A force is inclined at `60^(@)` to the horizontal. If its rectangular component in the horizontal direction is `50 N`, then magnitude of the vertical components of force is approximately

Options

A. `25 N`
B. `84 N`
C. `87 N`
D. `90 N`

Explanation

To solve the problem, we need to find the vertical component of a force that is inclined at an angle of \(60^\circ\) to the horizontal, given that the horizontal component is \(50 \, \text{N}\).

### Step-by-Step Solution:

1. **Identify the Components of the Force**:
The force can be resolved into two components:
- Horizontal component (\(F_x\))
- Vertical component (\(F_y\))

Given:
\[ F_x = 50 \, \text{N} \]

2. **Use Trigonometric Relationships**:
The horizontal and vertical components of the force can be expressed in terms of the angle of inclination (\(\theta\)):
\[ F_x = F \cdot \cos(\theta) \]
\[ F_y = F \cdot \sin(\theta) \]
where \(F\) is the magnitude of the force and \(\theta = 60^\circ\).

3. **Calculate the Magnitude of the Force**:
Rearranging the equation for the horizontal component:
\[ F = \frac{F_x}{\cos(\theta)} \]
Substituting the known values:
\[ F = \frac{50 \, \text{N}}{\cos(60^\circ)} \]
Since \(\cos(60^\circ) = \frac{1}{2}\):
\[ F = \frac{50 \, \text{N}}{\frac{1}{2}} = 100 \, \text{N} \]

4. **Calculate the Vertical Component**:
Now, we can find the vertical component using the sine function:
\[ F_y = F \cdot \sin(60^\circ) \]
We know that \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\):
\[ F_y = 100 \, \text{N} \cdot \frac{\sqrt{3}}{2} \]
\[ F_y = 50\sqrt{3} \, \text{N} \]

5. **Approximate the Value**:
To get an approximate numerical value:
\[ F_y \approx 50 \cdot 1.732 \approx 86.6 \, \text{N} \]
Rounding this gives:
\[ F_y \approx 87 \, \text{N} \]

### Final Answer:
The magnitude of the vertical component of the force is approximately \(87 \, \text{N}\).