, β, γ. Here γ = 90° (between the two strings). Then α = 150°, β = 120°, γ = 90°.
\(\frac{T_1}{\sin 150°} = \frac{T_2}{\sin 120°} = \frac{100}{\sin 90°}\) ⇒ \(T_1/0.5 = 100/1 = 100\) ⇒ \(T_1 = 50N\)
\(T_2/0.866 = 100\) ⇒ \(T_2 = 86.6N\)
✅ \(T_1 = 50N, T_2 = 86.6N\)
Q8 Three forces P, 2P, and 3P are in equilibrium. Find the cosine of the angle between the forces P and 2P.
Using triangle of forces: \((3P)^2 = P^2 + (2P)^2 + 2(P)(2P)\cos\theta\)
\(9P^2 = P^2 + 4P^2 + 4P^2\cos\theta\) ⇒ \(9P^2 = 5P^2 + 4P^2\cos\theta\) ⇒ \(4P^2\cos\theta = 4P^2\) ⇒ \(\cos\theta = 1\)
This is impossible as angle would be 0°, but that would mean collinear forces. For equilibrium, forces must be such that the largest force equals sum of others: 3P = P + 2P, so they are collinear. Thus angle between P and 2P is 180°.
\(\cos\theta = -1\)
\(9P^2 = P^2 + 4P^2 + 4P^2\cos\theta\) ⇒ \(9P^2 = 5P^2 + 4P^2\cos\theta\) ⇒ \(4P^2\cos\theta = 4P^2\) ⇒ \(\cos\theta = 1\)
This is impossible as angle would be 0°, but that would mean collinear forces. For equilibrium, forces must be such that the largest force equals sum of others: 3P = P + 2P, so they are collinear. Thus angle between P and 2P is 180°.
\(\cos\theta = -1\)
✅ \(\cos\theta = -1\) (angle = 180°)
Q9 A particle is in equilibrium under forces 4N, 5N, and 6N. Find the angle between the 4N and 5N forces.
Using triangle of forces: \(6^2 = 4^2 + 5^2 + 2(4)(5)\cos\theta\)
\(36 = 16 + 25 + 40\cos\theta\) ⇒ \(36 = 41 + 40\cos\theta\) ⇒ \(-5 = 40\cos\theta\) ⇒ \(\cos\theta = -0.125\) ⇒ \(\theta = \cos^{-1}(-0.125) \approx 97.18°\)
\(36 = 16 + 25 + 40\cos\theta\) ⇒ \(36 = 41 + 40\cos\theta\) ⇒ \(-5 = 40\cos\theta\) ⇒ \(\cos\theta = -0.125\) ⇒ \(\theta = \cos^{-1}(-0.125) \approx 97.18°\)
✅ \(\theta \approx 97.18°\)
Q10 Three forces acting at a point are in equilibrium. If two of them are 8N and 10N with an angle of 60° between them, find the third force.
\(R = \sqrt{8^2 + 10^2 + 2(8)(10)\cos60°} = \sqrt{64 + 100 + 160 × 0.5} = \sqrt{164 + 80} = \sqrt{244} = 2\sqrt{61} \approx 15.62N\)
✅ Third force = \(2\sqrt{61}N\)
Q11 Two forces of 12N and 16N act at a point with an angle of 90° between them. A third force keeps them in equilibrium. Find the magnitude and direction of the third force.
Resultant = \(\sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20N\)
Third force = 20N opposite to the resultant.
Direction: tanθ = 16/12 = 4/3 ⇒ θ = 53.13° from the 12N force.
Third force = 20N opposite to the resultant.
Direction: tanθ = 16/12 = 4/3 ⇒ θ = 53.13° from the 12N force.
✅ Third force = 20N at 233.13° from 12N
Q12 A weight of 200N is hung from a string. A horizontal force pulls it aside until the string makes 30° with vertical. Find tension and horizontal force.
Vertical: \(T\cos30° = 200\) ⇒ \(T × 0.866 = 200\) ⇒ \(T = 231N\)
Horizontal: \(H = T\sin30° = 231 × 0.5 = 115.5N\)
Horizontal: \(H = T\sin30° = 231 × 0.5 = 115.5N\)
✅ T = 231N, H = 115.5N
Q13 Three forces F₁, F₂, F₃ acting at a point are in equilibrium. If F₁ = 10N, F₂ = 15N, and the angle between F₁ and F₂ is 120°, find F₃.
\(F_3^2 = F_1^2 + F_2^2 + 2F_1F_2\cos120° = 100 + 225 + 2(10)(15)(-0.5) = 325 – 150 = 175\)
\(F_3 = \sqrt{175} = 5\sqrt{7} \approx 13.23N\)
\(F_3 = \sqrt{175} = 5\sqrt{7} \approx 13.23N\)
✅ \(F_3 = 5\sqrt{7}N\)
Q14 Using triangle of forces, prove Lami’s theorem.
For three forces in equilibrium, they form a triangle. Using sine law in the triangle: \(\frac{P}{\sin(180°-α)} = \frac{Q}{\sin(180°-β)} = \frac{R}{\sin(180°-γ)}\)
Since \(\sin(180°-θ) = \sinθ\), we get \(\frac{P}{\sinα} = \frac{Q}{\sinβ} = \frac{R}{\sinγ}\)
Since \(\sin(180°-θ) = \sinθ\), we get \(\frac{P}{\sinα} = \frac{Q}{\sinβ} = \frac{R}{\sinγ}\)
✅ Proved
Q15 A string is tied to a weight of 50N and passed over a smooth pulley. Another weight of 30N is hung. Find the tension and angle.
For a smooth pulley, tension is constant.
Horizontal: \(T\sinθ = T\sinφ\) ⇒ θ = φ
Vertical: \(2T\sinθ = 50 – 30 = 20\) and also \(2T\cosθ = ?\) Actually for equal angles, the horizontal components cancel and vertical equilibrium: \(2T\sinθ = 20\)
Also for the 30N weight: \(T = 30N\) (if hanging freely)? Actually if the string passes over a smooth pulley, tension is same on both sides. For weight 30N, T = 30N. Then \(2×30×\sinθ = 20\) ⇒ \(60\sinθ = 20\) ⇒ \(\sinθ = 1/3\) ⇒ θ = 19.47°
Horizontal: \(T\sinθ = T\sinφ\) ⇒ θ = φ
Vertical: \(2T\sinθ = 50 – 30 = 20\) and also \(2T\cosθ = ?\) Actually for equal angles, the horizontal components cancel and vertical equilibrium: \(2T\sinθ = 20\)
Also for the 30N weight: \(T = 30N\) (if hanging freely)? Actually if the string passes over a smooth pulley, tension is same on both sides. For weight 30N, T = 30N. Then \(2×30×\sinθ = 20\) ⇒ \(60\sinθ = 20\) ⇒ \(\sinθ = 1/3\) ⇒ θ = 19.47°
✅ T = 30N, θ = 19.47°
✏️ Practice Questions
P1 Three forces 6N, 8N, and 10N are in equilibrium. Find the angle between 6N and 8N.
✅ Answer: 90°
P2 A weight of 80N is suspended by two strings making 45° and 30° with horizontal. Find tensions.
✅ Answer: T₁ ≈ 41.4N, T₂ ≈ 58.6N
P3 Using Lami’s theorem, find force if P/sin30° = 20/sin60° = Q/sin90°.
✅ Answer: P = 11.55N, Q = 23.1N
P4 Two forces 10N and 10N act at 120°. Find the third force for equilibrium.
✅ Answer: 10N
P5 A particle is in equilibrium under three forces. Two are 9N and 12N at 90°. Find the third.
✅ Answer: 15N
📋 Multiple Choice Questions (HSEB Pattern)
MCQ 1 In triangle of forces, if three forces are in equilibrium, they form:
A) A closed triangle B) An open triangle C) A line D) A circle
✅ Answer: A) A closed triangle
MCQ 2 Lami’s theorem is applicable only for:
A) Two forces B) Three forces C) Four forces D) Any number of forces
✅ Answer: B) Three forces
MCQ 3 The angle between two forces in equilibrium with a third force of 10N when the other two are 6N and 8N is:
A) 30° B) 60° C) 90° D) 120°
✅ Answer: C) 90°
MCQ 4 The condition for equilibrium of a particle is:
A) ΣF = 0 B) ΣF > 0 C) ΣF < 0 D) ΣF ≠ 0
✅ Answer: A) ΣF = 0
MCQ 5 In Lami’s theorem, the angle used is:
A) Angle between forces B) Angle opposite to the force C) Angle adjacent D) Any angle
✅ Answer: B) Angle opposite to the force
MCQ 6 If three forces are in equilibrium, the resultant of any two is:
A) Equal and opposite to third B) Greater than third C) Less than third D) Zero
✅ Answer: A) Equal and opposite to third
MCQ 7 Forces that meet at a single point are called:
A) Coplanar forces B) Concurrent forces C) Parallel forces D) Collinear forces
✅ Answer: B) Concurrent forces
MCQ 8 The sine of the angle between two forces in Lami’s theorem is taken from:
A) Angle between the other two B) Angle opposite to the force C) 90° D) 180°
✅ Answer: B) Angle opposite to the force
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This PDF provides the solutions of every question from the 1st exercise of class 12 statics chapter.
This PDF provides the solutions of every question from the 1st exercise of class 12 statics chapter.
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📌 Key Statics Summary:
• Triangle of Forces: \(R^2 = P^2 + Q^2 + 2PQ\cos\theta\)
• Lami’s Theorem: \(\frac{F_1}{\sin\alpha} = \frac{F_2}{\sin\beta} = \frac{F_3}{\sin\gamma}\)
• Equilibrium Conditions: ΣFx = 0, ΣFy = 0, ΣM = 0
• Concurrent Forces: Forces whose lines of action intersect at a common point
• Coplanar Forces: Forces acting in the same plane
• Triangle of Forces: \(R^2 = P^2 + Q^2 + 2PQ\cos\theta\)
• Lami’s Theorem: \(\frac{F_1}{\sin\alpha} = \frac{F_2}{\sin\beta} = \frac{F_3}{\sin\gamma}\)
• Equilibrium Conditions: ΣFx = 0, ΣFy = 0, ΣM = 0
• Concurrent Forces: Forces whose lines of action intersect at a common point
• Coplanar Forces: Forces acting in the same plane