Problem 7 - Applied Mathematics

Question

A sphere of radius \(9 \mathrm{~cm}\) rests on a smooth inclined plane (angle \(30^{\circ}\) ). It is attached by a string fixed to a point on its surface to a point on the plane \(12 \mathrm{~cm}\) from the point of contact and on the same line of greatest slope. Find the tension. in the string if the weight of the sphere is \(100 \mathrm{~N}\).

Step-by-Step Solution

Answer

The tension in the string is approximately 62.5 N.

Step 1: Diagram and Forces

Start by drawing a diagram of the situation. Label the sphere of radius 9 cm, the inclined plane, the string, and the angles. Identify all the forces acting on the sphere: the weight (W = 100 N) acting vertically downward, the normal force (N) perpendicular to the plane, and the tension (T) in the string.

Step 2: Resolve Weight into Components

Resolve the weight of the sphere into two components: one parallel to the inclined plane (W \sin(30^\text{°})) and one perpendicular to the inclined plane (W \cos(30^\text{°}). Let's calculate these components:\[W \sin(30^\text{°}) = 100 \sin(30^\text{°}) = 100 \times 0.5 = 50 \text{ N} \]\[W \cos(30^\text{°}) = 100 \cos(30^\text{°}) = 100 \times \frac{\sqrt{3}}{2} \approx 86.6 \text{ N}\]

Step 3: Geometry Considerations

Determine the length and direction of the string. Given that the string is attached at 12 cm from the point of contact along the inclined plane, it forms a triangle with the inclined plane (base = 12 cm) and the radius (perpendicular height from the plane to center = 9 cm). Calculate the hypotenuse: \[\text{Hypotenuse} = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15 \text{ cm} \]

Step 4: Tension Components

Resolve the tension (T) into components along the incline and perpendicular to the incline. The tension components are: \[T \cos(\theta) \text{ parallel to plane and } T \sin(\theta) \text{ perpendicular to plane}.\]

Step 5: Balance Forces Perpendicular to Plane

For equilibrium, the sum of perpendicular forces must be zero. Hence, \[N = W \cos(30^\text{°}) + T \sin(\theta)\]

Step 6: Balance Forces Parallel to Plane

For equilibrium, the sum of parallel forces must be zero. Hence, \[T \cos(\theta) = W \sin(30^\text{°}) = 50 \text{ N}.\]From step 2 and 3, \[\cos(\theta) = \frac{12}{15} = 0.8,\theta = \text{ arccos}(0.8)\]Substitute \( \theta\)\[T = 50 N / \cos(\theta)\approx \frac{50}{0.8} = 62.5 \text{ N}\]

Key Concepts

physics of forces

Physics of forces deals with understanding how various forces interact and affect the motion or rest of objects. In the context of inclined plane problems, we encounter different forces like gravity, normal force, and tension. Gravity is always pulling the object downwards with a force equal to its weight, represented by the equation: \( W = mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity. Normal force, on the other hand, acts perpendicular to the surface of the inclined plane, counteracting a part of the gravitational force. Tension in a string also comes into play when objects are attached or restrained by strings or ropes. Understanding these forces is essential as they determine whether the object remains in equilibrium or moves.

tension in a string

Tension in a string refers to the pulling force transmitted along a string, rope, or chain when it is pulled tight by forces acting from opposite ends. In our problem, the sphere is attached to a string, which is fixed to a point on the inclined plane. To find the tension in the string, we focus on its components and the forces in equilibrium. The tension force can be split into two components: one along the incline and one perpendicular to it. Mathematically, the components are represented as follows:

• Parallel to the plane: \( T \cos(\theta) \)
• Perpendicular to the plane: \( T \sin(\theta) \)

By analyzing these components, we can determine how the tension in the string counteracts the gravitational force acting on the sphere.

geometry in mechanics

Geometry in mechanics involves using geometrical principles to understand and solve problems related to physical forces and motion. In our exercise, geometrical concepts help determine the relationship between the string length and the inclined plane. By understanding the triangle formed by the string, the inclined plane (base = 12 cm), and the radius of the sphere (perpendicular height = 9 cm), we calculate the hypotenuse using the Pythagorean theorem: \ \text{Hypotenuse} = \sqrt{12^2 + 9^2} = 15 \text{ cm} \. Geometry also helps us find angles and relationships that are crucial for resolving forces into components. It allows us to determine the angle \( \theta \) between the string and the inclined plane, which is essential for finding the correct tension force.

equilibrium of forces

Equilibrium of forces occurs when all the forces acting on an object balance each other, resulting in no net force and, consequently, no acceleration. In inclined plane problems, achieving equilibrium means both parallel and perpendicular forces must sum to zero. For the sphere at rest on the inclined plane, the forces include:

• Weight components: \( W \cos(30^\text{°}) \) and \( W \sin(30^\text{°}) \)
• Normal force: \( N \)
• Tension components: \( T \cos(\theta) \) and \( T \sin(\theta) \)

For the sphere to remain in equilibrium on the inclined plane, the equations we use are:

• For perpendicular forces: \( N = W \cos(30^\text{°}) + T \sin(\theta) \)
• For parallel forces: \( T \cos(\theta) = W \sin(30^\text{°}) \)

By solving these equations, we ensure the forces are balanced, allowing us to find the tension in the string accurately.