The vertical component of a force acting at `60^(@)` to the horizontal is `8.48` N . Find the force and the horizontal component .

Explanation

To solve the problem step by step, we will use the components of the force acting at an angle to the horizontal.

### Step 1: Identify the components of the force
Given that the vertical component of the force \( F_y \) is \( 8.48 \, \text{N} \) and the angle \( \theta \) is \( 60^\circ \), we can express the vertical component in terms of the total force \( F \):
\[ F_y = F \sin(60^\circ) \]

### Step 2: Substitute the known values
We know that \( F_y = 8.48 \, \text{N} \) and \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \). Therefore, we can write:
\[ 8.48 = F \cdot \frac{\sqrt{3}}{2} \]

### Step 3: Solve for the total force \( F \)
Rearranging the equation to solve for \( F \):
\[ F = \frac{8.48 \cdot 2}{\sqrt{3}} \]
Calculating this gives:
\[ F = \frac{16.96}{\sqrt{3}} \approx 9.8 \, \text{N} \]

### Step 4: Find the horizontal component of the force
Now, we need to find the horizontal component \( F_x \). The horizontal component can be expressed as:
\[ F_x = F \cos(60^\circ) \]
We know that \( \cos(60^\circ) = \frac{1}{2} \). Substituting the value of \( F \):
\[ F_x = 9.8 \cdot \frac{1}{2} = 4.9 \, \text{N} \]

### Final Results
Thus, the total force \( F \) is approximately \( 9.8 \, \text{N} \) and the horizontal component \( F_x \) is \( 4.9 \, \text{N} \).

### Summary
- Total Force \( F \approx 9.8 \, \text{N} \)
- Horizontal Component \( F_x = 4.9 \, \text{N} \)