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PROBLEM 6.90
Since the brace shown must remain in position even when the magnitude of P is
very small, a single safety spring is attached at D and E. The spring DE has a
constant of 50 lb/in. and an unstretched length of 7 in. Knowing that l = 10 in. and
that the magnitude of P is 800 lb, determine the force Q required to release the
brace.
PROBLEM 6.91
Determine the force P that must be applied to the toggle CDE to
maintain bracket ABC in the position shown.
SOLUTION
We note that CD and DE are two-force members.
Free body: Joint D:
Similarly,
It follows that
Also,
Free body: Member ABC:
()
() ()5()
30 150
CD y
CD x CD y CD x
F
FFF= =
()5()
DE y DE x
FF=
0: ( ) ( )
y DE y CD y
F FFΣ= =
( )( )
DE x CD x
FF
=
0: ( ) ( ) 0
x DE x CD x
F PF FΣ= − − =
1
( )( )2
DE x CD x
FF P= =
1
( ) ( ) 5 2.5
2
DE y CD y
F F PP
= = =
1
0: (2.5 )(300) (450) (910 N)(150) 0
2
A
M PP
Σ= + − =
(750 225) (910 N)(150)P+=
140.0 NP=
Dimensions in mm
PROBLEM 6.92
Determine the force P that must be applied to the toggle CDE to maintain
bracket ABC in the position shown.
SOLUTION
We note that CD and DE are two-force members.
Free body: Joint D:
Similarly,
It follows that
Also,
Free body: Member ABC:
()
() ()5()
30 150
CD y
CD x CD y CD x
F
FFF= =
()5()
DE y DE x
FF=
0: ( ) ( )
y DE y CD y
F FFΣ= =
( )( )
DE x CD x
FF
=
0: ( ) ( ) 0
x DE x CD x
F F FPΣ = + −=
1
( )( )2
DE x CD x
FF P
= =
1
( ) ( ) 5 2.5
2
DE y CD y
F F PP
= = =
1
0: (2.5 )(300) (450) (910 N)(150) 0
2
A
M PP
Σ= − − =
(750 225) (910 N)(150)P−=
260 NP=
Dimensions in
mm
PROBLEM 6.93
The telescoping arm ABC is used to provide an elevated
platform for construction workers. The workers and the
platform together have a mass of 200 kg and have a
combined center of gravity located directly above C. For the
position when
θ
= 20°, determine (a) the force exerted at B
by the single hydraulic cylinder BD, (b) the force exerted on
the supporting carriage at A.
SOLUTION
Geometry:
Free body: Arm ABC:
We note that BD is a two-force member.
(a)
(b)
(5 m)cos 20 4.6985 m
(2.4 m)cos 20 2.2553 m
(2.4 m)sin 20 0.8208 m
0.5 1.7553 m
0.9 1.7208 m
a
b
c
db
ec
= °=
= °=
= °=
=−=
=+=
1.7208
tan ; 44.43
1.7553
e
d
ββ
= = = °
2
(200 kg)(9.81 m/s ) 1.962 kNW= =
0: (1.962 kN)(4.6985 m) sin 44.43 (2.2553 m) cos 44.43(0.8208 m) 0
A BD BD
M FFΣ= − ° + =
9.2185 (0.9927) 0: 9.2867 kN
BD BD
FF−==
9.29 kN
BD
=F
44.4°
0: cos 0
x x BD
F AF
β
Σ= − =
(9.2867 kN)cos 44.43 6.632 kN
x
A= °=
6.632 kN
x
=A
0: 1.962 kN sin 0
y y BD
FA F
β
Σ= − + =
1.962 kN (9.2867 kN)sin 44.43 4.539 kN
y
A= − °=−
4.539 kN
y
=A
8.04 kN=A
34.4°
PROBLEM 6.94
The telescoping arm ABC of Prob. 6.93 can be lowered until
end C is close to the ground, so that workers can easily board
the platform. For the position when
θ
= −20°, determine
(a) the force exerted at B by the single hydraulic cylinder
BD, (b) the force exerted on the supporting carriage at A.
SOLUTION
Geometry:
Free body: Arm ABC:
(a)
(b)
(5 m)cos 20 4.6985 m
(2.4 m)cos 20 2.2552 m
(2.4 m)sin 20 0.8208 m
0.5 1.7553 m
0.9 0.0792 m
a
b
c
db
ec
= °=
= °=
= °=
=−=
= −=
0.0792
tan ; 2.584
1.7552
e
d
ββ
= = = °
2
(200 kg)(9.81 m/s )
1962 N 1.962 kN
W
W
=
= =
0: (1.962 kN)(4.6985 m) sin 2.584 (2.2553 m) cos 2.584 (0.8208 m) 0
A BD BD
M FFΣ= − ° − ° =
9.2185 (0.9216) 0 10.003 kN
BD BD
FF−==
10.00 kN
BD
=F
2.58°
0: cos 0
x x BD
F AF
β
Σ= − =
(10.003 kN)cos 2.583 9.993 kN
x
A= °=
9.993 kN
x
=A
0: 1.962 kN sin 0
y y BD
FA F
β
Σ= − + =
1.962 kN (10.003 kN)sin 2.583 1.5112 kN
y
A= − °=−
1.5112 kN
y=A
PROBLEM 6.95
The bucket of the front-end loader shown carries a
3200-lb load. The motion of the bucket is
controlled by two identical mechanisms, only
one of which is shown. Knowing that the
mechanism shown supports one-half of the
3200-lb load, determine the force exerted (a) by
cylinder CD, (b) by cylinder FH.
SOLUTION
Free body: Bucket: (one mechanism)
Note: There are two identical support mechanisms.
Free body: One arm BCE:
Free body: Arm DFG:
0: (1600 lb)(15 in.) (16 in.) 0
D AB
MFΣ= − =
1500 lb
AB
F=
8
tan 20
21.8
β
β
=
= °
0: (1500 lb)(23 in.) cos 21.8 (15 in.) sin 21.8 (5 in.) 0
E CD CD
M FFΣ= + ° − ° =
2858 lb
CD
F= −
2.86 kips
CD
FC=
0: (1600 lb)(75 in.) sin 45 (24 in.) cos 45 (6 in.) 0
G FH FH
M FFΣ= + ° − ° =
9.428 kips
FH
F= −
9.43 kips
FH
FC=
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw-Hill Education.
PROBLEM 6.96
The motion of the bucket of the front-end loader
shown is controlled by two arms and a linkage
that are pin-connected at D. The arms are located
symmetrically with respect to the central,
vertical, and longitudinal plane of the loader; one
arm AFJ and its control cylinder EF are shown.
The single linkage GHDB and its control cylinder
BC are located in the plane of symmetry. For the
position and loading shown, determine the force
exerted (a) by cylinder BC, (b) by cylinder EF.
SOLUTION
Free body: Bucket
Free body: Arm BDH
Free body: Entire mechanism
(Two arms and cylinders AFJE)
0: (4500 lb)(20 in.) (22 in.) 0
J GH
MF
Σ= − =
4091 lb
GH
F=
0: (4091 lb)(24 in.) (20 in.) 0
D BC
MFΣ=− − =
4909 lb
BC
F= −
4.91 kips
BC
FC=
PROBLEM 6.97
Determine the force in each member of the truss
shown.
SOLUTION
Reactions -Free Body: Entire Truss
0: 12 k N
0 : 0
0: 0
xx
Ey
y
F
M
F
= = ←
= =
= =
∑
∑
∑
D
D
E
Free Body: Joint D
0: cos30 12 kN 0
x CD
FF+= − =
∑
FC
13.86 kN, 13.86 kN
CD CD
F FT=+=
0: (13.86 kN)sin30 0
y AD
FF=+=
∑
6.93 kN, 6.93 kN
AD AD
F FC=−=
SOLUTION CONTINUED
PROBLEM 6.98
Determine the force in member DE and in each of the
members located to the left of DE for the inverted Howe
roof truss shown. State whether each member is in tension
or compression.
SOLUTION
Free body: Truss:
0: 0
xx
FΣ= =
A
0: (400 lb)(4 ) (800 lb)(3 ) (800 lb)(2 ) (800 lb) (4) 0
Hy
M d d d dA dΣ= + + + − =
1600 lb
y
A=
Angles:
6.72
tan 32.52
10.54
6.72
tan 16.26
23.04
aa
ββ
= = °
= = °
Free body: Joint A:
1200 lb
sin57.48 sin106.26 sin16.26
AC
AB
F
F= =
° °°
3613.8 lb
4114.3 lb
AB
AC
FC
FT
=
=
3610 lb , 4110 lb
AB AC
F CF T= =
consent of McGraw-Hill Education.
PROBLEM 6.91
Determine the force P that must be applied to the toggle CDE to
maintain bracket ABC in the position shown.
SOLUTION
We note that CD and DE are two-force members.
Free body: Joint D:
Similarly,
It follows that
Also,
Free body: Member ABC:
()
() ()5()
30 150
CD y
CD x CD y CD x
F
FFF= =
()5()
DE y DE x
FF=
0: ( ) ( )
y DE y CD y
F FFΣ= =
( )( )
DE x CD x
FF
=
0: ( ) ( ) 0
x DE x CD x
F PF FΣ= − − =
1
( )( )2
DE x CD x
FF P= =
1
( ) ( ) 5 2.5
2
DE y CD y
F F PP
= = =
1
0: (2.5 )(300) (450) (910 N)(150) 0
2
A
M PP
Σ= + − =
(750 225) (910 N)(150)P+=
140.0 NP=
Dimensions in mm
PROBLEM 6.92
Determine the force P that must be applied to the toggle CDE to maintain
bracket ABC in the position shown.
SOLUTION
We note that CD and DE are two-force members.
Free body: Joint D:
Similarly,
It follows that
Also,
Free body: Member ABC:
()
() ()5()
30 150
CD y
CD x CD y CD x
F
FFF= =
()5()
DE y DE x
FF=
0: ( ) ( )
y DE y CD y
F FFΣ= =
( )( )
DE x CD x
FF
=
0: ( ) ( ) 0
x DE x CD x
F F FPΣ = + −=
1
( )( )2
DE x CD x
FF P
= =
1
( ) ( ) 5 2.5
2
DE y CD y
F F PP
= = =
1
0: (2.5 )(300) (450) (910 N)(150) 0
2
A
M PP
Σ= − − =
(750 225) (910 N)(150)P−=
260 NP=
Dimensions in
mm
PROBLEM 6.93
The telescoping arm ABC is used to provide an elevated
platform for construction workers. The workers and the
platform together have a mass of 200 kg and have a
combined center of gravity located directly above C. For the
position when
θ
= 20°, determine (a) the force exerted at B
by the single hydraulic cylinder BD, (b) the force exerted on
the supporting carriage at A.
SOLUTION
Geometry:
Free body: Arm ABC:
We note that BD is a two-force member.
(a)
(b)
(5 m)cos 20 4.6985 m
(2.4 m)cos 20 2.2553 m
(2.4 m)sin 20 0.8208 m
0.5 1.7553 m
0.9 1.7208 m
a
b
c
db
ec
= °=
= °=
= °=
=−=
=+=
1.7208
tan ; 44.43
1.7553
e
d
ββ
= = = °
2
(200 kg)(9.81 m/s ) 1.962 kNW= =
0: (1.962 kN)(4.6985 m) sin 44.43 (2.2553 m) cos 44.43(0.8208 m) 0
A BD BD
M FFΣ= − ° + =
9.2185 (0.9927) 0: 9.2867 kN
BD BD
FF−==
9.29 kN
BD
=F
44.4°
0: cos 0
x x BD
F AF
β
Σ= − =
(9.2867 kN)cos 44.43 6.632 kN
x
A= °=
6.632 kN
x
=A
0: 1.962 kN sin 0
y y BD
FA F
β
Σ= − + =
1.962 kN (9.2867 kN)sin 44.43 4.539 kN
y
A= − °=−
4.539 kN
y
=A
8.04 kN=A
34.4°
PROBLEM 6.94
The telescoping arm ABC of Prob. 6.93 can be lowered until
end C is close to the ground, so that workers can easily board
the platform. For the position when
θ
= −20°, determine
(a) the force exerted at B by the single hydraulic cylinder
BD, (b) the force exerted on the supporting carriage at A.
SOLUTION
Geometry:
Free body: Arm ABC:
(a)
(b)
(5 m)cos 20 4.6985 m
(2.4 m)cos 20 2.2552 m
(2.4 m)sin 20 0.8208 m
0.5 1.7553 m
0.9 0.0792 m
a
b
c
db
ec
= °=
= °=
= °=
=−=
= −=
0.0792
tan ; 2.584
1.7552
e
d
ββ
= = = °
2
(200 kg)(9.81 m/s )
1962 N 1.962 kN
W
W
=
= =
0: (1.962 kN)(4.6985 m) sin 2.584 (2.2553 m) cos 2.584 (0.8208 m) 0
A BD BD
M FFΣ= − ° − ° =
9.2185 (0.9216) 0 10.003 kN
BD BD
FF−==
10.00 kN
BD
=F
2.58°
0: cos 0
x x BD
F AF
β
Σ= − =
(10.003 kN)cos 2.583 9.993 kN
x
A= °=
9.993 kN
x
=A
0: 1.962 kN sin 0
y y BD
FA F
β
Σ= − + =
1.962 kN (10.003 kN)sin 2.583 1.5112 kN
y
A= − °=−
1.5112 kN
y=A
PROBLEM 6.95
The bucket of the front-end loader shown carries a
3200-lb load. The motion of the bucket is
controlled by two identical mechanisms, only
one of which is shown. Knowing that the
mechanism shown supports one-half of the
3200-lb load, determine the force exerted (a) by
cylinder CD, (b) by cylinder FH.
SOLUTION
Free body: Bucket: (one mechanism)
Note: There are two identical support mechanisms.
Free body: One arm BCE:
Free body: Arm DFG:
0: (1600 lb)(15 in.) (16 in.) 0
D AB
MFΣ= − =
1500 lb
AB
F=
8
tan 20
21.8
β
β
=
= °
0: (1500 lb)(23 in.) cos 21.8 (15 in.) sin 21.8 (5 in.) 0
E CD CD
M FFΣ= + ° − ° =
2858 lb
CD
F= −
2.86 kips
CD
FC=
0: (1600 lb)(75 in.) sin 45 (24 in.) cos 45 (6 in.) 0
G FH FH
M FFΣ= + ° − ° =
9.428 kips
FH
F= −
9.43 kips
FH
FC=
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw-Hill Education.
PROBLEM 6.96
The motion of the bucket of the front-end loader
shown is controlled by two arms and a linkage
that are pin-connected at D. The arms are located
symmetrically with respect to the central,
vertical, and longitudinal plane of the loader; one
arm AFJ and its control cylinder EF are shown.
The single linkage GHDB and its control cylinder
BC are located in the plane of symmetry. For the
position and loading shown, determine the force
exerted (a) by cylinder BC, (b) by cylinder EF.
SOLUTION
Free body: Bucket
Free body: Arm BDH
Free body: Entire mechanism
(Two arms and cylinders AFJE)
0: (4500 lb)(20 in.) (22 in.) 0
J GH
MF
Σ= − =
4091 lb
GH
F=
0: (4091 lb)(24 in.) (20 in.) 0
D BC
MFΣ=− − =
4909 lb
BC
F= −
4.91 kips
BC
FC=
PROBLEM 6.97
Determine the force in each member of the truss
shown.
SOLUTION
Reactions -Free Body: Entire Truss
0: 12 k N
0 : 0
0: 0
xx
Ey
y
F
M
F
= = ←
= =
= =
∑
∑
∑
D
D
E
Free Body: Joint D
0: cos30 12 kN 0
x CD
FF+= − =
∑
FC
13.86 kN, 13.86 kN
CD CD
F FT=+=
0: (13.86 kN)sin30 0
y AD
FF=+=
∑
6.93 kN, 6.93 kN
AD AD
F FC=−=
SOLUTION CONTINUED
PROBLEM 6.98
Determine the force in member DE and in each of the
members located to the left of DE for the inverted Howe
roof truss shown. State whether each member is in tension
or compression.
SOLUTION
Free body: Truss:
0: 0
xx
FΣ= =
A
0: (400 lb)(4 ) (800 lb)(3 ) (800 lb)(2 ) (800 lb) (4) 0
Hy
M d d d dA dΣ= + + + − =
1600 lb
y
A=
Angles:
6.72
tan 32.52
10.54
6.72
tan 16.26
23.04
aa
ββ
= = °
= = °
Free body: Joint A:
1200 lb
sin57.48 sin106.26 sin16.26
AC
AB
F
F= =
° °°
3613.8 lb
4114.3 lb
AB
AC
FC
FT
=
=
3610 lb , 4110 lb
AB AC
F CF T= =
consent of McGraw-Hill Education.
