To solve the problem step by step, let's analyze the situation involving the mass, the string, and the forces acting on the system.
### Step 1: Understand the setup
We have a mass \( m = 10 \, \text{kg} \) hanging from a massless string of length \( L = 4 \, \text{m} \). A horizontal force \( F \) is applied at the midpoint of the string, causing the top half of the string to make an angle of \( 45^\circ \) with the vertical.
**Hint:** Visualize the setup by drawing a diagram showing the mass, the string, and the angle formed due to the applied force.
### Step 2: Analyze the forces acting on the mass
The mass experiences two forces:
1. The weight of the mass \( W = mg = 10 \times 9.81 \, \text{N} = 98.1 \, \text{N} \) acting downward.
2. The tension \( T \) in the string acting at an angle of \( 45^\circ \) with the vertical.
**Hint:** Remember that the tension in the string has both vertical and horizontal components.
### Step 3: Resolve the tension into components
The tension \( T \) can be resolved into two components:
- Vertical component: \( T \cos(45^\circ) \)
- Horizontal component: \( T \sin(45^\circ) \)
Since the angle is \( 45^\circ \), we know that \( \cos(45^\circ) = \sin(45^\circ) = \frac{1}{\sqrt{2}} \).
**Hint:** Use trigonometric identities to simplify the components of tension.
### Step 4: Set up the equations based on equilibrium
For vertical equilibrium:
\[ T \cos(45^\circ) = mg \]
Substituting the values:
\[ T \left(\frac{1}{\sqrt{2}}\right) = 98.1 \]
Thus,
\[ T = 98.1 \sqrt{2} \]
For horizontal equilibrium:
\[ F = T \sin(45^\circ) \]
Substituting the value of \( T \):
\[ F = (98.1 \sqrt{2}) \left(\frac{1}{\sqrt{2}}\right) \]
This simplifies to:
\[ F = 98.1 \]
**Hint:** Ensure you keep track of the units and the values used in calculations.
### Step 5: Calculate the final value of \( F \)
From the calculations above, we find:
\[ F = 98.1 \, \text{N} \]
**Hint:** Round the final answer to the nearest whole number if necessary, based on the context of the problem.
### Conclusion
The magnitude of the force \( F \) applied horizontally at the midpoint of the string is approximately \( 98.1 \, \text{N} \), which can be rounded to \( 100 \, \text{N} \) for practical purposes.
**Final Answer:** \( F \approx 100 \, \text{N} \)
