PROBLEM 2.111
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 840 lb, determine the vertical force
P exerted by the tower on the pin at A.
SOLUTION
0: 0
AB AC AD
PΣ= + + + =F TTT j
Free–Body Diagram at A:
PROBLEM 2.111 (Continued)
4 30 10
21 59 63
20 50 50
21 59 63
5 9 37 0
21 59 63
AB AC AD
AB AC AD
AB AC AD
TTT
T T TP
TTT
−+ −
+− − − +
+ +− =
i
j
k
PROBLEM 2.112
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 AC is 590 lb, determine the vertical force
P exerted by the tower on the pin at A.
SOLUTION
0: 0
AB AC AD
PΣ= + + + =F TTT j
Free–Body Diagram at A:
PROBLEM 2.112 (Continued)
4 30 10
21 59 63
20 50 50
21 59 63
5 9 37 0
21 59 63
AB AC AD
AB AC AD
AB AC AD
TTT
T T TP
TTT
−+ −
+− − − +
+ +− =
i
j
k
Setting the coefficients of
, , ,i jk
equal to zero:
Set
590 lb
AC
T=
in Eqs. (1) – (3):
(1′)
PROBLEM 2.113
In trying to move across a slippery icy surface, a 175-lb man uses
two ropes AB and AC. Knowing that the force exerted on the man
by the icy surface is perpendicular to that surface, determine the
tension in each rope.
SOLUTION
Free–Body Diagram at A
PROBLEM 2.113 (Continued)
Substituting the expressions obtained for
, ,,
AB AC
TTN
and W; factoring i, j, and k; and equating each of
the coefficients to zero gives the following equations:
PROBLEM 2.114
Solve Problem 2.113, assuming that a friend is helping the man
at A by pulling on him with a force P = −(45 lb)k.
PROBLEM 2.113 In trying to move across a slippery icy
surface, a 175–lb man uses two ropes AB and AC. Knowing that
the force exerted on the man by the icy surface is perpendicular
to that surface, determine the tension in each rope.
SOLUTION
Refer to Problem 2.113 for the figure and analysis leading to the following set of equations, Equation (3)
being modified to include the additional force
( 45 lb) .= −Pk
PROBLEM 2.115
For the rectangular plate of Problems 2.109 and 2.110,
determine the tension in each of the three cables knowing that
the weight of the plate is 792 N.
SOLUTION
See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and
(3) below. Setting
792 NP=
gives:
PROBLEM 2.116
For the cable system of Problems 2.107 and 2.108,
determine the tension in each cable knowing that
2880 NP=
and
0.Q=
SOLUTION
0: 0
A AB AC AD
Σ = + + ++ =F T T T PQ
Where
P=Pi
and
Q=Qj
PROBLEM 2.116 (Continued)
Solving the system of linear equations using conventional algorithms gives:
PROBLEM 2.117
For the cable system of Problems 2.107 and 2.108,
determine the tension in each cable knowing that
2880 NP=
and
576 N.Q=
SOLUTION
See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below:
48 12 48 0
53 13 61
AB AC AD
T T TP− − − +=
(1)
PROBLEM 2.118
For the cable system of Problems 2.107 and 2.108,
determine the tension in each cable knowing that
2880 NP=
and
576Q= −
N. (Q is directed downward).
SOLUTION
See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below:
48 12 48 0
53 13 61
AB AC AD
T T TP− − − +=
(1)
PROBLEM 2.119
For the transmission tower of Probs. 2.111 and 2.112,
determine the tension in each guy wire knowing that the
tower exerts on the pin at A an upward vertical force of
1800 lb.
PROBLEM 2.111 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 840 lb, determine
the vertical force P exerted by the tower on the pin at A.
SOLUTION
See Problem 2.111 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and
(3) below:
:i
4 30 10 0
21 59 63
AB AC AD
TTT−+ − =
(1)
4 30 10 0
TTT−+ − =
PROBLEM 2.120
Three wires are connected at point D, which is
located 18 in. below the T-shaped pipe support
ABC. Determine the tension in each wire when
a 180-lb cylinder is suspended from point D as
shown.
SOLUTION
Free–Body Diagram of Point D:
PROBLEM 2.120 (Continued)
and
Substituting the expressions obtained for
, , and
DA DB DC
TT T
and factoring i, j, and k:
( 0.70588 0.70588 )
(0.63324 0.52941 0.52941 )
(0.77397 0.47059 0.47059 )
DB DC
DA DB DC
DA DB DC
TT
T T TW
TTT
−+
++−
−−
i
j
k
Equating to zero the coefficients of i, j, k:
PROBLEM 2.121
A container of weight W is suspended from ring
A, to which cables AC and AE are attached. A
force P is applied to the end F of a third cable
that passes over a pulley at B and through ring A
and that is attached to a support at D. Knowing
that
1000 N,W=
determine the magnitude of P.
(Hint: The tension is the same in all portions of
cable FBAD.)
SOLUTION
The (vector) force in each cable can be written as the product of the (scalar) force and the unit vector
along the cable. That is, with
PROBLEM 2.121 (Continued)
Finally,
PROBLEM 2.122
Knowing that the tension in cable AC of the
system described in Problem 2.121 is 150 N,
determine (a) the magnitude of the force P, (b)
the weight W of the container.
PROBLEM 2.121 A container of weight W is
suspended from ring A, to which cables AC and
AE are attached. A force P is applied to the end F
of a third cable that passes over a pulley at B and
through ring A and that is attached to a support at
D. Knowing that
1000 N,W=
determine the
magnitude of P. (Hint: The tension is the same in
all portions of cable FBAD.)
SOLUTION
Here, as in Problem 2.121, the support of the container consists of the four cables AE, AC, AD, and AB,
PROBLEM 2.123
Cable BAC passes through a frictionless ring A and is
attached to fixed supports at B and C, while cables AD and
AE are both tied to the ring and are attached, respectively, to
supports at D and E. Knowing that a 200-lb vertical load P is
applied to ring A, determine the tension in each of the three
cables.
SOLUTION
Free Body Diagram at A:
Since
tension in
BAC
T=
cable BAC, it follows that
AB AC BAC
TTT= =
PROBLEM 2.123 (Continued)
Substituting into
0,
A
Σ=F
setting
( 200 lb) ,= −Pj
and setting the coefficients of i, j, k equal to
,
φ
we
obtain the following three equilibrium equations:
From
17.5 4
:0
62.5 5
BAC AD
TT− +=i
(1)