The figure below shows a pin-jointed fourbar mechanism. The crank AB (link 2) drives the mechanism. CCW at a constant speed of 10 rad/s. The weight of links 2 and 3 is negligible, and link 4 is 17 kg. The radius of gyration of link 4 relative to its center of gravity G is 45 mm. A torque of 50 N.mm is applied to link4. At this instant the angular position of links are 0₂-315°, 0,= 31.11°, 04-101.75%, determine: a) the angular velocity of links 3 and 4, b) the angular acceleration of link 4, c) the linear acceleration of the center of gravity G of link 4, d) the force in member BC (link3), e) the torque needed to drive the crank (AB) in this position. 315° A R2 25 mm 31.11 110m "R3 4④201.75 25 mm -120 mm mm 50 Nmm αω a) we awe sun (01-03) = -6.431 rad/s/③ c (04-03) -aw, am (04.0₂) = -1321 rady's / ③ b mi (03-09) Ws (09-02) b) A = 0.0392 C = 2.269 E=0.0942 B-00568 ; D = -0.0081 F3-2883 => 43=-35,474 Mads 9/19 Bu ; 4 = 6.451 nad/s/ - Acceleration off 0 = 85 (14) Set(of 25 (64) 5 (0) AG = (6.451) (25) (in (617) jus (101.75)_ 25(6431) "(600(101.795) AG = 52.661 +177.71 j (mm2 4 => +(161.75) 1961/52.66²+177.71 = 1.85,352 m/s 0-73.49°/ d) FBD of linkus hac + MD = ID A4 Teq.11 Эч ID= IG+ md? از 7825 ->(neglectul) Dy ②4 So N.mm DK IG=MKG = 17 (45) 2 =034425 kgm. / >>ID= 34425 + 17 (25)² = all 5050 Kg.mm² Eq.11 -30.25 + (FB)(60031.11) (604) (mm 7825) Qu + (3C) (min 31.11) (904) (Los 78.25)=45050 (6.451) -0.037738 FBC = 0.2406; 10/19 => FBC = 6.376 FBD inak 3 and dy TA Fxx C Бас 1450 Ax GMA =0.25 B TA-FBC (60731. 11) (0.05) in 45°) - TBC (in 3811) (0.025) Cox150 TA= 0.1547 Nm = 154.7 N. mm Qu
The figure below shows a pin-jointed fourbar mechanism. The crank AB (link 2) drives the mechanism. CCW at a constant speed of 10 rad/s. The weight of links 2 and 3 is negligible, and link 4 is 17 kg. The radius of gyration of link 4 relative to its center of gravity G is 45 mm. A torque of 50 N.mm is applied to link4. At this instant the angular position of links are 0₂-315°, 0,= 31.11°, 04-101.75%, determine: a) the angular velocity of links 3 and 4, b) the angular acceleration of link 4, c) the linear acceleration of the center of gravity G of link 4, d) the force in member BC (link3), e) the torque needed to drive the crank (AB) in this position. 315° A R2 25 mm 31.11 110m "R3 4④201.75 25 mm -120 mm mm 50 Nmm αω a) we awe sun (01-03) = -6.431 rad/s/③ c (04-03) -aw, am (04.0₂) = -1321 rady's / ③ b mi (03-09) Ws (09-02) b) A = 0.0392 C = 2.269 E=0.0942 B-00568 ; D = -0.0081 F3-2883 => 43=-35,474 Mads 9/19 Bu ; 4 = 6.451 nad/s/ - Acceleration off 0 = 85 (14) Set(of 25 (64) 5 (0) AG = (6.451) (25) (in (617) jus (101.75)_ 25(6431) "(600(101.795) AG = 52.661 +177.71 j (mm2 4 => +(161.75) 1961/52.66²+177.71 = 1.85,352 m/s 0-73.49°/ d) FBD of linkus hac + MD = ID A4 Teq.11 Эч ID= IG+ md? از 7825 ->(neglectul) Dy ②4 So N.mm DK IG=MKG = 17 (45) 2 =034425 kgm. / >>ID= 34425 + 17 (25)² = all 5050 Kg.mm² Eq.11 -30.25 + (FB)(60031.11) (604) (mm 7825) Qu + (3C) (min 31.11) (904) (Los 78.25)=45050 (6.451) -0.037738 FBC = 0.2406; 10/19 => FBC = 6.376 FBD inak 3 and dy TA Fxx C Бас 1450 Ax GMA =0.25 B TA-FBC (60731. 11) (0.05) in 45°) - TBC (in 3811) (0.025) Cox150 TA= 0.1547 Nm = 154.7 N. mm Qu
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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Hello, the question is solved, but I want a detailed explanation and explanation of the part D energy method because I do not understand anything from what is written. Are there several rules for the energy method? I want a plan on paper. Please, very important. I will give you a good feedback.
Transcribed Image Text:The figure below shows a pin-jointed fourbar mechanism. The crank AB (link 2) drives the mechanism.
CCW at a constant speed of 10 rad/s. The weight of links 2 and 3 is negligible, and link 4 is 17 kg. The
radius of gyration of link 4 relative to its center of gravity G is 45 mm. A torque of 50 N.mm is applied to
link4.
At this instant the angular position of links are 0₂-315°, 0,= 31.11°, 04-101.75%, determine:
a) the angular velocity of links 3 and 4,
b) the angular acceleration of link 4,
c) the linear acceleration of the center of gravity G of link 4,
d) the force in member BC (link3),
e) the torque needed to drive the crank (AB) in this position.
315°
A
R2
25 mm
31.11
110m
"R3
4④201.75
25 mm
-120 mm
mm
50 Nmm
αω
a) we awe sun (01-03) = -6.431 rad/s/③
c
(04-03)
-aw, am (04.0₂) = -1321 rady's / ③
b mi (03-09)
Ws (09-02)
b) A = 0.0392
C = 2.269
E=0.0942
B-00568
; D = -0.0081
F3-2883
=> 43=-35,474 Mads
9/19
Bu
; 4 = 6.451 nad/s/
Transcribed Image Text:- Acceleration off
0 = 85 (14) Set(of 25 (64) 5 (0)
AG = (6.451) (25) (in (617) jus (101.75)_ 25(6431) "(600(101.795)
AG
=
52.661 +177.71 j (mm2 4
=>
+(161.75)
1961/52.66²+177.71 = 1.85,352 m/s
0-73.49°/
d) FBD of linkus
hac
+ MD = ID A4 Teq.11
Эч
ID= IG+ md?
از
7825
->(neglectul)
Dy ②4
So N.mm
DK
IG=MKG = 17 (45) 2 =034425 kgm. /
>>ID= 34425 + 17 (25)² = all 5050 Kg.mm²
Eq.11 -30.25 + (FB)(60031.11) (604) (mm 7825)
Qu
+ (3C) (min 31.11) (904) (Los 78.25)=45050 (6.451)
-0.037738 FBC = 0.2406;
10/19
=> FBC = 6.376
FBD inak 3 and dy
TA
Fxx
C
Бас
1450
Ax
GMA =0.25
B
TA-FBC (60731. 11) (0.05) in 45°) - TBC (in 3811) (0.025) Cox150
TA= 0.1547 Nm = 154.7 N. mm
Qu
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