For a block kept on a table, the maximum force of static friction is `F_(1)` . If the set-up is taken to moon, the force is `F_(2)`. Then

Options

A. `F_(1)=F_(2)`
B. `F_(1)ltF_(2)`
C. `F_(2)ltF_(1)`
D. none of these

Explanation

To solve the problem, we need to understand the relationship between the maximum force of static friction and the gravitational force acting on the block. The maximum force of static friction can be calculated using the formula:

\[ F_{\text{max}} = \mu_s \cdot N \]

where:
- \( F_{\text{max}} \) is the maximum static friction force,
- \( \mu_s \) is the coefficient of static friction,
- \( N \) is the normal force acting on the object.

### Step-by-Step Solution:

1. **Identify the Forces on Earth:**
- When the block is on the table on Earth, the normal force \( N \) is equal to the weight of the block, which is given by \( N = mg \), where \( m \) is the mass of the block and \( g \) is the acceleration due to gravity on Earth (approximately \( 9.8 \, \text{m/s}^2 \)).
- Therefore, the maximum static friction force \( F_1 \) on Earth can be expressed as:
\[ F_1 = \mu_s \cdot mg \]

2. **Identify the Forces on the Moon:**
- When the same block is taken to the Moon, the normal force \( N \) will now be \( N = mg' \), where \( g' \) is the acceleration due to gravity on the Moon (approximately \( 1.6 \, \text{m/s}^2 \)).
- Thus, the maximum static friction force \( F_2 \) on the Moon can be expressed as:
\[ F_2 = \mu_s \cdot mg' \]

3. **Compare the Forces:**
- Now, we can compare \( F_1 \) and \( F_2 \):
\[ F_1 = \mu_s \cdot mg \quad \text{and} \quad F_2 = \mu_s \cdot mg' \]
- Since \( g > g' \) (because \( g \approx 9.8 \, \text{m/s}^2 \) and \( g' \approx 1.6 \, \text{m/s}^2 \)), we can conclude that:
\[ F_1 > F_2 \]

4. **Final Conclusion:**
- Therefore, the maximum force of static friction on Earth \( F_1 \) is greater than the maximum force of static friction on the Moon \( F_2 \):
\[ F_1 > F_2 \]