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At a given instant the center of the roller A on the bar has the velocity and acceleration shown. The length of the bar is 0.6 m. The radius of both the rollers is the same (and is in fact not required to solve the problem!) Determine the velocity and acceleration of the center of the roller B when the angle of the bar with the vertical, = 30⁰. Hint: Do not assume that the angle 0 that the bar makes with the vertical is always 30 degrees. It changes with time, and its derivative is not zero. It is 30 degrees only at this instant. However, the angle of the plane on which the roller B rolls with the horizontal is always 30 degrees. 4 m/s 6 m/s² A 30° 30° 0.6 m B

At a given instant the center of the roller A on the bar has the velocity and acceleration shown. The length of the bar is 0.6 m. The radius of both the rollers is the same (and is in fact not required to solve the problem!) Determine the velocity and acceleration of the center of the roller B when the angle of the bar with the vertical, = 30⁰. Hint: Do not assume that the angle 0 that the bar makes with the vertical is always 30 degrees. It changes with time, and its derivative is not zero. It is 30 degrees only at this instant. However, the angle of the plane on which the roller B rolls with the horizontal is always 30 degrees. 4 m/s 6 m/s² A 30° 30° 0.6 m B

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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Please help I am having the hardest time trying to single out and remove S_a from the 2nd derivitive equation for acceleration so I can solve for S_b double dot. And at the instant S_a=S_b

At a given instant the center of the roller A on the bar has the velocity and acceleration shown. The length of the bar is 0.6 m. The radius of both the rollers is the same (and is in fact not required to solve the problem!) Determine the velocity and acceleration of the center of the roller B when the angle of the bar with the vertical, 0 = 30º. Hint: Do not assume that the angle 0 that the bar makes with the vertical is always 30 degrees. It changes with time, and its derivative is not zero. It is 30 degrees only at this instant. However, the angle of the plane on which the roller B rolls with the horizontal is always 30 degrees. 4 m/s 6 m/s² 30⁰ 30° 0.6 m B
Transcribed Image Text:At a given instant the center of the roller A on the bar has the velocity and acceleration shown. The length of the bar is 0.6 m. The radius of both the rollers is the same (and is in fact not required to solve the problem!) Determine the velocity and acceleration of the center of the roller B when the angle of the bar with the vertical, 0 = 30º. Hint: Do not assume that the angle 0 that the bar makes with the vertical is always 30 degrees. It changes with time, and its derivative is not zero. It is 30 degrees only at this instant. However, the angle of the plane on which the roller B rolls with the horizontal is always 30 degrees. 4 m/s 6 m/s² 30⁰ 30° 0.6 m B
SA CA 120° SB 0.6m →OA=0.34641m Law of cosines C²=a²+b²-2abcos C 0.6²=SA+SB²-2SAS COSC t 6 DA Sin/20 Sin 30 - 0=2SASA +255-2cos/20 (SASB+ SASR) SASA+SESB-COS120(SASB+ SASB)) -S$1 = $8518-C05/20 $45B-CASTROSA SE SAŠA + COSIZO (SA5₂) = S(Sp-CisSA) B Z SB-COSIZO SA =-SASA + COS120 SASB SASA+SA³B-COS (SA³B+ SASA) -(SB+Cos/20SA) SA+S-Csiz SS-COSI20 SASA 0= SA (SA+SB-Cos/20SB-C05/205A) Cos/20SA-SA=S8 (1-COS/20) Cos/20SA-SAS-4 1-Cos/20 2SASA +25BS-2Cos/20SASB-2005/20 SBSA SASB+SASB 2 SASA +255 +255 +255B - 2005/20 (SASB + SASB + SASB+SASB) SASA+SA+SASB + S2-COS/20(SA58 +2SAS+SASA) SASA + SA² +SASB +5B - COSMOSA SB-COSI20 25A5B - COS/205A SA SA (SA+SB)-SA (COS120) SB-SA COSIZOSA SA(SA+SB)-SA (COSIOSB + Cos/20SA) +5² +SB²-C05/20 ZSASB SA(Cors+os1205A) = SA (SA+SB) +5² +53-C05 120 25/50 SA (SA+ SB)-(Custo Sp+cos/205A))+ $A²+ SB-CO5 120 255B SB (SA-COSIOSA) = SASA +57 +58-Cos120 25A5B-Cos 1205A SA SA-COS 120SA
Transcribed Image Text:SA CA 120° SB 0.6m →OA=0.34641m Law of cosines C²=a²+b²-2abcos C 0.6²=SA+SB²-2SAS COSC t 6 DA Sin/20 Sin 30 - 0=2SASA +255-2cos/20 (SASB+ SASR) SASA+SESB-COS120(SASB+ SASB)) -S$1 = $8518-C05/20 $45B-CASTROSA SE SAŠA + COSIZO (SA5₂) = S(Sp-CisSA) B Z SB-COSIZO SA =-SASA + COS120 SASB SASA+SA³B-COS (SA³B+ SASA) -(SB+Cos/20SA) SA+S-Csiz SS-COSI20 SASA 0= SA (SA+SB-Cos/20SB-C05/205A) Cos/20SA-SA=S8 (1-COS/20) Cos/20SA-SAS-4 1-Cos/20 2SASA +25BS-2Cos/20SASB-2005/20 SBSA SASB+SASB 2 SASA +255 +255 +255B - 2005/20 (SASB + SASB + SASB+SASB) SASA+SA+SASB + S2-COS/20(SA58 +2SAS+SASA) SASA + SA² +SASB +5B - COSMOSA SB-COSI20 25A5B - COS/205A SA SA (SA+SB)-SA (COS120) SB-SA COSIZOSA SA(SA+SB)-SA (COSIOSB + Cos/20SA) +5² +SB²-C05/20 ZSASB SA(Cors+os1205A) = SA (SA+SB) +5² +53-C05 120 25/50 SA (SA+ SB)-(Custo Sp+cos/205A))+ $A²+ SB-CO5 120 255B SB (SA-COSIOSA) = SASA +57 +58-Cos120 25A5B-Cos 1205A SA SA-COS 120SA
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