A solid sphere of mass `3 kg` is kept on a horizontal surface. The coefficient of static friction between the surfaces in contact is `2//7`. What maximum force can be applied at the highest point in the horizontal direction so that the sphere does not slip on the surface? (in `N`)

Explanation

To solve the problem, we will follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Mass of the sphere, \( m = 3 \, \text{kg} \)
- Coefficient of static friction, \( \mu = \frac{2}{7} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the Maximum Force of Static Friction:**
The maximum force of static friction \( F_f \) can be calculated using the formula:
\[ F_f = \mu \cdot m \cdot g \]
Substituting the values:
\[ F_f = \frac{2}{7} \cdot 3 \cdot 10 = \frac{60}{7} \, \text{N} \]

3. **Apply Newton's Second Law for Linear Motion:**
When a force \( F \) is applied at the highest point of the sphere, the net force acting on the sphere in the horizontal direction is:
\[ F - F_f = m \cdot a \]
Where \( a \) is the acceleration of the center of mass of the sphere.

4. **Apply Newton's Second Law for Rotational Motion:**
The torque \( \tau \) about the center of mass due to the applied force \( F \) and the friction force \( F_f \) can be expressed as:
\[ \tau = F \cdot r - F_f \cdot r = I \cdot \alpha \]
Here, \( I \) is the moment of inertia of the sphere, which is given by:
\[ I = \frac{2}{5} m r^2 \]
The angular acceleration \( \alpha \) is related to the linear acceleration \( a \) by:
\[ \alpha = \frac{a}{r} \]
Therefore, substituting \( \alpha \) into the torque equation gives:
\[ F \cdot r - F_f \cdot r = \frac{2}{5} m r^2 \cdot \frac{a}{r} \]
Simplifying this, we get:
\[ F - F_f = \frac{2}{5} m a \]

5. **Set Up the System of Equations:**
We now have two equations:
- From linear motion: \( F - F_f = m a \) (Equation 1)
- From rotational motion: \( F - F_f = \frac{2}{5} m a \) (Equation 2)

6. **Solve the Equations:**
From Equation 1, we can express \( a \):
\[ a = \frac{F - F_f}{m} \]
Substitute this into Equation 2:
\[ F - F_f = \frac{2}{5} m \cdot \frac{F - F_f}{m} \]
Simplifying gives:
\[ F - F_f = \frac{2}{5} (F - F_f) \]
Rearranging leads to:
\[ 5(F - F_f) = 2(F - F_f) \]
This implies:
\[ 3F = 3F_f \implies F = F_f \]

7. **Substituting the Maximum Friction Force:**
Since \( F \) must be less than or equal to \( F_f \):
\[ F \leq \frac{60}{7} \, \text{N} \]

### Final Answer:
Thus, the maximum force that can be applied at the highest point in the horizontal direction so that the sphere does not slip on the surface is:
\[ \boxed{\frac{60}{7} \, \text{N}} \approx 8.57 \, \text{N} \]