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At the instant shown, the arm OA of the conveyor belt is rotating about the z axis with a constant angular velocity w₁ = 6 rad/s, while at the same instant the arm is rotating upward at a constant rate ₂ = 4 rad/s. If the conveyor is running at a rate r = 5 ft/s, which is increasing at 7 = 8 ft/s², determine the velocity and acceleration of the package Pat the instant shown. Neglect the size of the package. SOLUTION = ₁ = {6k} rad/s = 0 To vo ao = 0 = = P/O= {41} rad/s Ωρο = 0 IPO = {4.243j + 4.243k) ft (VP/O)xyz (IP/O)xyz + P/O X IPIO = (5 cos 45°j + 5 sin 45°k) + (4i) x (4.243j + 4.243k) = {-13.44j + 20.51k) ft/s (ap/O)xyz = 8 cos 45j + 8 sin 45°k - 96.18j 39.60k = {-90.52j33.945k) ft/s² = w₁ = 6 rad/s r= 6 ft w₂ = 4 rad/s P 0 = 45° A

At the instant shown, the arm OA of the conveyor belt is rotating about the z axis with a constant angular velocity w₁ = 6 rad/s, while at the same instant the arm is rotating upward at a constant rate ₂ = 4 rad/s. If the conveyor is running at a rate r = 5 ft/s, which is increasing at 7 = 8 ft/s², determine the velocity and acceleration of the package Pat the instant shown. Neglect the size of the package. SOLUTION = ₁ = {6k} rad/s = 0 To vo ao = 0 = = P/O= {41} rad/s Ωρο = 0 IPO = {4.243j + 4.243k) ft (VP/O)xyz (IP/O)xyz + P/O X IPIO = (5 cos 45°j + 5 sin 45°k) + (4i) x (4.243j + 4.243k) = {-13.44j + 20.51k) ft/s (ap/O)xyz = 8 cos 45j + 8 sin 45°k - 96.18j 39.60k = {-90.52j33.945k) ft/s² = w₁ = 6 rad/s r= 6 ft w₂ = 4 rad/s P 0 = 45° A

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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20-51. At the instant shown, the arm OA of the conveyor belt is rotating about the z axis with a constant angular velocity w₁ = 6 rad/s, while at the same instant the arm is rotating upward at a constant rate ₂ = 4 rad/s. If the conveyor is running at a rate r = 5 ft/s, which is increasing at 7 = 8 ft/s², determine the velocity and acceleration of the package P at the instant shown. Neglect the size of the package. SOLUTION = = ₁ {6k) rad/s = 0 ro= vo ao = 0 P/O = {41} rad/s Ωρο = 0 IP/O= {4.243j + 4.243k) ft (VP/O)xyz = (İP/O)xyz + P/O X IP/O = (5 cos 45°j + 5 sin 45ºk) + (4i) × (4.243j + 4.243k) = {-13.44j + 20.51k} ft/s (ap/O)xyz = 8 cos 45j + 8 sin 45ºk - 96.18j - 39.60k = {-90.52j- 33.945k) ft/s² w₁ = 6 rad/s r=6 ft W₂ = 4 rad/s 0 = 45° A
Transcribed Image Text:20-51. At the instant shown, the arm OA of the conveyor belt is rotating about the z axis with a constant angular velocity w₁ = 6 rad/s, while at the same instant the arm is rotating upward at a constant rate ₂ = 4 rad/s. If the conveyor is running at a rate r = 5 ft/s, which is increasing at 7 = 8 ft/s², determine the velocity and acceleration of the package P at the instant shown. Neglect the size of the package. SOLUTION = = ₁ {6k) rad/s = 0 ro= vo ao = 0 P/O = {41} rad/s Ωρο = 0 IP/O= {4.243j + 4.243k) ft (VP/O)xyz = (İP/O)xyz + P/O X IP/O = (5 cos 45°j + 5 sin 45ºk) + (4i) × (4.243j + 4.243k) = {-13.44j + 20.51k} ft/s (ap/O)xyz = 8 cos 45j + 8 sin 45ºk - 96.18j - 39.60k = {-90.52j- 33.945k) ft/s² w₁ = 6 rad/s r=6 ft W₂ = 4 rad/s 0 = 45° A
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