A body of mass 8kg is in limiting equilibrium over an inclined plane of inclination `30^@`. If the inclination is made `60^@`, the minimum force required to prevent the body from sliding down is `(g = 10ms^(-2))`

Options

A. 80 N
B. `80/(sqrt3) N `
C. `(40 N)/(sqrt3)`
D. `40 sqrt(3) N`

Answer

To solve the problem step by step, we will analyze the forces acting on the body on the inclined plane and use the concept of limiting equilibrium.

### Step 1: Identify the forces acting on the body
When a body is on an inclined plane, the forces acting on it are:
- The weight of the body (mg), acting vertically downwards.
- The normal force (N), acting perpendicular to the inclined plane.
- The frictional force (f), acting parallel to the inclined plane, opposing the motion.

### Step 2: Resolve the weight into components
The weight of the body can be resolved into two components:
- Perpendicular to the incline: \( mg \cos \theta \)
- Parallel to the incline: \( mg \sin \theta \)

For our case, we have:
- Mass (m) = 8 kg
- Gravitational acceleration (g) = 10 m/s²
- Inclination angle (θ) = 30°

Calculating the weight:
\[ mg = 8 \, \text{kg} \times 10 \, \text{m/s}^2 = 80 \, \text{N} \]

### Step 3: Write the equilibrium condition for limiting friction
In limiting equilibrium, the frictional force (f) is given by:
\[ f = \mu N \]
Where \( \mu \) is the coefficient of friction. The normal force (N) is equal to the perpendicular component of the weight:
\[ N = mg \cos \theta \]

Thus, we have:
\[ f = \mu (mg \cos \theta) \]

### Step 4: Set up the equation for limiting equilibrium
At 30° inclination, the body is in limiting equilibrium, so:
\[ mg \sin 30° = f \]
Substituting for f:
\[ mg \sin 30° = \mu (mg \cos 30°) \]

### Step 5: Solve for the coefficient of friction (μ)
Substituting the values:
\[ 80 \sin 30° = \mu (80 \cos 30°) \]
\[ 80 \times \frac{1}{2} = \mu (80 \times \frac{\sqrt{3}}{2}) \]
Cancelling 80 from both sides:
\[ \frac{1}{2} = \mu \frac{\sqrt{3}}{2} \]
Thus:
\[ \mu = \frac{1}{\sqrt{3}} \]

### Step 6: Analyze the situation at 60° inclination
Now, when the inclination is changed to 60°, we need to find the minimum force (F) required to prevent the body from sliding down. The equation becomes:
\[ F + f = mg \sin 60° \]
Substituting for f:
\[ F + \mu (mg \cos 60°) = mg \sin 60° \]

### Step 7: Substitute the values
Substituting the known values:
\[ F + \frac{1}{\sqrt{3}} (80 \times \frac{1}{2}) = 80 \times \frac{\sqrt{3}}{2} \]
\[ F + \frac{40}{\sqrt{3}} = 40\sqrt{3} \]

### Step 8: Solve for F
Rearranging gives:
\[ F = 40\sqrt{3} - \frac{40}{\sqrt{3}} \]
To simplify, find a common denominator:
\[ F = \frac{40\sqrt{3} \cdot \sqrt{3}}{\sqrt{3}} - \frac{40}{\sqrt{3}} \]
\[ F = \frac{120 - 40}{\sqrt{3}} = \frac{80}{\sqrt{3}} \]

### Final Answer
The minimum force required to prevent the body from sliding down is:
\[ F = \frac{80}{\sqrt{3}} \, \text{N} \]