To determine the angle of inclination of the string to the vertical when a smooth sphere is supported against a vertical wall, we can analyze the forces acting on the sphere. The setup involves a sphere of weight \( w \) resting against a wall, with a string attached to it and the wall. The length of the string is equal to the radius \( r \) of the sphere. Let's break this down step by step.
Understanding the Forces at Play
In this scenario, we have three main forces acting on the sphere:
- Weight (w): This force acts vertically downward through the center of the sphere.
- Normal Force (N): This force is exerted by the wall on the sphere, acting horizontally towards the sphere's center.
- Tension (T): This force is exerted by the string, acting along the length of the string at an angle θ to the vertical.
Setting Up the Geometry
Since the length of the string is equal to the radius of the sphere, we can visualize the situation. The string forms a triangle with the vertical line from the center of the sphere to the ground. The radius of the sphere is also the vertical distance from the center of the sphere to the point of contact with the wall.
Applying Trigonometry
Let’s denote the angle of inclination of the string to the vertical as θ. The length of the string (which is also the radius \( r \)) can be expressed in terms of its vertical and horizontal components:
- The vertical component is \( r \cos(θ) \).
- The horizontal component is \( r \sin(θ) \).
Equilibrium Conditions
For the sphere to be in equilibrium, the sum of the forces in both the horizontal and vertical directions must equal zero.
- In the vertical direction: The weight of the sphere must be balanced by the vertical component of the tension in the string. Thus, we have:
w = T \cos(θ)
- In the horizontal direction: The normal force exerted by the wall must balance the horizontal component of the tension in the string. Therefore:
N = T \sin(θ)
Relating Tension and Weight
From the first equation, we can express tension \( T \) in terms of weight \( w \):
T = w / cos(θ)
Substituting into the Horizontal Force Equation
Now, substituting this expression for \( T \) into the horizontal force equation gives us:
N = (w / cos(θ)) \sin(θ)
Using the Geometry of the Sphere
Since the string length equals the radius, we can also use the geometry of the triangle formed by the radius, the vertical line, and the string. The relationship between the angle and the radius can be expressed as:
r = r \cos(θ) + r \sin(θ)
Dividing through by \( r \) (assuming \( r \neq 0 \)) gives:
1 = cos(θ) + sin(θ)
Finding the Angle θ
To solve for \( θ \), we can rearrange this equation:
sin(θ) + cos(θ) = 1
Using the identity \( sin(θ) + cos(θ) = \sqrt{2} \sin(θ + 45°) \), we can find the angle that satisfies this equation. Setting \( \sqrt{2} \sin(θ + 45°) = 1 \) leads us to:
θ + 45° = 45°
Thus, \( θ = 0° \) or \( θ = 45° \). However, since the string must be inclined, we find that:
θ = 45°
Final Thoughts
The angle of inclination of the string to the vertical is \( 45° \). This means that the string is positioned such that it equally balances the vertical and horizontal forces acting on the sphere, allowing it to remain in equilibrium against the wall. This analysis showcases the interplay between geometry and physics in solving problems involving forces and angles.
