Problem Statement
A smooth sphere of weight ‘W’ is supported in contact with a smooth vertical wall by a string fastened to a point (C) on its surface, the end being attached to a point (A) on the wall. If the length of the string (AC) is equal to the radius (R) of sphere, find the tension (T) in the string and the reaction (RB) on the wall.
Fig. : Forces acting on the sphere
Step-by-Step Solution
Step 1: Identify Forces and Equilibrium Point
The sphere is in equilibrium under the action of three forces. For equilibrium, their lines of action must be concurrent, intersecting at the center of the sphere (O):
- Weight (W): Acts vertically downwards through the center O.
- Wall Reaction (RB): Since the wall is smooth, the reaction at the contact point B is perpendicular (normal) to the wall, acting horizontally. Its line of action passes through O.
- String Tension (T): Acts along the string AC. For the forces to be concurrent at O, the line of action of the string must pass through O, meaning points A, C, and O are collinear.
Step 2: Geometric Analysis to Find Angles
Consider the right-angled triangle AOB formed by the wall anchor point (A), the sphere’s center (O), and the wall contact point (B). The right angle is at B.
- OB = Radius of the sphere = R.
- Since A, C, O are collinear, AC (string length) = R, and CO (radius) = R, the length AO = AC + CO = R + R = 2R.
Let \(\theta\) be the angle between the string direction (AO) and the horizontal reaction direction (OB), i.e., \( \theta = \angle AOB \). Using trigonometry in triangle AOB:
This means the string Tension (T) acts at an angle of \(60^\circ\) above the horizontal.
Step 3: Apply Lami’s Theorem at Center O
The three forces W, T, and RB are concurrent at O and in equilibrium. We need the angles between the forces when placed tail-to-tail, as used in the provided solution’s application of Lami’s Theorem:
- Angle between RB (horizontal) and W (vertical down) = \(90^\circ\). (This angle is opposite to T).
- Angle between W (vertical down) and T (acting 60° above horizontal) = \(90^\circ + (90^\circ – 60^\circ) = 120^\circ\). (This angle is opposite to RB, corresponding to \(180^\circ – \theta\) in the text’s formula structure).
- Angle between T (acting 60° above horizontal) and RB (horizontal) = \(180^\circ – 60^\circ = 120^\circ\). (This angle is opposite to W, corresponding to \(90^\circ + \theta\) in the text’s formula structure. *Correction*: text uses 180-theta opposite W, 90+theta opposite RB. Let’s match the text’s assignments).
Using the angles as assigned in the provided solution’s Lami’s application (\(\theta = 60^\circ\)):
- Angle opposite T = \(90^\circ\)
- Angle opposite RB = \(90^\circ + \theta = 90^\circ + 60^\circ = 150^\circ\)
- Angle opposite W = \(180^\circ – \theta = 180^\circ – 60^\circ = 120^\circ\)
Applying Lami’s theorem:
Step 4: Calculate Tension T and Reaction RB
Calculate Tension T:
Calculate Reaction RB:
Final Result
Reaction on the wall (RB) = \( \mathbf{\frac{W}{\sqrt{3}}} \approx 0.577 \, W \)
Conceptual Explanation & Applications
Core Concepts:
- Static Equilibrium: The sphere remains stationary, indicating that the net force acting on it is zero. The vector sum of the weight (W), the string tension (T), and the wall reaction (RB) equals zero: \( \vec{W} + \vec{T} + \vec{R_B} = \vec{0} \).
- Concurrent Forces: For equilibrium, the lines of action of the three forces acting on the sphere must intersect at a single point, which is the center (O) in this case.
- Smooth Surfaces: The term ‘smooth’ implies negligible friction. Therefore, the reaction force (\(R_B\)) exerted by the wall on the sphere is entirely normal (perpendicular) to the wall surface.
- Geometric Constraints: The condition that the string length equals the sphere’s radius fixes the geometry of the setup. This leads to a specific relationship (forming a 30-60-90 triangle AOB) and defines the angles between the forces.
- Trigonometry & Lami’s Theorem: Trigonometry is used to determine the angles from the geometric constraints. Lami’s Theorem then provides a direct relationship between the magnitudes of the three concurrent forces and the sines of the angles opposite them, allowing calculation of the unknowns T and RB based on the known weight W.
Real-World Applications:
- Mechanical Systems: Analyzing forces on rollers, spheres, or cylinders used in bearings or support structures where they contact other surfaces.
- Structural Analysis: Modeling forces in situations where curved elements (like arches or domes) are supported or restrained by tension members and abutments.
- Physics Education: Serving as a standard example problem to teach concepts of equilibrium, normal forces, tension, and the application of Lami’s Theorem or force resolution.
- Material Storage/Handling: Basic model for understanding forces when storing spherical objects (like tanks or pipes) against retaining walls or supports.
Why It Works:
The equilibrium of the sphere demands a precise balance of forces. The weight pulls vertically down. The wall pushes horizontally. The string pulls upwards and inwards. The specific geometry dictates the exact angle of the string tension needed to achieve this balance. The vertical component of the tension must equal the weight (W), while the horizontal component must equal the wall reaction (RB). Lami’s theorem encapsulates this force balance trigonometrically, leveraging the angles derived from the R=String Length condition to solve for the tension and reaction forces relative to the weight.
