A force is inclined at `30^(@)` to the horizontal. If its rectangular component in the horizontal direction is 50N, find the magnitude of the force and its vertical component.

Answer

To solve the problem step by step, we will follow the process of breaking down the force into its components and using trigonometric relationships.

### Step 1: Identify the given information
- The angle of inclination (θ) = 30 degrees
- The horizontal component of the force (F_horizontal) = 50 N

### Step 2: Use the relationship for the horizontal component
The horizontal component of the force can be expressed using the cosine of the angle:
\[ F_{\text{horizontal}} = F \cdot \cos(\theta) \]

Substituting the known values:
\[ 50 = F \cdot \cos(30^\circ) \]

### Step 3: Calculate cos(30 degrees)
We know that:
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \]

### Step 4: Substitute cos(30 degrees) into the equation
\[ 50 = F \cdot \frac{\sqrt{3}}{2} \]

### Step 5: Solve for the magnitude of the force (F)
Rearranging the equation:
\[ F = \frac{50 \cdot 2}{\sqrt{3}} \]
\[ F = \frac{100}{\sqrt{3}} \]

### Step 6: Calculate the vertical component of the force
The vertical component can be expressed using the sine of the angle:
\[ F_{\text{vertical}} = F \cdot \sin(\theta) \]

### Step 7: Calculate sin(30 degrees)
We know that:
\[ \sin(30^\circ) = \frac{1}{2} \]

### Step 8: Substitute F and sin(30 degrees) into the equation
\[ F_{\text{vertical}} = \left(\frac{100}{\sqrt{3}}\right) \cdot \sin(30^\circ) \]
\[ F_{\text{vertical}} = \left(\frac{100}{\sqrt{3}}\right) \cdot \frac{1}{2} \]
\[ F_{\text{vertical}} = \frac{100}{2\sqrt{3}} \]
\[ F_{\text{vertical}} = \frac{50}{\sqrt{3}} \]

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

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