A body of mass M is kept on a rough horizontal surface (friction coefficient `=μ). A person is trying to pull the body by applying a horizontal force but the body is not moving. The force by the horizontal surface on the surface of the body is F, where

Options

A. `F=mg`
B. `F=mu Mg`
C. `MgleFleMgsqrt(1+mu^2)`
D. `MggeFgeMgsqrt(1-mu^(2))`

Answer

To solve the problem step by step, we can summarize the situation and apply the principles of physics related to forces, friction, and equilibrium.

### Step 1: Identify the forces acting on the body
The body of mass \( M \) is resting on a rough horizontal surface. The forces acting on the body are:
- The weight of the body (\( W = mg \)), acting downwards.
- The normal force (\( N \)), acting upwards from the surface.
- The applied horizontal force (\( F' \)), trying to pull the body.
- The frictional force (\( F_R \)), opposing the motion, which acts horizontally in the opposite direction to \( F' \).

### Step 2: Analyze vertical forces
Since the body is at rest and not moving vertically, we can apply the equilibrium condition in the vertical direction:
\[ N = mg \]
This means the normal force \( N \) exerted by the surface is equal to the weight of the body.

### Step 3: Analyze horizontal forces
Since the body is not moving horizontally, the applied force \( F' \) must be equal to the frictional force \( F_R \):
\[ F' = F_R \]
The maximum static frictional force can be expressed as:
\[ F_R \leq \mu N \]
Substituting \( N \) from Step 2:
\[ F_R \leq \mu mg \]
Thus, we can say:
\[ F' \leq \mu mg \]

### Step 4: Determine the resultant force from the surface
The force \( F \) exerted by the surface on the body is the vector sum of the normal force and the frictional force. Since these forces are perpendicular to each other, we can use the Pythagorean theorem:
\[ F = \sqrt{F_R^2 + N^2} \]
Substituting the expressions for \( F_R \) and \( N \):
\[ F = \sqrt{(\mu mg)^2 + (mg)^2} \]
This simplifies to:
\[ F = mg \sqrt{\mu^2 + 1} \]

### Step 5: Establish the limits for \( F \)
From the analysis, we know:
- The minimum value of \( F \) occurs when \( F' = 0 \) (no applied force), which gives \( F = mg \).
- The maximum value of \( F \) occurs when \( F' \) is at its maximum static friction limit, which leads to:
\[ F \leq mg \sqrt{1 + \mu^2} \]

### Final Result
Thus, we can conclude:
\[ mg \leq F \leq mg \sqrt{1 + \mu^2} \]

### Answer
The force \( F \) by the horizontal surface on the surface of the body is such that:
\[ F \text{ is greater than or equal to } mg \text{ and less than or equal to } mg \sqrt{1 + \mu^2} \]