PROBLEM 2.99
Using two ropes and a roller chute, two workers are unloading a 200–lb
cast–iron counterweight from a truck. Knowing that at the instant
shown the counterweight is kept from moving and that the positions of
Points A, B, and C are, respectively, A(0, –20 in., 40 in.), B(–40 in.,
50 in., 0), and C(45 in., 40 in., 0), and assuming that no friction exists
between the counterweight and the chute, determine the tension in each
rope. (Hint: Since there is no friction, the force exerted by the chute on
the counterweight must be perpendicular to the chute.)
SOLUTION
From the geometry of the chute:
(2 )
5
(0.8944 0.4472 )
N
N
= +
= +
N jk
jk
222
(40 in.) (70 in.) (40 in.)
(40 in.) (70 in.) (40 in.)
90 in.
[( 40 in.) (70 in.) (40 in.) ]
90 in. 474
999
AB AB AB
AB
AB AB
AB
AB
AB
TT
AB
T
T
=+−
= ++
=
= =
= −+ −
= −+−
ijk
Tλ
ijk
T ijk
and
222
(45 in.) (60 in.) (40 in.)
(45 in.) (60 in.) (40 in.) 85 in.
AC
AC
=+−
= ++ =
ijk
[(45 in.) (60 in.) (40 in.) ]
85 in.
AC AC AC
AC
AC
TT
AC
T
= =
= +−
Tλ
ijk
SOLUTION Continued
With
200 lb,W=
and equating the factors of i, j, and k to zero, we obtain the linear algebraic equations:
:i
490
9 17
AB AC
TT−+ =
(1)
:j
7 12 2 200 lb 0
9 17 5
AB AC
TT+ +− =
(2)
:k
48 1
0
9 17 5
AB AC
TT N−− + =
(3)
65.6 lb
AB
T=
55.1 lb
AC
T=
consent of McGraw–Hill Education.
PROBLEM 2.100
Collars A and B are connected by a 25–in.-long wire and can slide
freely on frictionless rods. If a 60–lb force Q is applied to collar
B as shown, determine (a) the tension in the wire when
9 in.,x=
(b) the corresponding magnitude of the force P
required to maintain the equilibrium of the system.
SOLUTION
Free–Body Diagrams of Collars:
A: B:
(20 in.)
25 in.
AB AB x z
AB
−− +
= = i jk
λ
Collar A:
0: 0
y z AB AB
PN N TΣ= + + + =F i jk λ
AB
0
25 in.
AB
Tx
P−=
(1)
Collar B:
0: (60 lb) 0
x y AB AB
NNT
′′
Σ= + + − =F kijλ
AB
60 lb 0
25 in.
AB
Tz
−=
(2)
(a)
2 22 2
9 in. (9 in.) (20 in.) (25 in.)
12 in.
xz
z
= + +=
=
From Eq. (2):
60 lb (12 in.)
25 in.
AB
T−
125.0 lb
AB
T=
(b) From Eq. (1):
(125.0 lb)(9 in.)
25 in.
P=
45.0 lbP=
consent of McGraw–Hill Education.
PROBLEM 2.101
Collars A and B are connected by a 25–in.–long wire and can slide
freely on frictionless rods. Determine the distances x and z for
which the equilibrium of the system is maintained when
120 lbP=
and
60 lb.Q=
SOLUTION
See Problem 2.100 for the diagrams and analysis leading to Equations (1) and (2) below:
0
25 in.
AB
Tx
P= =
(1)
60 lb 0
25 in.
AB
Tz
−=
(2)
PROBLEM 2.102
Collars A and B are connected by a 525–mm–long wire and
can slide freely on frictionless rods. If a force
(341 N)=Pj
is applied to collar A, determine (a) the
tension in the wire when y
155 mm,=
(b) the magnitude
of the force Q required to maintain the equilibrium of the
system.
SOLUTION
For both Problems 2.102 and 2.103: Free–Body Diagrams of Collars:
2222
()AB x y z=++
Here
2 222
(0.525 m) (0.20 m) yz= ++
or
22 2
0.23563 myz+=
Thus, when y given, z is determined,
Now
1(0.20 )m
0.525 m
0.38095 1.90476 1.90476
AB
AB
AB
yz
yz
=
= −+
=−+
λ
ijk
ijk
Where y and z are in units of meters, m.
PROBLEM 2.103
Solve Problem 2.102 assuming that
275 mm.y=
PROBLEM 2.102 Collars A and B are connected by a
525-mm–long wire and can slide freely on frictionless
rods. If a force
(341 N)=Pj
is applied to collar A,
determine (a) the tension in the wire when y
155 mm,=
(b) the magnitude of the force Q required to maintain the
equilibrium of the system.
SOLUTION
From the analysis of Problem 2.102, particularly the results:
22 2
0.23563 m
341 N
1.90476
341 N
AB
yz
Ty
Qz
y
+=
=
=
With
275 mm 0.275 m,y= =
we obtain:
2 22
0.23563 m (0.275 m)
0.40 m
z
z
= −
=
and
(a)
341 N 651.00
(1.90476)(0.275 m)
AB
T= =
or
651 N
AB
T=
and
(b)
341 N(0.40 m)
(0.275 m)
Q=
or
496 NQ=
PROBLEM 2.104
Two structural members A and B are bolted to a bracket as
shown. Knowing that both members are in compression and
that the force is 15 kN in member A and 10 kN in member B,
determine by trigonometry the magnitude and direction of the
resultant of the forces applied to the bracket by members A and
B.
SOLUTION
Using the force triangle and the laws of cosines and sines,
we have
180 (40 20 )
120
γ
= °− °+ °
= °
Then
222
2
(15 kN) (10 kN)
2(15 kN)(10 kN)cos120
475 kN
21.794 kN
R
R
= +
−°
=
=
and
10 kN 21.794 kN
sin sin120
10 kN
sin sin120
21.794 kN
0.39737
23.414
α
α
α
=°
= °
=
=
Hence:
50 73.414
φα
= + °=
21.8 kN
=R
73.4°
consent of McGraw–Hill Education.
PROBLEM 2.105
Determine the x and y components of each of the forces shown.
SOLUTION
Compute the following distances:
22
22
22
(24 in.) (45 in.)
51.0 in.
(28 in.) (45 in.)
53.0 in.
(40 in.) (30 in.)
50.0 in.
OA
OB
OC
= +
=
= +
=
= +
=
102-lb Force:
24 in.
102 lb 51.0 in.
x
F= −
48.0 lb
x
F= −
45 in.
102 lb 51.0 in.
y
F= +
90.0 lb
y
F= +
106-lb Force:
28 in.
106 lb 53.0 in.
x
F= +
56.0 lb
x
F= +
45 in.
106 lb 53.0 in.
y
F= +
90.0 lb
y
F= +
200-lb Force:
40 in.
200 lb 50.0 in.
x
F= −
160.0 lb
x
F= −
30 in.
200 lb 50.0 in.
y
F= −
120.0 lb
y
F= −
consent of McGraw–Hill Education.
PROBLEM 2.106
The hydraulic cylinder BC exerts on member AB a force P
directed along line BC. Knowing that P must have a 600–N
component perpendicular to member AB, determine (a) the
magnitude of the force P, (b) its component along line AB.
SOLUTION
180 45 90 30
180 45 90 30
15
α
α
°= °+ + °+ °
= °− °− °− °
= °
(a)
cos
cos
600 N
cos15
621.17 N
x
x
P
P
P
P
α
α
=
=
=°
=
621 NP=
(b)
tan
tan
(600 N)tan15
160.770 N
y
x
yx
P
P
PP
α
α
=
=
= °
=
160.8 N
y
P=
consent of McGraw–Hill Education.
PROBLEM 2.99
Using two ropes and a roller chute, two workers are unloading a 200–lb
cast–iron counterweight from a truck. Knowing that at the instant
shown the counterweight is kept from moving and that the positions of
Points A, B, and C are, respectively, A(0, –20 in., 40 in.), B(–40 in.,
50 in., 0), and C(45 in., 40 in., 0), and assuming that no friction exists
between the counterweight and the chute, determine the tension in each
rope. (Hint: Since there is no friction, the force exerted by the chute on
the counterweight must be perpendicular to the chute.)
SOLUTION
From the geometry of the chute:
(2 )
5
(0.8944 0.4472 )
N
N
= +
= +
N jk
jk
222
(40 in.) (70 in.) (40 in.)
(40 in.) (70 in.) (40 in.)
90 in.
[( 40 in.) (70 in.) (40 in.) ]
90 in. 474
999
AB AB AB
AB
AB AB
AB
AB
AB
TT
AB
T
T
=+−
= ++
=
= =
= −+ −
= −+−
ijk
Tλ
ijk
T ijk
and
222
(45 in.) (60 in.) (40 in.)
(45 in.) (60 in.) (40 in.) 85 in.
AC
AC
=+−
= ++ =
ijk
[(45 in.) (60 in.) (40 in.) ]
85 in.
AC AC AC
AC
AC
TT
AC
T
= =
= +−
Tλ
ijk
SOLUTION Continued
With
200 lb,W=
and equating the factors of i, j, and k to zero, we obtain the linear algebraic equations:
:i
490
9 17
AB AC
TT−+ =
(1)
:j
7 12 2 200 lb 0
9 17 5
AB AC
TT+ +− =
(2)
:k
48 1
0
9 17 5
AB AC
TT N−− + =
(3)
65.6 lb
AB
T=
55.1 lb
AC
T=
consent of McGraw–Hill Education.
PROBLEM 2.100
Collars A and B are connected by a 25–in.-long wire and can slide
freely on frictionless rods. If a 60–lb force Q is applied to collar
B as shown, determine (a) the tension in the wire when
9 in.,x=
(b) the corresponding magnitude of the force P
required to maintain the equilibrium of the system.
SOLUTION
Free–Body Diagrams of Collars:
A: B:
(20 in.)
25 in.
AB AB x z
AB
−− +
= = i jk
λ
Collar A:
0: 0
y z AB AB
PN N TΣ= + + + =F i jk λ
AB
0
25 in.
AB
Tx
P−=
(1)
Collar B:
0: (60 lb) 0
x y AB AB
NNT
′′
Σ= + + − =F kijλ
AB
60 lb 0
25 in.
AB
Tz
−=
(2)
(a)
2 22 2
9 in. (9 in.) (20 in.) (25 in.)
12 in.
xz
z
= + +=
=
From Eq. (2):
60 lb (12 in.)
25 in.
AB
T−
125.0 lb
AB
T=
(b) From Eq. (1):
(125.0 lb)(9 in.)
25 in.
P=
45.0 lbP=
consent of McGraw–Hill Education.
PROBLEM 2.101
Collars A and B are connected by a 25–in.–long wire and can slide
freely on frictionless rods. Determine the distances x and z for
which the equilibrium of the system is maintained when
120 lbP=
and
60 lb.Q=
SOLUTION
See Problem 2.100 for the diagrams and analysis leading to Equations (1) and (2) below:
0
25 in.
AB
Tx
P= =
(1)
60 lb 0
25 in.
AB
Tz
−=
(2)
PROBLEM 2.102
Collars A and B are connected by a 525–mm–long wire and
can slide freely on frictionless rods. If a force
(341 N)=Pj
is applied to collar A, determine (a) the
tension in the wire when y
155 mm,=
(b) the magnitude
of the force Q required to maintain the equilibrium of the
system.
SOLUTION
For both Problems 2.102 and 2.103: Free–Body Diagrams of Collars:
2222
()AB x y z=++
Here
2 222
(0.525 m) (0.20 m) yz= ++
or
22 2
0.23563 myz+=
Thus, when y given, z is determined,
Now
1(0.20 )m
0.525 m
0.38095 1.90476 1.90476
AB
AB
AB
yz
yz
=
= −+
=−+
λ
ijk
ijk
Where y and z are in units of meters, m.
PROBLEM 2.103
Solve Problem 2.102 assuming that
275 mm.y=
PROBLEM 2.102 Collars A and B are connected by a
525-mm–long wire and can slide freely on frictionless
rods. If a force
(341 N)=Pj
is applied to collar A,
determine (a) the tension in the wire when y
155 mm,=
(b) the magnitude of the force Q required to maintain the
equilibrium of the system.
SOLUTION
From the analysis of Problem 2.102, particularly the results:
22 2
0.23563 m
341 N
1.90476
341 N
AB
yz
Ty
Qz
y
+=
=
=
With
275 mm 0.275 m,y= =
we obtain:
2 22
0.23563 m (0.275 m)
0.40 m
z
z
= −
=
and
(a)
341 N 651.00
(1.90476)(0.275 m)
AB
T= =
or
651 N
AB
T=
and
(b)
341 N(0.40 m)
(0.275 m)
Q=
or
496 NQ=
PROBLEM 2.104
Two structural members A and B are bolted to a bracket as
shown. Knowing that both members are in compression and
that the force is 15 kN in member A and 10 kN in member B,
determine by trigonometry the magnitude and direction of the
resultant of the forces applied to the bracket by members A and
B.
SOLUTION
Using the force triangle and the laws of cosines and sines,
we have
180 (40 20 )
120
γ
= °− °+ °
= °
Then
222
2
(15 kN) (10 kN)
2(15 kN)(10 kN)cos120
475 kN
21.794 kN
R
R
= +
−°
=
=
and
10 kN 21.794 kN
sin sin120
10 kN
sin sin120
21.794 kN
0.39737
23.414
α
α
α
=°
= °
=
=
Hence:
50 73.414
φα
= + °=
21.8 kN
=R
73.4°
consent of McGraw–Hill Education.
PROBLEM 2.105
Determine the x and y components of each of the forces shown.
SOLUTION
Compute the following distances:
22
22
22
(24 in.) (45 in.)
51.0 in.
(28 in.) (45 in.)
53.0 in.
(40 in.) (30 in.)
50.0 in.
OA
OB
OC
= +
=
= +
=
= +
=
102-lb Force:
24 in.
102 lb 51.0 in.
x
F= −
48.0 lb
x
F= −
45 in.
102 lb 51.0 in.
y
F= +
90.0 lb
y
F= +
106-lb Force:
28 in.
106 lb 53.0 in.
x
F= +
56.0 lb
x
F= +
45 in.
106 lb 53.0 in.
y
F= +
90.0 lb
y
F= +
200-lb Force:
40 in.
200 lb 50.0 in.
x
F= −
160.0 lb
x
F= −
30 in.
200 lb 50.0 in.
y
F= −
120.0 lb
y
F= −
consent of McGraw–Hill Education.
PROBLEM 2.106
The hydraulic cylinder BC exerts on member AB a force P
directed along line BC. Knowing that P must have a 600–N
component perpendicular to member AB, determine (a) the
magnitude of the force P, (b) its component along line AB.
SOLUTION
180 45 90 30
180 45 90 30
15
α
α
°= °+ + °+ °
= °− °− °− °
= °
(a)
cos
cos
600 N
cos15
621.17 N
x
x
P
P
P
P
α
α
=
=
=°
=
621 NP=
(b)
tan
tan
(600 N)tan15
160.770 N
y
x
yx
P
P
PP
α
α
=
=
= °
=
160.8 N
y
P=
consent of McGraw–Hill Education.