### Problem 3: Friction and Equilibrium Analysis**Problem Statement:**The coefficient of friction between the 100 lb block (shown in the figure below) and the incline plane is 0.25 and that between the cord and cylindrical support is 0.3. Determine the range of cylinder weight \( W \) for which the system shown below will be in equilibrium.**Figure:**The figure accompanying this problem depicts the following setup:- A 100-pound block rests on an inclined plane angled at 25 degrees. There is a coefficient of friction (\(\mu\)) between the block and the plane, which is 0.25.- The block is connected by a cord that goes over a cylindrical support.- The cylindrical support has a coefficient of friction of 0.3 between the cord and the support.- The other end of the cord is attached to a hanging weight \( W \).**Diagram Explanation:**1. **Inclined Plane** - The inclined plane is at an angle of 25 degrees. - The 100 lb block is placed on this inclined plane.2. **Friction** - Friction between the block and the inclined plane is given by the coefficient \(\mu = 0.25\). - Friction between the cord and the cylindrical support is given by \(\mu = 0.3\).3. **Cylindrical Support** - The cylinder through which the cord passes helps in redirecting the force applied by weight \( W \).4. **Block and Weight** - The 100 lb block applies a force down the incline due to gravity. - The hanging weight \( W \) exerts a downward force through the cord.**Goal:**The objective is to determine the range of the cylinder weight \( W \) that will keep this system in equilibrium, where equilibrium implies that all forces are balanced and there is no net motion in the system.### Equilibrium ConditionsTo solve this problem, the following must be accounted for:- The force of gravity acting on the 100 lb block and its components parallel and perpendicular to the inclined plane.- The frictional force opposing the motion of the block along the plane.- The tension in the cord due to the weight \( W \) and its effect on the 100 lb block.- The frictional force acting on the cord as it passes over the cylindrical support.Applying

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Description: ### Problem 3: Friction and Equilibrium Analysis**Problem Statement:**The coefficient of friction between the 100 lb block (shown in the figure below) and the incline plane is 0.25 and that between the cord and cylindrical support is 0.3. Determine

According to the information weight of the box is 100 lb coefficient of friction is 0.25 at an angle…

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Q.3) The coefficient of friction between the 100 lb block (shown in figure below) and the incline plane is 0.25 and that between the cord and cylindrical support is 0.3. Determine the range of cylinder weight W for which the system shown below will be in equilibrium. H = 0.3 100 lb p = 0.25 25° W

Q.3) The coefficient of friction between the 100 lb block (shown in figure below) and the incline plane is 0.25 and that between the cord and cylindrical support is 0.3. Determine the range of cylinder weight W for which the system shown below will be in equilibrium. H = 0.3 100 lb p = 0.25 25° W

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### Problem 3: Friction and Equilibrium Analysis

**Problem Statement:**
The coefficient of friction between the 100 lb block (shown in the figure below) and the incline plane is 0.25 and that between the cord and cylindrical support is 0.3. Determine the range of cylinder weight \( W \) for which the system shown below will be in equilibrium.

**Figure:**

The figure accompanying this problem depicts the following setup:

- A 100-pound block rests on an inclined plane angled at 25 degrees. There is a coefficient of friction (\(\mu\)) between the block and the plane, which is 0.25.
- The block is connected by a cord that goes over a cylindrical support.
- The cylindrical support has a coefficient of friction of 0.3 between the cord and the support.
- The other end of the cord is attached to a hanging weight \( W \).

**Diagram Explanation:**

1. **Inclined Plane**
- The inclined plane is at an angle of 25 degrees.
- The 100 lb block is placed on this inclined plane.

2. **Friction**
- Friction between the block and the inclined plane is given by the coefficient \(\mu = 0.25\).
- Friction between the cord and the cylindrical support is given by \(\mu = 0.3\).

3. **Cylindrical Support**
- The cylinder through which the cord passes helps in redirecting the force applied by weight \( W \).

4. **Block and Weight**
- The 100 lb block applies a force down the incline due to gravity.
- The hanging weight \( W \) exerts a downward force through the cord.

**Goal:**
The objective is to determine the range of the cylinder weight \( W \) that will keep this system in equilibrium, where equilibrium implies that all forces are balanced and there is no net motion in the system.

### Equilibrium Conditions

To solve this problem, the following must be accounted for:
- The force of gravity acting on the 100 lb block and its components parallel and perpendicular to the inclined plane.
- The frictional force opposing the motion of the block along the plane.
- The tension in the cord due to the weight \( W \) and its effect on the 100 lb block.
- The frictional force acting on the cord as it passes over the cylindrical support.

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