A boy wants to hold a 50 kg box on a snow covered hill. The hill makes an angle of `30^(@)` with the horizontal. What force the boy must exert parallel to the slope ?

Options

A. 25 N
B. 245 N
C. 424 N
D. 490 N

Answer

To solve the problem, we need to determine the force that the boy must exert parallel to the slope to hold the box in place. We can break this down into steps:

### Step 1: Identify the Forces Acting on the Box
The forces acting on the box include:
1. The gravitational force acting downward (weight of the box).
2. The normal force acting perpendicular to the slope.
3. The force exerted by the boy acting parallel to the slope.

### Step 2: Calculate the Weight of the Box
The weight (W) of the box can be calculated using the formula:
\[ W = m \cdot g \]
where:
- \( m = 50 \, \text{kg} \) (mass of the box)
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity)

Calculating the weight:
\[ W = 50 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 490.5 \, \text{N} \]

### Step 3: Resolve the Weight into Components
Since the hill is inclined at an angle of \( 30^\circ \), we need to resolve the weight into two components:
1. Component parallel to the slope (\( W_{\parallel} \))
2. Component perpendicular to the slope (\( W_{\perpendicular} \))

The component of weight acting parallel to the slope is given by:
\[ W_{\parallel} = W \cdot \sin(\theta) \]
where \( \theta = 30^\circ \).

Calculating \( W_{\parallel} \):
\[ W_{\parallel} = 490.5 \, \text{N} \cdot \sin(30^\circ) \]
Since \( \sin(30^\circ) = 0.5 \):
\[ W_{\parallel} = 490.5 \, \text{N} \cdot 0.5 = 245.25 \, \text{N} \]

### Step 4: Determine the Force Exerted by the Boy
To hold the box in place, the boy must exert a force equal to the component of the weight acting down the slope. Therefore, the force exerted by the boy (\( F \)) is:
\[ F = W_{\parallel} = 245.25 \, \text{N} \]

### Conclusion
The force that the boy must exert parallel to the slope to hold the box in place is approximately \( 245.25 \, \text{N} \).

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