Let `F,F_N and f` denote the magnitudes of the contact force, normal force and the friction exerted by one surface on the other kept in contact. If none of these is zero,

Options

A. `FgtF_N`
B. `Fgt1`
C. F_N > f`
D. `F_N-fltFltF_N+f`.

Answer

To solve the problem, we need to analyze the relationships between the contact force (F), normal force (Fn), and frictional force (f) when none of these forces is zero.

### Step-by-Step Solution:

1. **Understanding the Forces**:
- The contact force (F) is the resultant of the normal force (Fn) and the frictional force (f).
- The normal force acts perpendicular to the surfaces in contact, while the frictional force acts parallel to the surfaces.

2. **Using Vector Addition**:
- Since Fn and f are perpendicular to each other, we can use the Pythagorean theorem to find the magnitude of the contact force:
\[ F = \sqrt{F_n^2 + f^2} \]

3. **Analyzing the Magnitudes**:
- From the equation \( F = \sqrt{F_n^2 + f^2} \), we can deduce that:
- \( F \) is always greater than both \( F_n \) and \( f \) because both \( F_n \) and \( f \) are squared and added together.

4. **Frictional Force Relation**:
- The frictional force can be expressed as:
\[ f = \mu F_n \]
- Here, \( \mu \) is the coefficient of friction, which can vary depending on the surfaces in contact.

5. **Comparing Normal Force and Frictional Force**:
- The relationship between \( F_n \) and \( f \) depends on the value of \( \mu \):
- If \( \mu > 1 \), then \( f \) can be greater than \( F_n \).
- If \( \mu < 1 \), then \( F_n \) can be greater than \( f \).
- Therefore, it is not always true that \( F_n > f \).

6. **Verifying the Options**:
- **Option 1**: \( F > F_n \) (True)
- **Option 2**: \( F > f \) (True)
- **Option 3**: \( F_n > f \) (Not always true)
- **Option 4**: \( F_n - f < F < F_n + f \) (True)
- This can be shown by manipulating the inequalities derived from the Pythagorean theorem.

### Conclusion:
The correct options are 1, 2, and 4.