C 3 m B 1 m = 10° A C B T cos 52,98° -0,4Nc1 = 500a No GT sin 52,98° = 0 -G(1,5 cos 10°) + sin(52,98° -10°) 3 = -500a (0,2605) T cos 52,98° +0,4Nc = -500a - -Nc GT sin 52,98° = 0 -G (1,5 cos 10°) + sin(52,98° -10°) 3 = 500a (0,2605) D Ꭰ T sin 52,98° -0,4Nc = 0 NC - G + T cos(52,98° -10°) = 0 -G (1,5 sin 10°) + cos(52,98° -10°) 3 = 500a (0,2605) T cos 52,98° 0,4NC = 500a No G cos 10° + T sin 52,98° = 0 - -G cos 10° (1,5 cos 10°) + sin(52,98° 10°) 3 = -500a (0,2605)
C 3 m B 1 m = 10° A C B T cos 52,98° -0,4Nc1 = 500a No GT sin 52,98° = 0 -G(1,5 cos 10°) + sin(52,98° -10°) 3 = -500a (0,2605) T cos 52,98° +0,4Nc = -500a - -Nc GT sin 52,98° = 0 -G (1,5 cos 10°) + sin(52,98° -10°) 3 = 500a (0,2605) D Ꭰ T sin 52,98° -0,4Nc = 0 NC - G + T cos(52,98° -10°) = 0 -G (1,5 sin 10°) + cos(52,98° -10°) 3 = 500a (0,2605) T cos 52,98° 0,4NC = 500a No G cos 10° + T sin 52,98° = 0 - -G cos 10° (1,5 cos 10°) + sin(52,98° 10°) 3 = -500a (0,2605)
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
Related questions
Question
The length of the pipe AC is 3 m and the mass m 500 kg. It is attached to the tail hook of the pick-up truck with a 0.6 m long chain AB. The car and the pipe are in a steadily accelerating motion. The coefficient of motion friction between the pipe and the ground at point C is μk = 0.4.
b) Choose the correct group of equations of motion
Determine the acceleration a [m/s2] of the center of mass of the pipe.
Transcribed Image Text:C
3 m
B
1 m
= 10°
Transcribed Image Text:A
C
B
T cos 52,98° -0,4Nc1 = 500a
No GT sin 52,98° = 0
-G(1,5 cos 10°) + sin(52,98° -10°) 3 = -500a (0,2605)
T cos 52,98° +0,4Nc = -500a
-
-Nc GT sin 52,98° = 0
-G (1,5 cos 10°) + sin(52,98° -10°) 3 = 500a (0,2605)
D
Ꭰ
T sin 52,98° -0,4Nc = 0
NC - G + T cos(52,98° -10°) = 0
-G (1,5 sin 10°) + cos(52,98° -10°) 3 = 500a (0,2605)
T cos 52,98° 0,4NC = 500a
No G cos 10° + T sin 52,98° = 0
-
-G cos 10° (1,5 cos 10°) + sin(52,98° 10°) 3 = -500a (0,2605)
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