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

Options

A. 25 N
B. 75 N
C. 87 N
D. 100 N

Explanation

To solve the problem step by step, we will use the concepts of vector components, specifically for a force inclined at an angle.

### Step 1: Identify the components of the force
Given:
- The angle of inclination, θ = 60°.
- The horizontal component of the force, \( F_H = 50 \, \text{N} \).

### Step 2: Use the relationship between the components and the angle
The horizontal component of the force can be expressed using the cosine of the angle:
\[ F_H = F \cos(θ) \]
Substituting the known values:
\[ 50 = F \cos(60°) \]

### Step 3: Calculate \( \cos(60°) \)
We know that:
\[ \cos(60°) = \frac{1}{2} \]
Substituting this value into the equation:
\[ 50 = F \cdot \frac{1}{2} \]

### Step 4: Solve for the magnitude of the force \( F \)
To find \( F \), multiply both sides by 2:
\[ F = 50 \cdot 2 = 100 \, \text{N} \]

### Step 5: Calculate the vertical component of the force
The vertical component of the force can be expressed using the sine of the angle:
\[ F_V = F \sin(θ) \]
Substituting the known values:
\[ F_V = 100 \sin(60°) \]
We know that:
\[ \sin(60°) = \frac{\sqrt{3}}{2} \]
So,
\[ F_V = 100 \cdot \frac{\sqrt{3}}{2} = 50\sqrt{3} \, \text{N} \]

### Final Answers:
- Magnitude of the force \( F = 100 \, \text{N} \)
- Vertical component of the force \( F_V = 50\sqrt{3} \, \text{N} \)