3. Two shafts 30 in. apart, each carrying a four-step pulley, are to be connected by an open belt. The driving shaft is to turn 150 rpm while the driven shaft is to turn 50, 150, 250, and 600 rpm. The smallest step of the dover is 10 in, in diameter. Find the diameters of all the steps.
3. Two shafts 30 in. apart, each carrying a four-step pulley, are to be connected by an open belt. The driving shaft is to turn 150 rpm while the driven shaft is to turn 50, 150, 250, and 600 rpm. The smallest step of the dover is 10 in, in diameter. Find the diameters of all the steps.
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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Question
Please solve ellaborately and include the untis in every step, show diagram and write your solutions clearly and readable if your solution is in written form. Your work will be appreaciated much. Thank You.
Transcribed Image Text:3. Two shafts 30 in. apart, each carrying a four-step pulley, are to be connected by
shaft is to turn 150 rpm while the driven shaft is to turn 50, 150, 250, and 600 rpm. The smallest step of the
daver is 10 in. in diameter. Find the diameters of all the steps.
an open belt. The driving
Given:
Solution:
C= 30 in.
Na = 150 rpm
N2 = 50 rpm
N = 150 rpm
N = 250 rpm
Ng = 600 rpm
N,D =N,D,
150(10) = 50 (D,)
D2 = 30 in.
Lo = (D, + D2) + 2C + (D2
D,)?
Lo = (10 + 30) + 2(30) +
4(30)
(30 – 10)²
D = 10 in.
Required:
D; , D2, D, , D4 , D, , D, , D7 , D;
Lo = 126.1652 in.
Transcribed Image Text:3. Two shafts 30 in. apart, each carrying a four-step pulley, are to be connected by an open belt. The driving
shaft is to turn 150 rpm while the driven shaft is to turn 50, 150, 250, and 600 rpm. The smallest step of the
daver is 10 in. in diameter. Find the diameters of all the steps.
Solution
(Ds + D6) + 2C + (D; - D6)? = 126.1652
(D, + D3) + 2C + – (D4 – D3)² = 126.1652
4C
2
40
N;D; = N,D,
150D, = 250D,
D, = (3/5)(D;)
N;D; = N,D,
150D, = 150D4
D; = D,
%3D
D3 = D, = 21.061 in.
D; = 25.9685 in.
D = 15.5811 in.
%3D
(D, + Da) + 2C + (D, - Da)? = 126.1652
N;D, = NºD8
D, = 31.3512 in.
%3D
150D, = 600DS
D, = 4Ds
D3 = 7.8378 in.
%3D
%3D
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