Problem Statement
A machine weighing 1500 N is supported by two chains attached to some point on the machine. One of these ropes (OA) goes to an eye bolt in the wall and is inclined \(30^\circ\) to the horizontal and the other (OB) goes to a hook in the ceiling and is inclined at \(45^\circ\) to the horizontal. Find the tensions in the two chains.
O
Step-by-Step Solution
Forces and Angles in Equilibrium
- The junction point O is in equilibrium under three concurrent forces.
- Force 1 (W): Weight of the machine = 1500 N (acting vertically downwards).
- Force 2 (\(T_1\)): Tension in chain OA, inclined \(30^\circ\) to the horizontal.
- Force 3 (\(T_2\)): Tension in chain OB, inclined \(45^\circ\) to the horizontal.
- Angles between forces (for Lami’s Theorem, based on diagram analysis and provided solution text):
- Angle between \(T_1\) and \(T_2\) (opposite W) = \( (180^\circ – 30^\circ – 45^\circ) = 105^\circ \).
- Angle between \(T_2\) and W (opposite \(T_1\)) = \( 90^\circ + (90^\circ – 45^\circ) = 90^\circ + 45^\circ = 135^\circ \).
- Angle between \(T_1\) and W (opposite \(T_2\)) = \( 90^\circ + 30^\circ = 90^\circ + 30^\circ = 120^\circ \).
- Lami’s theorem is applicable.
Step 1: Apply Lami’s Theorem
At point O, the three forces W, \(T_1\), and \(T_2\) are concurrent and in equilibrium. Applying Lami’s theorem as presented in the provided solution:
Step 2: Calculate Tension T₁
From Lami’s theorem relation:
Substitute approximate values from the provided text (\( \sin(135^\circ) \approx 0.707 \), \( \sin(105^\circ) \approx 0.965 \)):
Step 3: Calculate Tension T₂
Similarly, from Lami’s theorem relation:
Substitute approximate values from the provided text (\( \sin(120^\circ) \approx 0.866 \), \( \sin(105^\circ) \approx 0.965 \)):
Final Result
Tension in chain OB (\(T_2\)) = \( \mathbf{1346.11 \, N} \)
Conceptual Explanation & Applications
Core Concepts:
- Static Equilibrium: The junction point O, where the chains meet the machine, is stationary. This implies that the vector sum of all forces acting at this point (the downward weight W, and the upward tensions T₁ and T₂) is zero: \( \vec{W} + \vec{T_1} + \vec{T_2} = \vec{0} \).
- Concurrent Forces: All three forces (W, T₁, T₂) act through the single point O.
- Lami’s Theorem: A theorem applicable for determining unknown forces when exactly three concurrent forces are in equilibrium. It relates the magnitude of each force to the sine of the angle directly opposite to it (i.e., the angle between the other two forces): \( \frac{F_1}{\sin \alpha} = \frac{F_2}{\sin \beta} = \frac{F_3}{\sin \gamma} \).
- Angle Determination: Correctly finding the angles between the pairs of forces is essential for applying Lami’s theorem. These angles are derived from the known inclinations of the chains relative to the horizontal and the vertical direction of the weight.
Real-World Applications:
- Structural Engineering: Analyzing forces in simple cable-stayed structures, suspension systems for walkways or platforms.
- Mechanical Engineering: Designing lifting equipment like engine hoists, crane supports where a load is shared between multiple angled cables or chains.
- Construction & Rigging: Planning lifts of heavy machinery or building components, ensuring support cables are not overloaded.
- Physics Education: Demonstrating vector equilibrium principles using force tables or simulations.
Why It Works:
The equilibrium condition at point O means that the upward components of the tensions \(T_1\) and \(T_2\) must balance the downward weight W, and the horizontal components of \(T_1\) and \(T_2\) must cancel each other out. Lami’s theorem elegantly encapsulates these balance conditions through trigonometric relationships derived from the geometry of the force vectors forming a closed triangle. It allows for a direct calculation of the unknown tensions without explicitly resolving each force into horizontal and vertical components.
