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PROBLEM 2.108
Knowing that
20 ,
α
= °
determine the tension (a) in cable
AC, (b) in rope BC.
SOLUTION
Free-Body Diagram Force Triangle
Law of sines:
1200 lb
sin 110 sin 5 sin 65
AC BC
TT
= =
°° °
(a)
1200 lb sin 110
sin 65
AC
T= °
°
1244 lb
AC
T=
(b)
1200 lb sin 5
sin 65
BC
T= °
°
115.4 lb
BC
T=
PROBLEM 2.109
For W = 800 N, P = 200 N, and d = 600 mm, determine
the value of h consistent with equilibrium.
SOLUTION
Free-Body Diagram
800 N
AC BC
TT= =
( )
22
AC BC h d= = +
22
0: 2(800 N) 0
y
h
FP
hd
Σ = −=
+
2
800 1
2
Pd
h
= +
PROBLEM 2.110
Three forces are applied to a bracket as shown. The directions of the two 150-N
forces may vary, but the angle between these forces is always 50°. Determine
the range of values of α for which the magnitude of the resultant of the forces
acting at A is less than 600 N.
SOLUTION
Combine the two 150-N forces into a resultant force Q:
2(150 N)cos25
271.89 N
Q= °
=
Equivalent loading at A:
Using the law of cosines:
22 2
(600 N) (500 N) (271.89 N) 2(500 N)(271.89 N)cos(55 )
cos(55 ) 0.132685
α
α
= + + °+
°+ =
Two values for
:
α
55 82.375
27.4
α
α
°+ =
= °
or
55 82.375
55 360 82.375
222.6
α
α
α
°+ =− °
°+ = °− °
= °
For 600 lb:R<
27.4 222.6
α
°< <
consent of McGraw-Hill Education.
PROBLEM 2.111
Cable AB is 65 ft long, and the tension in that cable is 3900 lb.
Determine (a) the x, y, and z components of the force exerted by the
cable on the anchor B, (b) the angles
,
x
θ
,
y
θ
and
z
θ
defining the
direction of that force.
SOLUTION
From triangle AOB:
56 ft
cos 65 ft
0.86154
30.51
y
y
θ
θ
=
=
= °
(a)
sin cos20
(3900 lb)sin30.51 cos20
xy
FF
θ
=−°
=− °°
1861 lb
x
F= −
cos (3900 lb)(0.86154)
yy
FF
θ
=+=
3360 lb
y
F= +
(3900 lb)sin 30.51° sin 20°
z
F= +
677 lb
z
F= +
(b)
1861 lb
cos 0.4771
3900 lb
x
x
F
F
θ
==−=−
118.5
x
θ
= °
From above:
30.51
y
θ
= °
30.5
y
θ
= °
677 lb
cos 0.1736
3900 lb
z
z
F
F
θ
==+=+
80.0
z
θ
= °
consent of McGraw-Hill Education.
PROBLEM 2.112
Three cables are used to tether a balloon as shown. Determine the
vertical force P exerted by the balloon at A knowing that the tension
in cable AB is 259 N.
SOLUTION
The forces applied at A are:
, , , and
AB AC AD
TTT P
(4.20 m) (5.60 m) 7.00 m
(2.40 m) (5.60 m) (4.20 m) 7.40 m
(5.60 m) (3.30 m) 6.50 m
AB AB
AC AC
AD AD
=−− =
=−+ =
=−− =
ij
ijk
jk
C
C
C
and
( 0.6 0.8 )
(0.32432 0.75676 0.56757 )
( 0.86154 0.50769 )
AB AB AB AB AB
AC AC AC AC AC
AD AD AD AD AD
AB
TT T
AB
AC
TT T
AC
AD
TT T
AD
= = =−−
===−+
===−−
Tλ ij
Tλ jk
Tλ jk
C
C
C
SOLUTION Continued
Equilibrium condition
0: 0
AB AC AD
FP
Σ= + + + =TTT j
Substituting the expressions obtained for
, , and
AB AC AD
TT T
and factoring i, j, and k:
( 0.6 0.32432 ) ( 0.8 0.75676 0.86154 )
(0.56757 0.50769 ) 0
AB AC AB AC AD
AC AD
T T T T TP
TT
− + +− − − +
+− =
ij
k
0.6 0.32432 0
AB AC
TT−+ =
(1)
0.8 0.75676 0.86154 0
AB AC AD
T T TP− − − +=
(2)
0.56757 0.50769 0
AC AD
TT−=
(3)
AB
479.15 N
535.66 N
AC
AD
T
T
=
=
1031 N=P
consent of McGraw-Hill Education.
PROBLEM 2.113
Three cables are used to tether a balloon as shown. Determine the
vertical force P exerted by the balloon at A knowing that the tension in
cable AC is 444 N.
SOLUTION
See Problem 2.112 for the figure and the analysis leading to the linear algebraic Equations (1), (2),
and (3) below:
0.6 0.32432 0
AB AC
TT−+ =
(1)
0.8 0.75676 0.86154 0
AB AC AD
T T TP− − − +=
(2)
0.56757 0.50769 0
AC AD
TT−=
(3)
AC
equations using conventional algorithms gives
240 N
496.36 N
AB
AD
T
T
=
=
956 N=P
consent of McGraw-Hill Education.
PROBLEM 2.114
A transmission tower is held by three guy wires attached to a
pin at A and anchored by bolts at B, C, and D. If the tension in
wire AB is 630 lb, determine the vertical force P exerted by the
tower on the pin at A.
SOLUTION
Free Body A:
0: 0
AB AC AD
PΣ= + + + =F TTT j
= 45 90 30 105 ft
30 90 65 115 ft
20 90 60 110 ft
AB AB
AC AC
AD AD
−− + =
=−+ =
=−− =
ijk
ijk
ijk
C
C
C
We write
362
777
AB AB AB AB
AB
AB
TT
AB
T
= =
=−−+
Tλ
ijk
C
6 18 13
23 23 23
AC AC AC AC
AC
AC
TT
AC
T
= =
= −+
Tλ
ijk
C
296
11 11 11
AD AD AD AD
AD
AD
TT
AD
T
= =
= −−
Tλ
ijk
C
Substituting into the Eq.
0Σ=F
and factoring
, , :i jk
36 2
7 23 11
6 18 9
7 23 11
2 13 6 0
7 23 11
AB AC AD
AB AC AD
AB AC AD
TTT
T T TP
TTT
−+ +
+− − − +
++− =
i
j
k
PROBLEM 2.114 (Continued)
Setting the coefficients of
, , ,i jk
equal to zero:
:i
36 20
7 23 11
AB AC AD
TTT−+ + =
(1)
:j
6 18 9 0
7 23 11
AB AC AD
T T TP− − − +=
(2)
:k
2 13 6 0
7 23 11
AB AC AD
TTT+−=
(3)
Set
630 lb
AB
T=
in Eqs. (1) – (3):
62
270 lb 0
23 11
AC AD
TT−+ + =
(1′)
18 9
540 lb 0
23 11
AC AD
T TP− − − +=
(2′)
13 6
180 lb 0
23 11
AC AD
TT+−=
(3′)
Solving,
467.42 lb 814.35 lb 1572.10 lb
AC AD
TTP
= = =
1572 lbP=
consent of McGraw-Hill Education.
PROBLEM 2.115
A rectangular plate is supported by three cables as shown. Knowing
that the tension in cable AD is 520 N, determine the weight of the plate.
SOLUTION
See Problem 2.114 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and
(3) below:
85
0.6 0
17 13
AB AC AD
TTT−+ + =
(1)
12 9.6
0.64 0
17 13
AB AC AD
T T TP
− + − +=
(2)
9 7.2
0.48 0
17 13
AB AC AD
TTT+ −=
(3)
Making
520 N
AD
T=
in Eqs. (1) and (3):
80.6 200 N 0
17
AB AC
TT−+ +=
(1′)
90.48 288 N 0
17 AB AC
TT+ −=
(3′)
Multiply (1′) by 9, (3′) by 8, and add:
9.24 504 N 0 54.5455 N
AC AC
TT−= =
Substitute into (1′) and solve for
:
AB
T
17 (0.6 54.5455 200) 494.545 N
8
AB AB
TT=×+ =
Substitute for the tensions in Eq. (2) and solve for P:
12 9.6
(494.545 N) 0.64(54.5455 N) (520 N)
17 13
768.00 N
P=++
=
Weight of plate
768 NP= =
consent of McGraw-Hill Education.
PROBLEM 2.108
Knowing that
20 ,
α
= °
determine the tension (a) in cable
AC, (b) in rope BC.
SOLUTION
Free-Body Diagram Force Triangle
Law of sines:
1200 lb
sin 110 sin 5 sin 65
AC BC
TT
= =
°° °
(a)
1200 lb sin 110
sin 65
AC
T= °
°
1244 lb
AC
T=
(b)
1200 lb sin 5
sin 65
BC
T= °
°
115.4 lb
BC
T=
PROBLEM 2.109
For W = 800 N, P = 200 N, and d = 600 mm, determine
the value of h consistent with equilibrium.
SOLUTION
Free-Body Diagram
800 N
AC BC
TT= =
( )
22
AC BC h d= = +
22
0: 2(800 N) 0
y
h
FP
hd
Σ = −=
+
2
800 1
2
Pd
h
= +
PROBLEM 2.110
Three forces are applied to a bracket as shown. The directions of the two 150-N
forces may vary, but the angle between these forces is always 50°. Determine
the range of values of α for which the magnitude of the resultant of the forces
acting at A is less than 600 N.
SOLUTION
Combine the two 150-N forces into a resultant force Q:
2(150 N)cos25
271.89 N
Q= °
=
Equivalent loading at A:
Using the law of cosines:
22 2
(600 N) (500 N) (271.89 N) 2(500 N)(271.89 N)cos(55 )
cos(55 ) 0.132685
α
α
= + + °+
°+ =
Two values for
:
α
55 82.375
27.4
α
α
°+ =
= °
or
55 82.375
55 360 82.375
222.6
α
α
α
°+ =− °
°+ = °− °
= °
For 600 lb:R<
27.4 222.6
α
°< <
consent of McGraw-Hill Education.
PROBLEM 2.111
Cable AB is 65 ft long, and the tension in that cable is 3900 lb.
Determine (a) the x, y, and z components of the force exerted by the
cable on the anchor B, (b) the angles
,
x
θ
,
y
θ
and
z
θ
defining the
direction of that force.
SOLUTION
From triangle AOB:
56 ft
cos 65 ft
0.86154
30.51
y
y
θ
θ
=
=
= °
(a)
sin cos20
(3900 lb)sin30.51 cos20
xy
FF
θ
=−°
=− °°
1861 lb
x
F= −
cos (3900 lb)(0.86154)
yy
FF
θ
=+=
3360 lb
y
F= +
(3900 lb)sin 30.51° sin 20°
z
F= +
677 lb
z
F= +
(b)
1861 lb
cos 0.4771
3900 lb
x
x
F
F
θ
==−=−
118.5
x
θ
= °
From above:
30.51
y
θ
= °
30.5
y
θ
= °
677 lb
cos 0.1736
3900 lb
z
z
F
F
θ
==+=+
80.0
z
θ
= °
consent of McGraw-Hill Education.
PROBLEM 2.112
Three cables are used to tether a balloon as shown. Determine the
vertical force P exerted by the balloon at A knowing that the tension
in cable AB is 259 N.
SOLUTION
The forces applied at A are:
, , , and
AB AC AD
TTT P
(4.20 m) (5.60 m) 7.00 m
(2.40 m) (5.60 m) (4.20 m) 7.40 m
(5.60 m) (3.30 m) 6.50 m
AB AB
AC AC
AD AD
=−− =
=−+ =
=−− =
ij
ijk
jk
C
C
C
and
( 0.6 0.8 )
(0.32432 0.75676 0.56757 )
( 0.86154 0.50769 )
AB AB AB AB AB
AC AC AC AC AC
AD AD AD AD AD
AB
TT T
AB
AC
TT T
AC
AD
TT T
AD
= = =−−
===−+
===−−
Tλ ij
Tλ jk
Tλ jk
C
C
C
SOLUTION Continued
Equilibrium condition
0: 0
AB AC AD
FP
Σ= + + + =TTT j
Substituting the expressions obtained for
, , and
AB AC AD
TT T
and factoring i, j, and k:
( 0.6 0.32432 ) ( 0.8 0.75676 0.86154 )
(0.56757 0.50769 ) 0
AB AC AB AC AD
AC AD
T T T T TP
TT
− + +− − − +
+− =
ij
k
0.6 0.32432 0
AB AC
TT−+ =
(1)
0.8 0.75676 0.86154 0
AB AC AD
T T TP− − − +=
(2)
0.56757 0.50769 0
AC AD
TT−=
(3)
AB
479.15 N
535.66 N
AC
AD
T
T
=
=
1031 N=P
consent of McGraw-Hill Education.
PROBLEM 2.113
Three cables are used to tether a balloon as shown. Determine the
vertical force P exerted by the balloon at A knowing that the tension in
cable AC is 444 N.
SOLUTION
See Problem 2.112 for the figure and the analysis leading to the linear algebraic Equations (1), (2),
and (3) below:
0.6 0.32432 0
AB AC
TT−+ =
(1)
0.8 0.75676 0.86154 0
AB AC AD
T T TP− − − +=
(2)
0.56757 0.50769 0
AC AD
TT−=
(3)
AC
equations using conventional algorithms gives
240 N
496.36 N
AB
AD
T
T
=
=
956 N=P
consent of McGraw-Hill Education.
PROBLEM 2.114
A transmission tower is held by three guy wires attached to a
pin at A and anchored by bolts at B, C, and D. If the tension in
wire AB is 630 lb, determine the vertical force P exerted by the
tower on the pin at A.
SOLUTION
Free Body A:
0: 0
AB AC AD
PΣ= + + + =F TTT j
= 45 90 30 105 ft
30 90 65 115 ft
20 90 60 110 ft
AB AB
AC AC
AD AD
−− + =
=−+ =
=−− =
ijk
ijk
ijk
C
C
C
We write
362
777
AB AB AB AB
AB
AB
TT
AB
T
= =
=−−+
Tλ
ijk
C
6 18 13
23 23 23
AC AC AC AC
AC
AC
TT
AC
T
= =
= −+
Tλ
ijk
C
296
11 11 11
AD AD AD AD
AD
AD
TT
AD
T
= =
= −−
Tλ
ijk
C
Substituting into the Eq.
0Σ=F
and factoring
, , :i jk
36 2
7 23 11
6 18 9
7 23 11
2 13 6 0
7 23 11
AB AC AD
AB AC AD
AB AC AD
TTT
T T TP
TTT
−+ +
+− − − +
++− =
i
j
k
PROBLEM 2.114 (Continued)
Setting the coefficients of
, , ,i jk
equal to zero:
:i
36 20
7 23 11
AB AC AD
TTT−+ + =
(1)
:j
6 18 9 0
7 23 11
AB AC AD
T T TP− − − +=
(2)
:k
2 13 6 0
7 23 11
AB AC AD
TTT+−=
(3)
Set
630 lb
AB
T=
in Eqs. (1) – (3):
62
270 lb 0
23 11
AC AD
TT−+ + =
(1′)
18 9
540 lb 0
23 11
AC AD
T TP− − − +=
(2′)
13 6
180 lb 0
23 11
AC AD
TT+−=
(3′)
Solving,
467.42 lb 814.35 lb 1572.10 lb
AC AD
TTP
= = =
1572 lbP=
consent of McGraw-Hill Education.
PROBLEM 2.115
A rectangular plate is supported by three cables as shown. Knowing
that the tension in cable AD is 520 N, determine the weight of the plate.
SOLUTION
See Problem 2.114 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and
(3) below:
85
0.6 0
17 13
AB AC AD
TTT−+ + =
(1)
12 9.6
0.64 0
17 13
AB AC AD
T T TP
− + − +=
(2)
9 7.2
0.48 0
17 13
AB AC AD
TTT+ −=
(3)
Making
520 N
AD
T=
in Eqs. (1) and (3):
80.6 200 N 0
17
AB AC
TT−+ +=
(1′)
90.48 288 N 0
17 AB AC
TT+ −=
(3′)
Multiply (1′) by 9, (3′) by 8, and add:
9.24 504 N 0 54.5455 N
AC AC
TT−= =
Substitute into (1′) and solve for
:
AB
T
17 (0.6 54.5455 200) 494.545 N
8
AB AB
TT=×+ =
Substitute for the tensions in Eq. (2) and solve for P:
12 9.6
(494.545 N) 0.64(54.5455 N) (520 N)
17 13
768.00 N
P=++
=
Weight of plate
768 NP= =
consent of McGraw-Hill Education.
