-
8/3/2019 Derrick Problem
1/13
Derrick Problem
Quizz #8 by John Willden, P.Eng.
-
8/3/2019 Derrick Problem
2/13
T = 8 300.0N
25
2
1
A
C
B
D
2m
2m
4m3m
The 3 metreshere correspond
to AD
The 4 metres herecorrespond to AC
The dimensions tell us the slope of DC
3
4
That is 4:3
Force Slope
-
8/3/2019 Derrick Problem
3/13
T = -8300N
25
2
1
A
C
B
D
This includes the two T forces and theunknowns CV, CH and
AH.
We use CV and CH instead ofDV and DHbecause we need to show
where the
forces acting on D are coming from.
Freebody at D
AHCH
CV
T = 8300N
8300 cos 25= -7522N8300 sin 25
= -3508N
First we do a freebody about D since
that is where the loads are applied.
Then we decompose the angled forces
to vertical and horizontal.
-
8/3/2019 Derrick Problem
4/13
T = -8300N
25
2
1
A
C
B
D
Freebody at D
AHCH
CV
T = 8300N
8300 cos 25= -7522N8300 sin 25
= -3508N
In the vertical direction there is only one
unknown, CV. Well do a vertical balance at D first.
VD = 0 = - 3 508N 8 300N + CV
- CV = - 3 508N 8 300N
CV = 11 808N
Now we copy CV up to its location on the
freebody.
CV = 11 808N
-
8/3/2019 Derrick Problem
5/13
T = -8300N
25
2
1
A
C
B
D
Freebody at D
AHCH
CV
T = 8300N
8300 cos 25= -7522N8300 sin 25
= -3508N
Now we can find CH from CV by the
slope.. CH = CV.= 11 808N = 8 856N
CV = 11 808N
Now copy CH up to the drawing.
CH = 8 856N
-
8/3/2019 Derrick Problem
6/13
T = -8300N
25
2
1
A
C
B
D
Freebody at D
AHCH
CV
T = 8300N
8300 cos 25= -7522N8300 sin 25
= -3508N
Now we can findAH from a horizontal
force balance. HD = 0 = - 7 522N + CH + AH
0 = - 7 522N + 8 856N + AH
- AH = 1 334N AH = -1 334N
CV = 11 808N
Now copyAH up to the drawing.
CH = 8 856N AH = -1 334N
-
8/3/2019 Derrick Problem
7/13
T = -8300N
25
2
1
A
C
B
D
Freebody at D
AHCH
CV
T = 8300N
SinceAH is negative we need to switch
the arrow around.
CV = 11 808N
8300 cos 25= -7522N8300 sin 25
= -3508N
CH = 8 856N AH = -1 334N
-
8/3/2019 Derrick Problem
8/13
T = -8300N
25
2
1
A
B
D
Freebody at C
CH
T = 8300N
Now we need to do a freebody at C
Our first step is to find DH and DV fromCH and CV
DH = - CH = -11 808N by observation
And DV = - CV = - 8 856N also
Now draw DH and DV on the diagram
CV = 11 808N
8300 cos 25= -7522N8300 sin 25
= -3508N
CH = 8 856N
DH = - 8 856N
DV =
- 11 808N
-
8/3/2019 Derrick Problem
9/13
25
2
-1
A
B
D
Freebody at CDH = - 8 856N
DV =
- 11 808N
BH =
- 4 428N
+ 8 856N
C
BV =
Now we need to find BH fromDH B
H
= - DH
= - - 8 856N = + 8 856N
by observation
Now draw BH on the diagram
Now BV = - BH
BV = - 8 856N = - 4 428Nalso by observation
Now draw BV on the diagram
-
8/3/2019 Derrick Problem
10/13
25
2
-1
A
B
D
Freebody at CDH = - 8 856N
DV =
- 11 808N
BH =
- 4 428N
+ 8 856N
C
BV =
BV and BH can be relocated down to B
to become the reactions at B
A vertical balance at C can find AV from
BV
and DV
VC = 0 = BV + DV + AV0 = - 4 428N 11 808N + AV
- AV = - 4 428N 11 808N
+ AV = +16 236N Now draw AV on the diagram
- 4 428N
+ 8 856NBH =
BV =
AV = 16 236N
-
8/3/2019 Derrick Problem
11/13
25
2
-1
B
D
Freebody at CDH = - 8 856N
DV =
- 11 808N
BH =
- 4 428N
+ 8 856N
C
BV =
So now all the reaction forces areknown
- 4 428N
+ 8 856NBH =
BV =
AV = 16 236N
AH = -1 334N
T = 8300N
T = 8300N
-
8/3/2019 Derrick Problem
12/13
-
8/3/2019 Derrick Problem
13/13
T = 8 300.0N
25
2
1
A
C
B
D
2m
2m
4m3m
