uniform wheel of 50 cm diameter and 1kN weight rests against a rigid rectangular block of thickness 20 cm. Considering all the surfaces smooth, determine the least pull to be applied through the centre of wheel to get just turn it over the corner of the block reaction of the block using lamis theorem

Given the problem, we need to determine the least pull \( P \) required to just turn the wheel over the corner of the block using Lami's theorem. Here's a step-by-step approach:

### Step 1: Understand the problem
- **Wheel diameter:** 50 cm (so radius \( r = 25 \) cm = 0.25 m)
- **Weight of the wheel:** 1 kN (which acts vertically downward at the center)
- **Block thickness:** 20 cm = 0.2 m

The wheel is resting against the corner of the block, and we need to calculate the pull \( P \) that should be applied horizontally through the center to just turn the wheel over the corner.

### Step 2: Identify forces acting on the wheel
1. **Weight \( W = 1 \) kN:** Acts vertically downward through the center of the wheel.
2. **Pull \( P \):** Acts horizontally through the center of the wheel.
3. **Reaction \( R \):** Acts along the line connecting the point of contact at the corner of the block to the center of the wheel.

### Step 3: Geometry of the situation
- **Distance from center to the edge of the block:** The distance \( d \) between the center of the wheel and the edge of the block is the difference between the radius of the wheel and the thickness of the block:
  \[
  d = r - 0.2 = 0.25 - 0.2 = 0.05 \text{ m}
  \]

### Step 4: Apply Lami's Theorem
Lami's theorem states that for a body in equilibrium, the ratio of the magnitude of each force to the sine of the angle opposite to it is constant:

\[
\frac{P}{\sin(90^\circ + \theta)} = \frac{W}{\sin(90^\circ)} = \frac{R}{\sin(\theta)}
\]

Here, \( \theta \) is the angle between \( W \) and \( R \).

1. **Angle \( \theta \):** The angle \( \theta \) can be found using the geometry of the problem:
   \[
   \theta = \tan^{-1}\left(\frac{d}{r}\right) = \tan^{-1}\left(\frac{0.05}{0.25}\right) = \tan^{-1}\left(0.2\right)
   \]
   Thus, \( \theta = 11.31^\circ \).

2. **Substitute values into Lami's Theorem:**

   For \( W = 1 \) kN and \( \theta = 11.31^\circ \):
   \[
   \frac{P}{\sin(90^\circ + 11.31^\circ)} = \frac{1 \text{ kN}}{\sin(90^\circ)} = \frac{R}{\sin(11.31^\circ)}
   \]

   Since \( \sin(90^\circ) = 1 \) and \( \sin(90^\circ + \theta) = \cos(\theta) \), we get:
   \[
   \frac{P}{\cos(11.31^\circ)} = 1 \text{ kN}
   \]
   Therefore:
   \[
   P = 1 \times \cos(11.31^\circ) \text{ kN}
   \]
   \[
   P = 0.981 \text{ kN}
   \]

### Step 5: Conclusion
The least pull \( P \) required to just turn the wheel over the corner of the block is approximately **0.981 kN**.

Would you like more details or have any questions about the solution?

### 5 Related Questions:
1. How does the thickness of the block affect the pull required to move the wheel?
2. What would happen if the surfaces were not smooth? How would friction be accounted for?
3. How would the problem change if the pull was applied at a different angle?
4. Can you use another method, like the principle of moments, to solve this problem?
5. How would the solution change if the weight of the wheel was doubled?

### Tip:
When applying Lami's theorem, always ensure that the angles used are correctly identified relative to the forces, as errors here can lead to incorrect solutions.

Learn how to determine the least horizontal pull required to just turn a uniform wheel of 50 cm diameter and 1 kN weight over a rigid rectangular block of 20 cm thickness using Lami's theorem. Includes step-by-step calculation and explanation.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation