A force is inclined at an angle of `60^(@)` from the horizontal. If the horizontal component of the force is 40N,calculate the vertical component.

Explanation

To solve the problem step by step, we will use the concepts of vector components and trigonometry.

### Step 1: Understand the components of the force
The force \( F \) can be broken down into two components:
- Horizontal component \( F_x = F \cos \theta \)
- Vertical component \( F_y = F \sin \theta \)

Where \( \theta \) is the angle of inclination from the horizontal.

### Step 2: Identify the given values
From the problem, we know:
- The angle \( \theta = 60^\circ \)
- The horizontal component \( F_x = 40 \, \text{N} \)

### Step 3: Relate the horizontal component to the force
Using the formula for the horizontal component:
\[ F_x = F \cos(60^\circ) \]
Substituting the known values:
\[ 40 = F \cos(60^\circ) \]
Since \( \cos(60^\circ) = \frac{1}{2} \), we can rewrite the equation:
\[ 40 = F \cdot \frac{1}{2} \]

### Step 4: Solve for the magnitude of the force \( F \)
To find \( F \), we multiply both sides by 2:
\[ F = 40 \times 2 = 80 \, \text{N} \]

### Step 5: Calculate the vertical component
Now, we can find the vertical component using the formula:
\[ F_y = F \sin(60^\circ) \]
Substituting the value of \( F \):
\[ F_y = 80 \sin(60^\circ) \]
Since \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \), we have:
\[ F_y = 80 \cdot \frac{\sqrt{3}}{2} \]
\[ F_y = 40\sqrt{3} \, \text{N} \]

### Final Answer
The vertical component of the force is \( 40\sqrt{3} \, \text{N} \).

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