(5) Determine the components of reaction at the fixed support A. The 400 N, 500 N, and 600 N forces are parallel to the x, y, z axes, respectively. Be sure to draw the unknown support reactions at fixed support A.
(5) Determine the components of reaction at the fixed support A. The 400 N, 500 N, and 600 N forces are parallel to the x, y, z axes, respectively. Be sure to draw the unknown support reactions at fixed support A.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
Correct the assignment and wrote two to three sentences explaining the error
![ENGR&214
Spring 2017
1. (5) Determine the components of reaction at the fixed support A. The 400 N, 500 N, and 600 N
forces are parallel to the x, y, z axes, respectively. Be sure to draw the unknown support reactions at
fixed support A.
EFx=0: Ax - 400N = 0
400 N
Ax =
EFY:0. Ay.
1
AY
1+ EF2 = 0: Az
A z
500 N = 0
500 N
F
600 N 0
= 600 N
+2
:
400 N
(HA)2 = 0. (HA)2 = 0
(HA) y = 0: (MA)Y - 600 (0.75 m)
(HA) y = 600N (0.75m)
(HA) Y = 450 N. m/
xx₂
+3.5/5
M²
Ax
✓✓
(EMA)X = 0 (MA)x - 600N (075) - LIDON (0. Samt = 0
(MA)X
(MA)x=
DO
Initials
A
-1
600 N (0.75m) + 400 N (0.5m)
650 N.m
0
1m
0.5 m
0.75 m
0.75 A 500
N
AY (MAY
600 N
8½/2
1/2
there is another applied
force causing moment
about yaxs
Ax = 400 N
Ay
Az
“
= 500 NV
600 N 1,
(MA)x= 650m
(MA)Y = 450.m
(MA) z = 0 N.m
Page 2 of 5](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6e7ba281-dd87-4688-be97-9abf26130e05%2Faf8d8510-d90d-48df-bcae-748667372e01%2Fj6y13lwq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:ENGR&214
Spring 2017
1. (5) Determine the components of reaction at the fixed support A. The 400 N, 500 N, and 600 N
forces are parallel to the x, y, z axes, respectively. Be sure to draw the unknown support reactions at
fixed support A.
EFx=0: Ax - 400N = 0
400 N
Ax =
EFY:0. Ay.
1
AY
1+ EF2 = 0: Az
A z
500 N = 0
500 N
F
600 N 0
= 600 N
+2
:
400 N
(HA)2 = 0. (HA)2 = 0
(HA) y = 0: (MA)Y - 600 (0.75 m)
(HA) y = 600N (0.75m)
(HA) Y = 450 N. m/
xx₂
+3.5/5
M²
Ax
✓✓
(EMA)X = 0 (MA)x - 600N (075) - LIDON (0. Samt = 0
(MA)X
(MA)x=
DO
Initials
A
-1
600 N (0.75m) + 400 N (0.5m)
650 N.m
0
1m
0.5 m
0.75 m
0.75 A 500
N
AY (MAY
600 N
8½/2
1/2
there is another applied
force causing moment
about yaxs
Ax = 400 N
Ay
Az
“
= 500 NV
600 N 1,
(MA)x= 650m
(MA)Y = 450.m
(MA) z = 0 N.m
Page 2 of 5
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