If the 1. The double pulley shown is attached to a 8-mm-radius shaft that fits loosely in a fixed bearing. Ms = 0.45 and M = 0.40 and the weight of the pulley is 25 N, (a) determine the magnitude of the force P required to start raising the load, and (b) determine the magnitude of the force P required to start lowering the load a constant speed. Fr=8mm x 0145 = 3.6 mm = 01366 M х1 фотин 01036m) HP - 20,036 = 0 P 20,036 100 mn 90 mm A 45 mm, 20kg +90361 р. 20,036 MIM 20 kg P CH 23,99 kg Jag

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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### Problem Statement:

1. The double pulley shown is attached to an 8-mm-radius shaft that fits loosely in a fixed bearing. If the static friction coefficient \((\mu_s)\) is 0.45 and the kinetic friction coefficient \((\mu_k)\) is 0.40, and the weight of the pulley is 25 N, determine:
   - (a) the magnitude of the force \(P\) required to start raising the load.
   - (b) the magnitude of the force \(P\) required to start lowering the load at a constant speed.

### Diagram Explanation:

The diagram shows a double pulley system with the following details:
- A point \(A\) marks the shaft's position.
- The radius of the pulley from point \(A\) to the outer edge is 90 mm.
- The radius of the shaft is 8 mm.
- A weight of 20 kg is suspended from the pulley, with the rope going over the pulley and being pulled by force \(P\) on the other side.
- The distance from the center of the pulley to the point of application of force \(P\) is 45 mm.

### Calculations Shown:

- The frictional force (\(F_r\)) is calculated using the radius of the shaft:
  \[
  F_r = 8\, \text{mm} \times 0.45 = 3.6\, \text{mm} = 0.0366\, \text{m}
  \]

- Using the distances and weights given:
  \[
  \left( \frac{90\, \text{mm}}{100\, \text{mm}} \right) \rightarrow 0.036\, \text{m} - 20\, \text{kg}
  \]
  
- Equations attempted to solve for \(P\):
  \[
  P - 20.036 = 0
  \]
  \[
  90.036\: P = 20.036
  \]
  \[
  P = 2.3 \text{ (calculated value)}
  \]

*Note: The transcription captures the essence and content of the material for educational presentation.*
Transcribed Image Text:### Problem Statement: 1. The double pulley shown is attached to an 8-mm-radius shaft that fits loosely in a fixed bearing. If the static friction coefficient \((\mu_s)\) is 0.45 and the kinetic friction coefficient \((\mu_k)\) is 0.40, and the weight of the pulley is 25 N, determine: - (a) the magnitude of the force \(P\) required to start raising the load. - (b) the magnitude of the force \(P\) required to start lowering the load at a constant speed. ### Diagram Explanation: The diagram shows a double pulley system with the following details: - A point \(A\) marks the shaft's position. - The radius of the pulley from point \(A\) to the outer edge is 90 mm. - The radius of the shaft is 8 mm. - A weight of 20 kg is suspended from the pulley, with the rope going over the pulley and being pulled by force \(P\) on the other side. - The distance from the center of the pulley to the point of application of force \(P\) is 45 mm. ### Calculations Shown: - The frictional force (\(F_r\)) is calculated using the radius of the shaft: \[ F_r = 8\, \text{mm} \times 0.45 = 3.6\, \text{mm} = 0.0366\, \text{m} \] - Using the distances and weights given: \[ \left( \frac{90\, \text{mm}}{100\, \text{mm}} \right) \rightarrow 0.036\, \text{m} - 20\, \text{kg} \] - Equations attempted to solve for \(P\): \[ P - 20.036 = 0 \] \[ 90.036\: P = 20.036 \] \[ P = 2.3 \text{ (calculated value)} \] *Note: The transcription captures the essence and content of the material for educational presentation.*
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