A weight of 20 KN supported by two cords, one 3 m long and the other 4m long with points of support 5 m apart find the tensions T1 and T2 in the cords.

Problem Statement

A weight of 20 kN is supported by two cords from points A and B, 5 m apart. Cord AC is 3 m long and cord BC is 4 m long. Find the tensions $T_1$ (in cord AC) and $T_2$ (in cord BC).

Weight supported by two cords forming a right triangle

Diagram showing support points A, B and suspension point C

Step-by-Step Solution

Step 1: Geometric Analysis of Support Triangle ABC

Given lengths: AC = 3 m, BC = 4 m, AB = 5 m.

Check if triangle ABC is right-angled using the converse of Pythagoras theorem:

$$ AC^2 + BC^2 = 3^2 + 4^2 = 9 + 16 = 25 $$ $$ AB^2 = 5^2 = 25 $$ $$ \text{Since } AC^2 + BC^2 = AB^2 \text{, the triangle ABC is right-angled at C (} \angle C = 90^\circ \text{).} $$

Find angles A and B using the Sine Rule or basic trigonometry:

$$ \frac{AC}{\sin(\angle B)} = \frac{BC}{\sin(\angle A)} = \frac{AB}{\sin(\angle C)} $$ $$ \frac{3}{\sin(\angle B)} = \frac{4}{\sin(\angle A)} = \frac{5}{\sin(90^\circ)} = \frac{5}{1} = 5 $$ $$ \sin(\angle B) = \frac{3}{5} = 0.6 \implies \angle B = \arcsin(0.6) \approx 36.87^\circ = 36^\circ 52′ $$ $$ \sin(\angle A) = \frac{4}{5} = 0.8 \implies \angle A = \arcsin(0.8) \approx 53.13^\circ = 53^\circ 8′ $$

Step 2: Determine Angles for Lami’s Theorem at Point C

Let $T_1$ be the tension in cord AC and $T_2$ be the tension in cord BC. Let W = 20 kN be the weight acting vertically down from C.

Define angles relative to the vertical ($\alpha$ for T1, $\beta$ for T2) as specified in the provided solution:

$$ \alpha = 90^\circ – \angle B = 90^\circ – 36^\circ 52′ = 53^\circ 8′ $$

$$ \beta = 90^\circ – \angle A = 90^\circ – 53^\circ 8′ = 36^\circ 52′ $$

These represent the angles the cords make with the vertical line through C.

The angle between the two cords T1 and T2 is $ \alpha + \beta = 53^\circ 8′ + 36^\circ 52′ = 90^\circ $. This is consistent with $\angle C = 90^\circ$.

Step 3: Apply Lami’s Theorem at Point C

The forces acting at point C are T1, T2, and W. The angles opposite to these forces are:

Angle opposite W (between T1 and T2): $ \alpha + \beta = 90^\circ $
Angle opposite T1 (between T2 and W): $ 180^\circ – \beta = 180^\circ – 36^\circ 52′ = 143^\circ 8′ $
Angle opposite T2 (between T1 and W): $ 180^\circ – \alpha = 180^\circ – 53^\circ 8′ = 126^\circ 52′ $

Applying Lami’s theorem using the specific formulation from the provided solution text:

$$ \frac{T_1}{\sin(53^\circ 8′)} = \frac{T_2}{\sin(36^\circ 52′)} = \frac{W}{\sin(90^\circ)} $$ $$ \frac{T_1}{\sin(53^\circ 8′)} = \frac{T_2}{\sin(36^\circ 52′)} = \frac{20}{\sin(90^\circ)} $$

Step 4: Calculate Tensions T₁ and T₂

Calculate $T_1$:

$$ \frac{T_1}{\sin(53^\circ 8′)} = \frac{20}{\sin(90^\circ)} $$ $$ T_1 = \frac{20}{\sin(90^\circ)} \times \sin(53^\circ 8′) $$ $$ T_1 = \frac{20}{1} \times 0.8 $$ $$ T_1 = 16 \, \text{kN} $$

Calculate $T_2$:

$$ \frac{T_2}{\sin(36^\circ 52′)} = \frac{20}{\sin(90^\circ)} $$ $$ T_2 = \frac{20}{\sin(90^\circ)} \times \sin(36^\circ 52′) $$ $$ T_2 = \frac{20}{1} \times 0.6 $$ $$ T_2 = 12 \, \text{kN} $$

Final Result

Tension in 3m cord AC ($T_1$) = $ \mathbf{16 \, kN} $
Tension in 4m cord BC ($T_2$) = $ \mathbf{12 \, kN} $

Conceptual Explanation & Applications

Core Concepts:

Static Equilibrium: Point C, where the weight is attached, is stationary. Therefore, the vector sum of the three forces acting on it (Weight W downwards, Tension T₁ along AC, Tension T₂ along BC) must be zero: $ \vec{W} + \vec{T_1} + \vec{T_2} = \vec{0} $.
Geometric Constraints: The fixed lengths of the supporting cords (AC=3m, BC=4m) and the distance between the anchor points (AB=5m) dictate the geometry of the system.
Pythagorean Theorem: The specific lengths (3, 4, 5) form a Pythagorean triple ($3^2 + 4^2 = 5^2$), proving that the angle between the two cords at point C ($\angle ACB$) is exactly $90^\circ$.
Trigonometry: Used to determine the angles ($\angle CAB, \angle CBA$) within the support triangle ABC, which subsequently help find the angles the cords make with the vertical or horizontal.
Lami’s Theorem: Provides a relationship between the magnitudes of three concurrent forces in equilibrium and the sines of the angles opposite them. It’s applied at point C to relate W, T₁, and T₂.

Real-World Applications:

Structural Analysis: Calculating tensions in simple cable structures, suspension bridges elements, or supports for hanging objects where the geometry is fixed.
Load Distribution: Understanding how a load is distributed between multiple supports based on the angles of attachment.
Safety Engineering: Ensuring cables or ropes used in lifting or suspension are strong enough for the calculated tensions based on the setup geometry.
Art Installations / Decorations: Designing stable ways to hang heavy objects like chandeliers or sculptures using multiple cords.

Why It Works:
The system is in static equilibrium, meaning all forces balance. The specific 3-4-5 geometry fixes the angles involved. Since $\angle ACB = 90^\circ$, the tension $T_1$ must support the component of the weight resolved perpendicular to $T_2$, and $T_2$ must support the component perpendicular to $T_1$. Lami’s theorem mathematically formalizes this balance. By first determining the angles using geometry and trigonometry, we can then apply Lami’s theorem (or resolve forces into components) at point C to solve for the unknown tensions required to balance the 20 kN weight.