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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 driver is 10 in. in diameter. Find the diameters of all the steps. rpm. The smallest step of the

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 driver is 10 in. in diameter. Find the diameters of all the steps. rpm. The smallest step of the

Elements Of Electromagnetics
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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From the example problem, assume this is to be connected by a CROSS BELT. A. Determine the diameters of all the step.

B. Angle of contact for each pair of pulleys,

C. belt length. 

Example: 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 driver is 10 in. in diameter. Find the diameters of all the steps. Given: Solution: C = 30 in. = 150 N,D¡ = N,D2 150(10) = 50 (D2) D2 = 30 in. Na rpm N2 = 50 rpm N4 = 150 rpm No = 250 rpm Lo = (D1 + D2) + 2C + ÷(D2 D1)? 2 N3 = 600 rpm Lo = -(10 + 30) + 2(30) + 4(30) (30 – 10)2 2. = 10 in. Lo = 126.1652 in. Required: D1 , D2 , D3 , D4 , D; , D6 , D7 , D3
Transcribed Image Text:Example: 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 driver is 10 in. in diameter. Find the diameters of all the steps. Given: Solution: C = 30 in. = 150 N,D¡ = N,D2 150(10) = 50 (D2) D2 = 30 in. Na rpm N2 = 50 rpm N4 = 150 rpm No = 250 rpm Lo = (D1 + D2) + 2C + ÷(D2 D1)? 2 N3 = 600 rpm Lo = -(10 + 30) + 2(30) + 4(30) (30 – 10)2 2. = 10 in. Lo = 126.1652 in. Required: D1 , D2 , D3 , D4 , D; , D6 , D7 , D3
Example: 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 driver is 10 in. in diameter. Find the diameters of all the steps. Solution (Ds + D6) + 2C + (D5 – D6)² = 126.1652 (D4 + D3) + 2C + (D4 – D3)² = 126.1652 N;D; = N¿D6 150D; = 250D6 D6 = (3/5)(D5) N;D3 = N4D4 150D, = 150D4 D; = D4 D; = 25.9685 in. D3 = D4 = 21.061 in. D6 = 15.5811 in. (D, + Dg) + 2C + (D, – D3)² = 126.1652 N,D, = N3D8 D, = 31.3512 in. 150D, = 600D8 D, = 4D8 D3 = 7.8378 in.
Transcribed Image Text:Example: 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 driver is 10 in. in diameter. Find the diameters of all the steps. Solution (Ds + D6) + 2C + (D5 – D6)² = 126.1652 (D4 + D3) + 2C + (D4 – D3)² = 126.1652 N;D; = N¿D6 150D; = 250D6 D6 = (3/5)(D5) N;D3 = N4D4 150D, = 150D4 D; = D4 D; = 25.9685 in. D3 = D4 = 21.061 in. D6 = 15.5811 in. (D, + Dg) + 2C + (D, – D3)² = 126.1652 N,D, = N3D8 D, = 31.3512 in. 150D, = 600D8 D, = 4D8 D3 = 7.8378 in.
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