A solid sphere of mass 0.50 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?

Explanation

To solve the problem of determining the maximum force that can be applied at the highest point of a solid sphere without causing it to slip, we can follow these steps:

### Step 1: Identify the parameters
- Mass of the sphere (m) = 0.50 kg
- Coefficient of static friction (μ) = 2/7
- Gravitational acceleration (g) = 9.81 m/s² (for calculations, we can use g ≈ 10 m/s² for simplicity)

### Step 2: Calculate the normal force (N)
The normal force acting on the sphere is equal to its weight:
\[ N = mg = 0.50 \, \text{kg} \times 10 \, \text{m/s}^2 = 5 \, \text{N} \]

### Step 3: Calculate the maximum static friction force (f_max)
The maximum static friction force can be calculated using the formula:
\[ f_{\text{max}} = \mu N \]
Substituting the values:
\[ f_{\text{max}} = \frac{2}{7} \times 5 \, \text{N} = \frac{10}{7} \, \text{N} \approx 1.43 \, \text{N} \]

### Step 4: Set up the equations of motion
When a force \( F \) is applied at the highest point of the sphere, it will cause both translational and rotational motion. The translational acceleration \( a \) of the center of mass and the angular acceleration \( \alpha \) are related through the radius \( r \) of the sphere:
\[ a = \alpha r \]

### Step 5: Apply Newton's second law for translation
The net force acting on the sphere in the horizontal direction is:
\[ F - f = ma \]
Where \( f \) is the friction force opposing the motion.

### Step 6: Apply the torque equation
The torque \( \tau \) about the center of mass due to the applied force \( F \) and the friction force \( f \) is:
\[ \tau = F \cdot r - f \cdot r = I \alpha \]
For a solid sphere, the moment of inertia \( I \) is:
\[ I = \frac{2}{5} m r^2 \]
Thus, we can write:
\[ F \cdot r - f \cdot r = \frac{2}{5} m r^2 \cdot \frac{a}{r} \]
This simplifies to:
\[ F - f = \frac{2}{5} ma \]

### Step 7: Solve the equations
Now we have two equations:
1. \( F - f = ma \) (1)
2. \( F - f = \frac{2}{5} ma \) (2)

From (1) and (2), we can equate the right-hand sides:
\[ ma = \frac{2}{5} ma \]
This implies:
\[ F - f = \frac{2}{5} ma \]

### Step 8: Relate the forces
From the static friction condition, we know that:
\[ f \leq f_{\text{max}} \]
Substituting \( f_{\text{max}} = \frac{10}{7} \, \text{N} \):
\[ F - \frac{10}{7} = \frac{2}{5} ma \]

### Step 9: Substitute \( a \)
Using \( a = \frac{5}{7} F \) from the earlier equations, we can substitute back:
\[ F - \frac{10}{7} = \frac{2}{5} m \left(\frac{5}{7} F\right) \]
This leads to:
\[ F - \frac{10}{7} = \frac{2}{7} F \]

### Step 10: Solve for \( F \)
Rearranging gives:
\[ F - \frac{2}{7} F = \frac{10}{7} \]
\[ \frac{5}{7} F = \frac{10}{7} \]
\[ F = 2 \, \text{N} \]

### Conclusion
The maximum force that can be applied at the highest point in the horizontal direction without causing the sphere to slip is:
\[ F_{\text{max}} = 2 \, \text{N} \]