1. A mail sack with a mass of 85 kg hangs from a vertical rope 2.5 m long.
the worker moves the mail sack to the side 0.9 m (measured horizontally) while keeping the length of the rope
time is tight. (The attachment point of the upper part of the rope remains in place.)
a. What horizontal force is needed to hold the mail sack in the new location?
b. When the sack is moved to this location, how much work does the worker do?

To solve these problems, we need to use concepts from physics, specifically equilibrium and work done by a force.

---

### **(a) Horizontal Force Calculation**

The situation involves a static equilibrium where the mail sack is held at rest at an angle. The forces acting are:
- The **tension** in the rope (T),
- The **weight** of the sack (W = mg),
- The **horizontal force** (F) exerted by the worker.

The rope forms a triangle with the horizontal displacement \( x = 0.9 \, \text{m} \) and the rope length \( L = 2.5 \, \text{m} \). The vertical displacement \( y \) can be found using the Pythagorean theorem:

\[
y = \sqrt{L^2 - x^2} = \sqrt{(2.5)^2 - (0.9)^2} = \sqrt{6.25 - 0.81} = \sqrt{5.44} = 2.33 \, \text{m}.
\]

The angle \( \theta \) between the rope and the vertical can now be found using:

\[
\sin\theta = \frac{x}{L}, \quad \cos\theta = \frac{y}{L}.
\]

\[
\sin\theta = \frac{0.9}{2.5} = 0.36, \quad \cos\theta = \frac{2.33}{2.5} = 0.932.
\]

At equilibrium, the horizontal force balances the horizontal component of tension:

\[
F = T \sin\theta.
\]

From the vertical forces, tension balances the weight:

\[
T \cos\theta = W \quad \text{or} \quad T = \frac{W}{\cos\theta}.
\]

Substituting \( W = mg = 85 \cdot 9.8 = 833 \, \text{N} \) and \( \cos\theta = 0.932 \):

\[
T = \frac{833}{0.932} = 893.3 \, \text{N}.
\]

Now substitute \( T \) and \( \sin\theta = 0.36 \) into the horizontal force equation:

\[
F = T \sin\theta = 893.3 \cdot 0.36 = 321.6 \, \text{N}.
\]

---

### **(b) Work Done**

The work \( W \) done is the horizontal force multiplied by the horizontal displacement:

\[
W = F \cdot x.
\]

Using \( F = 321.6 \, \text{N} \) and \( x = 0.9 \, \text{m} \):

\[
W = 321.6 \cdot 0.9 = 289.44 \, \text{J}.
\]

---

### **Final Answers**
1. **Horizontal Force:** \( F = 321.6 \, \text{N} \)
2. **Work Done:** \( W = 289.44 \, \text{J} \)

---

Would you like more detailed steps or have any specific questions?  
Here are some related questions to explore further:  
1. How does the angle affect the magnitude of the horizontal force?  
2. What would happen if the rope length were shorter?  
3. How is work affected if the horizontal displacement increases?  
4. What is the potential energy change in the sack during this process?  
5. How can we calculate tension directly using force components?  

**Tip:** When solving static equilibrium problems, always break forces into components and carefully analyze the geometry.

A mail sack with a mass of 85 kg hangs from a vertical rope 2.5 m long. The worker moves the mail sack to the side 0.9 m (measured horizontally) while keeping the length of the rope. (The attachment point of the upper part of the rope remains in place.) a. What horizontal force is needed to hold the mail sack in the new location? b. When the sack is moved to this location, how much work does the worker do?

Calculate the horizontal force needed to hold an 85 kg mail sack displaced 0.9 m horizontally on a 2.5 m rope and the work done to move it. This problem uses static equilibrium, trigonometry, and energy concepts. Suitable for high school physics students in Grades 11-12.

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