561
6–69.
Determine the reactions at supports Aand B.
SOLUTION
Member DB:
a
Ans.
Member AC:
Ans.
Ay=5.10 kip
:
+©Fx=0; Ax–3.50a4
5b=0
Bx=2.80 kip
+©MB=0; 3.15 (6) –FCD a3
5b(9) =0
6ft
500lb/ft
6ft
8ft
9ft
700lb/ft
6ft
AC
D
B
Ans:
A
y
=5.10 kip
562
SOLUTION
Free Body Diagram. The solution will be very much simplified if one realizes that
member AB is a two force member. Also, the tension in the cable is equal to the
weight of the cylinder and is constant throughout the cable.
Equations of Equilibrium. Consider the equilibrium of member BC by referring to
its FBD, Fig. a,
a+
ΣMC=0;
FAB
(2) +75(9.81)(0.3) –75(9.81)(2.8) =0
S
+ Σ
F
x=
0; 1532.81
–75(9.81) –Cx=0
Thus,
FB=FAB =1532.81 N
6–70.
Determine the horizontal and vertical components of force
at pins B and C. The suspended cylinder has a mass of 75 kg.
A
B
C
1.5 m
0.3 m
2 m
0.5 m
563
SOLUTION
Free Body Diagram. The compound beam is being dismembered into members AB,
BD and DE of which their respective FBDs are shown in Fig. a, b and c.
Equations of Equilibrium. Equilibrium of member DE will be considered first by
referring to Fig. c.
NE=18.0 kN
Next, member BD, Fig. b.
Finally, member AB, Fig. a
6–71.
Determine the reactions at the supports A, C, and E of the
compound beam.
4 m 3 m
3 m 6 m
2 m
AD
BE
C
3 kN/m
12 kN
Ans:
NE=18.0 kN
564
SOLUTION
Free Body Diagram. The frame is being dismembered into members AC and BC of
which their respective FBDs are shown in Fig. a and b.
Equations of Equilibrium. Write the moment equation of equilibrium about point A
for member AC, Fig. a and point B for member BC, Fig. b.
Solving Eqs. (1) and (2)
The negative sign indicates that
C
y acts in the sense opposite to that shown in the
FBD write the force equation of equilibrium for member AC, Fig. a,
S
+ Σ
F
x=
0; Ax+200
sin 60°–461.88 =0
Ax=61.88 N
Also, for member BC, Fig. b
S
+ Σ
F
x=
0;
Bx+461.88 –800 =0
Bx=338.12 N
Thus,
FC
=2
Cx
2
+
Cy
2
=2
461.88
2
+
(
–
338.12)2
=
572.41 N
=
572 N Ans.
=2
+
=2
*6–72.
Determine the resultant force at pins A, B, and C on the
three-member frame.
200 N/m
60fi
2 m
800 N
2 m
B
C
A
565
SOLUTION
Free Body Diagram. The solution will be very much simplified if one realizes that
member CD is a two force member.
Equation of Equilibrium. Consider the equilibrium of member BD, Fig. b
S
+ Σ
NE=3600 N =3.60 kN
NB=900 N
Then the equilibrium of member AC gives
S
+ Σ
F
x=
0;
Ax=0
Ans.
6–73.
Determine the reactions at the supports at A, E, and B of
the compound beam.
3 m
900 N/m 900 N/m
4 m3 m
ACD
B
3 m
3 m
E
Ans:
NE=3.60 kN
NB=900 N
566
6–74.
The wall crane supports a load of 700 lb. Determine the
horizontal and vertical components of reaction at the pins A
and D.Also, what is the force in the cable at the winch W?
SOLUTION
Pulley E:
Member ABC:
a
T
BD =2409 lb
T
BD sin 45°(4) –350 sin 60°(4) #700 (8) =0+©MA=0;
4 ft
D
AB
C
W
4 ft
700 lb
4 ft
Ans:
567
6–75.
The wall crane supports a load of 700 lb. Determine the
horizontal and vertical components of reaction at the pins A
and D.Also, what is the force in the cable at the winch W?
The jib ABC has a weight of 100 lb and member BD has a
weight of 40 lb. Each member is uniform and has a center of
gravity at its center.
4ft
D
AB
C
E
4ft
700 lb
60
4ft
SOLUTION
Pulley E:
Member ABC:
a
Member DB:
a
From Eq. (1)
Ans.A
x
=2.00 kip
+©M
D
=0; –40 (2) –1803.1 (4) +B
x
(4) =0
+©M
A
=0; B
y
(4) –700 (8) –100 (4) –350 sin 60° (4) =0
+c©F
y
=0; 2T–700 =0
568
SOLUTION
Free Body Diagram. The frame will be dismembered into members AD, EF, CD
Equations of Equilibrium. Write the moment equations of equilibrium about point
A for member AD, Fig. a, and point B for member BC, Fig. b.
a+
ΣMA=0;
FEF
a4
5b
(3) –FCD (4.5) –400 (4.5)(2.25) =0 (1)
ΣMB=0;
Solving Eqs. (1) and (2)
FEF =3375 N
FCD =900 N
Write the force equation of equilibrium for member AD, Fig. a,
S
+
x=
Also, for member BC, Fig. b
S
+ Σ
F
x=
0; 3375
a
4
5b
–900 –Bx=0
Bx=1800 N =1.80 kN
Ans.
*6–76.
Determine the horizontal and vertical components of force
which the pins at A and B exert on the frame. 400 N/m
1.5 m
2 m
3 m
3 m
1.5 m
F
E
D
C
569
6–77.
SOLUTION
Member AC:
a
Member BDE:
a
+c©Fy=0; –500 –120 a3
5b+2180 –Ey=0
+©Mg=0; 500 (8) +120a3
5b(5) –Dy(2) =0
NC=120 lb
+©MA=0; NC(5) –600 =0
The two-member structure is connected at Cby a pin, which
is fixed to BDE and passes through the smooth slot in
member AC. Determine the horizontal and vertical
components of reaction at the supports.
3ft3ft2ft
4ft
A
B
CD
E
600 lb ft
500 lb
Ans:
Ax=96
lb
A
y
=72
lb
570
SOLUTION
Free Body Diagram. The compound beam will be dismembered into members ABD
and CD of which their respective FBD are shown in Fig. a and b.
Equations of Equilibrium. First, consider the equilibrium of member CD, Fig. b,
a+
ΣMD=0;
NC (3) –2(3)(1.5) =0
NC=3.00 kN
Ans.
Dx=0
Next, the equilibrium of member ABD gives,
a+
ΣMB=0;
2(9)(1.5) –3.00(3) –NA(6) =0
NA=3.00 kN
Ans.
ΣMA=0;
Bx=0
6–78.
The compound beam is pin supported at B and supported
by rockers at A and C. There is a hinge (pin) at D. Determine
the reactions at the supports. C
D
B
A
6 m 3 m
2 kN/m
3 m
Ans:
NC=3.00 kN
NA=3.00 kN
Bx=0
6–79.
The toggle clamp is subjected to a force Fat the handle.
Determine the vertical clamping force acting at E.
1.5 a
1.5 a
60
a/2
a/2
a/2
E
C
D
F
B
A
SOLUTION
Free Body Diagram: The solution for this problem will be simplified if one realizes
that member CD is a two force member.
Equations of Equilibrium: From FBD (a),
a
From (b),
F
CD
cos 30°aa
CD
sin 30°aa
B
=0;
572
*6–80.
573
6–81.
C
D
E
F
G
H
2m
1m
1m
2m1m
2m
AB
The hoist supports the 125-kg engine. Determine the force
the load creates in member DB and in member FB, which
contains the hydraulic cylinder H.
SOLUTION
Free Body Diagrams: The solution for this problem will be simplified if one realizes
that members FB and DB are two-force members.
Equations of Equilibrium: For FBD(a),
a
From FBD (b),
:
+©Fx=0; Ex–1938.87
¢
1
210
≤
=0
+©ME=0; 1226.25(3) –FFB
¢
3
210
≤
(2) =0
Ans:
6–82.
A 5-lb force is applied to the handles of the vise grip.
Determine the compressive force developed on the smooth
bolt shank Aat the jaws.
5lb
5lb
3 in.1 in.1.5 in.
20fi
A
B
E
C
D
1in.
0.75 in.
SOLUTION
From FBD (a)
a
+©ME=0;
5(4) –FCD sin 30.26°(1) =0FCD =39.693 lb
Ans:
575
SOLUTION
Free Body Diagram. The assembly will be dismembered into members GFE, EDC,
Equations of Equilibrium. First, consider the equilibrium of member GFE, Fig. a,
a+
ΣME=0;
6(3) –FFD
ΣMF=0;
E
(1) =0 F
FD =
9
2
5 kN
=
20.1 kN Ans.
Next, for member EDC, Fig. b,
a+
ΣMC=0;
9.00(3) –
a
9
2
5
b
a
1
25b
(1) –FBD
a
1
22b
(1) =0
ΣMD=0;
Finally, for member ABC, Fig. c
a+
ΣMA=0;
C″
y (3) –
a
18
2
2
b
a1
22b
(2) =0
C
y
″=12.0 kN
Ans.
6–83.
Determine the force in members FD and DB of the frame.
Also, find the horizontal and vertical components of
reaction the pin at C exerts on member ABC and
member EDC.
B
A
G
F
E
D
C
2 m
6 kN
1 m
2 m1 m
576
SOLUTION
Free Body Diagram. The FBDs of the entire assembly, member CDB and the
cylinder are shown in Figs. a, b and c, respectively.
Equations of Equilibrium. First consider the equilibrium of the entire assembly,
Fig.a,
Next, for member CDB, Fig. b
Finally for the cylinder, Fig. c
*6–84.
Determine the force that the smooth 20-kg cylinder exerts
on members AB and CDB. Also, what are the horizontal
and vertical components of reaction at pin A?
C
1
m
1.5 m
A
2 m
BE
D
577
6–85.
The three power lines exert the forces shown on the truss
joints, which in turn are pin-connected to the poles AH and
EG.Determine the force in the guy cable AI and the pin
reaction at the support H.
125ft
20 ft
AE
B
800lb
800lb800 lb
IF
CD
HG
20 ft20 ft
40 ft40 ft
30 ft
30 ft 30 ft 30 ft 50 ft50 ft
SOLUTION
Joint B:
Joint C:
Joint A:
Solving,
:
+©F
x
=0; –T
AI
sin 21.801° –F
H
cos 76.504° +1264.91 cos 18.435° +1131.37 cos 45° =0
Ans:
578
6–86.
G
w
G
C
D
AB
E
C
M
SOLUTION
Free-Body Diagram:The solution for this problem will be simplified if one realizes
Equations of Equilibrium:From FBD (a),
a
From (b),
F
AD
sin 70°152–60162–250172=0+©M
B
=0;
A
6–87.
Determine the force that the jaws Jof the metal cutters
exert on the smooth cable Cif 100-N forces are applied to
the handles.The jaws are pinned at Eand A, and Dand B.
There is also a pin at F.
F
15$
15$
20 mm
20 mm
30 mm 80 mm
B
J
C
D
E
A
15$
15$
400 mm
400 mm
100 N
100 N
SOLUTION
Free Body Diagram:The solution for this problem will be simplified if one realizes
that member EDis a two force member.
Equations of Equilibrium:From FBD (b),
From FBD (b),
Ans:
580
*6–88.
Themachine shown is used for forming metal plates.It
consists of two toggles ABCandDEF,which are operated by
the hydraulic cylinder H.The toggles push the movable bar G
forward, pressing the plate pinto the cavity.Ifthe force which
the plate exerts on the head is determine the
force Fin the hydraulic cylinder when u=30°.
P=12 kN,
200mm
fiF
F
Pfi12 kN
H
F
C
G
A
D
E
Bp
200mm
200mm
200mm
ufi30
ufi30
SOLUTION
Member EF:
Joint E: