In children's park there was a slide to be made by contract. By mistake, the person who had taken the contract made the coefficient of friction of the slide as high as `1//3`. Now, the fun is that the child expecting to slide down the incline will stop somewhere in between. Find the angle `theta` with the horizontal at which he will stop on the incline.(Assume negligible frictional losses)

Options

A. `45^@`
B. `37^@`
C. `53^@`
D. `60^@`

Answer

To solve the problem, we need to find the angle \( \theta \) at which a child will stop on an incline due to the effects of friction. Here's a step-by-step solution:

### Step 1: Understand the Forces Acting on the Child
When a child is on the slide, two main forces act on him:
1. The gravitational force \( \mathbf{F_g} = mg \) acting downward.
2. The normal force \( \mathbf{N} \) acting perpendicular to the surface of the slide.

### Step 2: Break Down the Gravitational Force
The gravitational force can be resolved into two components:
- Parallel to the incline: \( F_{\text{parallel}} = mg \sin \theta \)
- Perpendicular to the incline: \( F_{\text{perpendicular}} = mg \cos \theta \)

### Step 3: Apply the Condition for Equilibrium
For the child to stop on the incline, the frictional force must balance the component of gravity acting down the incline. The frictional force \( F_{\text{friction}} \) can be expressed as:
\[ F_{\text{friction}} = \mu N \]
where \( \mu \) is the coefficient of friction.

### Step 4: Express the Normal Force
The normal force \( N \) is equal to the perpendicular component of the gravitational force:
\[ N = mg \cos \theta \]

### Step 5: Set Up the Equation
Since the child is not sliding down, we set the frictional force equal to the gravitational component acting down the incline:
\[ \mu N = mg \sin \theta \]

Substituting the expression for \( N \):
\[ \mu (mg \cos \theta) = mg \sin \theta \]

### Step 6: Simplify the Equation
We can cancel \( mg \) from both sides (assuming \( m \neq 0 \)):
\[ \mu \cos \theta = \sin \theta \]

### Step 7: Rearrange to Find \( \tan \theta \)
Rearranging gives us:
\[ \tan \theta = \frac{\mu}{1} = \mu \]

### Step 8: Substitute the Given Coefficient of Friction
We are given that \( \mu = \frac{1}{3} \):
\[ \tan \theta = \frac{1}{3} \]

### Step 9: Calculate \( \theta \)
To find \( \theta \), we take the arctangent:
\[ \theta = \tan^{-1} \left( \frac{1}{3} \right) \]

Using a calculator:
\[ \theta \approx 18.43^\circ \]

### Final Answer
The angle \( \theta \) with the horizontal at which the child will stop on the incline is approximately \( 18.43^\circ \).

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