If the normal force is doubled and limiting frictional force is maintained at the same value, the coefficient of friction

Options

A. remains uncharged
B. becomes halved
C. gets doubled
D. gets quadrupled

Answer

To solve the problem, we need to understand the relationship between the normal force, frictional force, and the coefficient of friction. Let's break it down step by step.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The frictional force (F_r) can be expressed in terms of the coefficient of friction (μ) and the normal force (N) using the formula:
\[ F_r = \mu \cdot N \]
This is our initial equation.

2. **Given Conditions**:
- The normal force is doubled: \( N' = 2N \)
- The limiting frictional force remains constant: \( F_r' = F_r \)

3. **Expressing the New Coefficient of Friction**:
We need to find the new coefficient of friction (μ') when the normal force is doubled. We can express the new coefficient of friction as:
\[ \mu' = \frac{F_r'}{N'} \]

4. **Substituting the Known Values**:
Since \( F_r' = F_r \) and \( N' = 2N \), we can substitute these into the equation:
\[ \mu' = \frac{F_r}{2N} \]

5. **Relating to the Original Coefficient of Friction**:
From our initial equation, we know that:
\[ \mu = \frac{F_r}{N} \]
Now, substituting this into the equation for μ':
\[ \mu' = \frac{F_r}{2N} = \frac{\mu \cdot N}{2N} = \frac{\mu}{2} \]

6. **Conclusion**:
Thus, the new coefficient of friction (μ') when the normal force is doubled while keeping the limiting frictional force constant is:
\[ \mu' = \frac{\mu}{2} \]

### Final Answer:
The coefficient of friction becomes half of the original coefficient of friction.

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