A body of weight 20 N is suspended by two strings 5 m and 12 m long and other ends being fastened to the extremities of a rod of length 13 m. If the rod be so held that the bodyhangs immediately below its middle point, find out the tensions in the strings.

Problem Statement

A body of weight 20 N is suspended by two strings, OA (5 m long) and OB (12 m long), with the other ends being fastened to the extremities (A and B) of a rod of length 13 m. If the rod is held such that the body (at O) hangs immediately below its middle point, find out the tensions T₁ (in string OA) and T₂ (in string OB).

Weight suspended from a rod by two strings

Fig. 2.36: Equilibrium of the suspended body

Step-by-Step Solution

Step 1: Geometric Analysis of Triangle AOB

The suspension point O and the rod ends A and B form triangle AOB.

Given lengths: OA = 5 m, OB = 12 m, AB = 13 m.

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

$$ OA^2 + OB^2 = 5^2 + 12^2 = 25 + 144 = 169 $$ $$ AB^2 = 13^2 = 169 $$ $$ \text{Since } OA^2 + OB^2 = AB^2 \text{, triangle AOB is right-angled at O (} \angle AOB = 90^\circ \text{).} $$

Define angle $\theta = \angle ABO$ (the angle at vertex B in triangle AOB). Calculate its sine and cosine:

$$ \sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{OA}{AB} = \frac{5}{13} $$ $$ \cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{OB}{AB} = \frac{12}{13} $$

Step 2: Apply Lami’s Theorem at Point O

Point O is in equilibrium under the action of three concurrent forces:

Weight W = 20 N (acting vertically downwards).
Tension T₁ (in string OA).
Tension T₂ (in string OB).

The angle between T₁ and T₂ is $\angle AOB = 90^\circ$.

Applying Lami’s theorem as formulated in the provided solution (relating forces to the sines of the angles opposite them, using $\theta = \angle ABO$)):

$$ \frac{T_1}{\sin(\text{angle opp } T_1)} = \frac{T_2}{\sin(\text{angle opp } T_2)} = \frac{W}{\sin(\text{angle opp } W)} $$ $$ \frac{T_1}{\sin(180^\circ – \theta)} = \frac{T_2}{\sin(90^\circ + \theta)} = \frac{W}{\sin(90^\circ)} $$

Simplify using trigonometric identities ($ \sin(180^\circ – x) = \sin x $, $ \sin(90^\circ + x) = \cos x $, and $ \sin(90^\circ) = 1 $):”

$$ \frac{T_1}{\sin \theta} = \frac{T_2}{\cos \theta} = \frac{W}{1} $$ $$ \frac{T_1}{\sin \theta} = \frac{T_2}{\cos \theta} = W = 20 $$

Step 3: Calculate Tensions T₁ and T₂

Calculate T₁ using $ \sin \theta = 5/13 $:

$$ \frac{T_1}{\sin \theta} = 20 $$ $$ T_1 = 20 \sin \theta $$ $$ T_1 = 20 \times \frac{5}{13} = \frac{100}{13} $$ $$ T_1 \approx 7.69 \, \text{N} $$

Calculate T₂ using $ \cos \theta = 12/13 $:

$$ \frac{T_2}{\cos \theta} = 20 $$ $$ T_2 = 20 \cos \theta $$ $$ T_2 = 20 \times \frac{12}{13} = \frac{240}{13} $$ $$ T_2 \approx 18.46 \, \text{N} $$

Final Result

Tension in 5m string OA ($T_1$) = $ \mathbf{\frac{100}{13} \, N} \approx \mathbf{7.69 \, N} $
Tension in 12m string OB ($T_2$) = $ \mathbf{\frac{240}{13} \, N} \approx \mathbf{18.46 \, N} $

Conceptual Explanation & Applications

Core Concepts:

Static Equilibrium: The body suspended at point O is not accelerating, meaning the vector sum of the forces acting on it (Weight W, Tension T₁, Tension T₂) is zero: $ \vec{W} + \vec{T_1} + \vec{T_2} = \vec{0} $.
Geometric Constraints: The specified lengths of the strings (OA=5m, OB=12m) and the rod (AB=13m) define the geometry of the suspension system.
Pythagorean Theorem: The 5-12-13 relationship is a Pythagorean triple ($5^2 + 12^2 = 13^2$), confirming that the angle formed by the two strings at the suspension point O ($\angle AOB$) is precisely $90^\circ$.
Trigonometry: Used to determine the sine and cosine of the angles within the right-angled triangle AOB, which are needed for applying Lami’s theorem as formulated.
Lami’s Theorem: Applicable because three concurrent forces (W, T₁, T₂) are in equilibrium at point O. It relates the magnitude of each force to the sine of the angle between the other two forces. The condition that “O hangs immediately below the middle point” likely ensures the orientation needed for the specific application of Lami’s theorem shown in the solution.

Real-World Applications:

Structural Engineering: Analyzing forces in simple cable-stayed or suspension systems, like hanging a sign or equipment from two points on a beam.
Lifting and Rigging: Calculating tensions in slings when using a spreader bar (the rod) to lift an object, especially when the sling lengths create specific angles.
Art Installations/Displays: Designing suspension systems for exhibits or artworks hanging from fixed structures.
Physics Education: A classic problem demonstrating equilibrium, vector forces, and the application of geometric principles (Pythagorean theorem) and equilibrium theorems (Lami’s theorem).

Why It Works:
The fixed lengths establish a right-angle configuration between the strings at the point of suspension. For the body to hang in equilibrium, the upward pulling forces from the two strings (T₁ and T₂) must perfectly counteract the downward force of gravity (W). The 5-12-13 geometry determines the exact angles involved. Lami’s theorem provides a mathematical shortcut to find the necessary tensions in each string by relating them to the weight and the sines of the angles between the forces. The tension is distributed unevenly because the strings have different lengths and therefore different angles relative to the vertical (when O is below the midpoint).