2 sphere each of mass 8 root 3 kg and radius 5 cm lie in contact in a hemispherical. Bowl of radius 40 cm. There is no friction. The contact force exerted by one sphere to another

To determine the contact force exerted by one sphere on the other in this scenario, we need to analyze the forces acting on the spheres due to gravity and their arrangement within the hemispherical bowl. Let's break this down step by step.

Understanding the Setup

We have two spheres, each with a mass of \(8\sqrt{3}\) kg and a radius of 5 cm, resting in a hemispherical bowl with a radius of 40 cm. Since the spheres are in contact and there is no friction, the only forces acting on them are gravitational forces and the contact force between them.

Calculating the Weight of Each Sphere

The weight of an object is calculated using the formula:

• Weight (W) = mass (m) × gravitational acceleration (g)

Here, the gravitational acceleration \(g\) is approximately \(9.81 \, \text{m/s}^2\). Thus, the weight of each sphere can be calculated as follows:

• Weight of one sphere = \(8\sqrt{3} \, \text{kg} \times 9.81 \, \text{m/s}^2\)
• Weight of one sphere ≈ \(8 \times 1.732 \times 9.81 \approx 137.3 \, \text{N}\)

Analyzing the Forces at Play

Since the spheres are in contact and resting in a bowl, we need to consider the vertical and horizontal components of the forces acting on them. The weight of each sphere acts downwards, while the contact force acts along the line connecting the centers of the two spheres.

Geometry of the Arrangement

The centers of the spheres are separated by a distance equal to the sum of their radii, which is:

• Distance between centers = \(5 \, \text{cm} + 5 \, \text{cm} = 10 \, \text{cm} = 0.1 \, \text{m}\)

Calculating the Contact Force

In the absence of friction, the contact force must balance the vertical components of the weight of the spheres. Since both spheres are identical, they will exert equal and opposite forces on each other. The contact force \(F_c\) can be derived from the equilibrium of forces acting on one sphere:

• Let \(F_c\) be the contact force.
• In vertical equilibrium, the sum of the vertical forces must equal zero:
• Weight of the sphere = \(F_c \cdot \sin(\theta)\), where \(\theta\) is the angle of the line connecting the centers with respect to the vertical.

To find \(\theta\), we can use the geometry of the situation. The vertical distance from the center of the sphere to the bottom of the bowl is \(40 \, \text{cm} - 5 \, \text{cm} = 35 \, \text{cm} = 0.35 \, \text{m}\). The horizontal distance is \(0.1 \, \text{m}\). Thus, we can find \(\tan(\theta)\):

• \(\tan(\theta) = \frac{0.1}{0.35}\)
• \(\theta \approx \tan^{-1}(0.2857) \approx 15.9^\circ\)

Now, substituting back into the equilibrium equation:

• Weight = \(F_c \cdot \sin(15.9^\circ)\)
• Solving for \(F_c\): \(F_c = \frac{W}{\sin(15.9^\circ)}\)

Final Calculation

Using the weight we calculated earlier:

• Weight ≈ \(137.3 \, \text{N}\)
• Calculating \(F_c\): \(F_c = \frac{137.3}{\sin(15.9^\circ)}\)
• Using \(\sin(15.9^\circ) \approx 0.273\): \(F_c \approx \frac{137.3}{0.273} \approx 503.3 \, \text{N}\)

Thus, the contact force exerted by one sphere on the other is approximately \(503.3 \, \text{N}\). This force ensures that the spheres remain in equilibrium within the bowl, balancing their weights and maintaining their positions.