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
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
Problem 1.1MA
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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.*](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1e4fcbe0-ed74-4447-b38a-a2af13f9fabc%2Fcc65dedc-7ca6-4aa3-b626-6de8123a5bd6%2F61v94xc_processed.jpeg&w=3840&q=75)
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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