SOLUTION Continued
Setting the coefficients of the unit vectors equal to zero:
( )
: 450 2.2283 0
CE
F−+ =k
201.95 lb
CE
F=
or
202 lb
CE
F=
: 1.11417(201.95) ( ) 0
A
M− +=j
PROBLEM 4.72
Solve Prob. 4.69, assuming that the hinge at A has been removed and
that the hinge at B can exert couples about axes parallel to the x and y
axes.
PROBLEM 4.69 A 10–kg storm window measuring 900 × 1500 mm is
held by hinges at A and B. In the position shown, it is held away from
the side of the house by a 600-mm stick CD. Assuming that the hinge at
A does not exert any axial thrust, determine the magnitude of the force
exerted by the stick and the components of the reactions at A and B.
SOLUTION
Free–Body Diagram: Since CD is a two–force member,
CD
F
2
(10 kg)(9.81 m/s ) (98.10 N)=−=−Wj
(0.75 m)sin 23.074 (0.75 m)cos23.074 (0.45 m)
(0.29394 m) (0.690 m) (0.45 m)
(cos11.5370 sin11.5370 )
(0.97980 0.20 )
G
G
CD CD
CD CD
F
F
=−+
= −+
= +
= +
r i jk
r i jk
F ij
F ij
0:() () 0
B B x B y G C CD
M MMΣ = + +×+× =i jr Wr F
( ) ( ) (0.29394 0.690 0.45 ) ( 98.10)
(1.5 0.9 ) (0.97980 0.20 ) 0
Bx By
CD
MM
F
+ + − + ×−
++ × + =
i j i jk j
jk i j
( ) ( ) 28.836 44.145 1.46970 0.88182 0.180 0
B x B y CD CD CD
MM F F F+− ++ + − =ijki k j i
consent of McGraw–Hill Education.
SOLUTION Continued
Equating the coefficients of the unit vectors to zero,
28.836 1.46970 0; 19.6203 N
CD CD
FF−+ = =k:
( ) 0.88182(19.6203) 0; ( ) 17.3016 N m
By By
MM+ = =−⋅j:
( ) 44.145 0.180(19.6203) 0; ( ) 40.613 N m
Bx Bx
MM+− = =− ⋅i:
19.62 N
CD
F=
(40.6 N m) (17.30 N m)
B
=− ⋅− ⋅
M ij
0: cos11.5370 0
19.6203cos11.5370 0
19.22 N
x x CD
x
x
F BF
B
B
Σ = + °=
+=
= −
0: sin11.5370 0
19.6230sin11.5370 98.1 0
94.2 N
y y CD
y
y
F BF W
B
B
Σ= + −=
+ −=
=
0: 0
zz
FB
Σ= =
(19.22 N) (94.2 N)=−+B ij
PROBLEM 4.73
The assembly shown is welded to collar A that fits on the
vertical pin shown. The pin can exert couples about the x
and z axes but does not prevent motion about or along the y–
axis. For the loading shown, determine the tension in each
cable and the reaction at A.
SOLUTION
Free–Body Diagram:
First note:
22
22
(0.08 m) (0.06 m)
(0.08) (0.06) m
( 0.8 0.6 )
(0.12 m) (0.09 m)
(0.12) (0.09) m
(0.8 0.6 )
CF CF CF CF
CF
DE DE DE DE
DE
TT
T
TT
T
−+
= = +
= −+
−
= = +
= −
ij
Tλ
ij
jk
Tλ
jk
From F.B.D. of assembly:
0: 0.6 0.8 480 N 0
y CF DE
F TTΣ= + − =
or
0.6 0.8 480 N
CF DE
TT
+=
(1)
0: (0.8 )(0.135 m) (0.6 )(0.08 m) 0
y CF DE
MT TΣ= − + =
or
2.25
DE CF
TT=
(2)
Substituting Equation (2) into Equation (1),
0.6 0.8[(2.25) ] 480 N
CF CF
TT+=
200.00 N
CF
T=
or
200 N
CF
T=
and from Equation (2):
2.25(200.00 N) 450.00
DE
T= =
or
450 N
DE
T=
SOLUTION Continued
From F.B.D. of assembly:
0: (0.6)(450.00 N) 0 270.00 N
zz z
FA AΣ= − = =
0: (0.8)(200.00 N) 0 160.000 N
xx x
FA AΣ= − = =
or
(160.0 N) (270 N)= +A ik
0: (480 N)(0.135 m) [(200.00 N)(0.6)](0.135 m)
[(450 N)(0.8)](0.09 m) 0
x
xA
MMΣ= + −
−=
16.2000 N m
x
A
M=−⋅
0: (480 N)(0.08 m) [(200.00 N)(0.6)](0.08 m)
[(450 N)(0.8)](0.08 m) 0
z
zA
MMΣ= − +
+=
0
z
A
M=
or
(16.20 N m)
A
=−⋅
Mi
PROBLEM 4.74
Three rods are welded together to form a “corner” that is supported
by three eyebolts. Neglecting friction, determine the reactions at
A, B, and C when
240P=
lb,
12 in., 8 in.,ab= =
and
10 in.c=
SOLUTION
From F.B.D. of weldment:
///
0: 0
O AO BO CO
Σ = ×+ ×+ ×=M r Ar Br C
12 0 0 0 8 0 0 0 10 0
000
yz x z xy
AA B B CC
++ =
i j k i jk i j k
( 12 12 ) (8 8 ) ( 10 10 ) 0
z y zx y x
A A BB C C− + + − +− + =j k ik i j
From i–coefficient:
8 10 0
zy
BC−=
or
1.25
zy
BC=
(1)
j–coefficient:
12 10 0
zx
AC−+ =
or
1.2
xz
CA=
(2)
k–coefficient:
12 8 0
yx
AB−=
or
1.5
xy
BA=
(3)
0: 0Σ= + + − =F ABCP
or
( ) ( 240 lb) ( ) 0
xx yy zz
BC AC AB+ + +− + + =i jk
From i–coefficient:
0
xx
BC+=
or
xx
CB= −
(4)
j–coefficient:
240 lb 0
yy
AC+− =
or
240 lb
yy
AC+=
(5)
consent of McGraw–Hill Education.
SOLUTION Continued
PROBLEM 4.75
Determine whether the block shown is in equilibrium and find the magnitude and
direction of the friction force when
θ
= 25° and P = 750 N.
SOLUTION
Assume equilibrium:
0: (1200 N)sin 25° (750 N)cos 25° 0
x
FFΣ= + − =
172.6 NF= +
172.6 N=F
0: (1200 N)cos 25° (750 N)sin 25° 0
y
FNΣ= − − =
1404.5 NN=
Maximum friction force:
0.35(1404.5 N) 491.6 N
ms
FN
m
= = =
Since
,
m
FF<
block is in equilibrium
Friction force:
172.6 N=F
25.0°
consent of McGraw–Hill Education.
PROBLEM 4.76
Determine whether the block shown is in equilibrium and find the magnitude and
direction of the friction force when
θ
= 30° and P = 150 N.
SOLUTION
Assume equilibrium:
0: (1200 N)sin 30° (150 N)cos 30° 0
x
FFΣ= + − =
470.1 NF= −
470.1 N=F
0: (1200 N)cos 30 (150 N)sin 30 0
y
FNΣ = − °− °=
1114.2 NN=
(a) Maximum friction force:
0.35(1114.2 N)
390.0 N
ms
FN
m
=
=
=
Since F is and
,
m
FF>
block moves down
(b) Actual friction force:
0.25(1114.2 N) 279 N
kk
FF N
m
= = = =
279 N=F
30.0°
consent of McGraw–Hill Education.
PROBLEM 4.77
Determine whether the block shown is in equilibrium and find the
magnitude and direction of the friction force when P = 120 lb.
SOLUTION
Assume equilibrium:
0: (50 lb)sin 30 (120 lb)cos 40 0
x
FFΣ = + °− °=
66.925 lbF= +
0: (50 lb)cos 30 (120 lb)sin 40 0
y
FNΣ = − °− °=
120.436 lbN= +
Maximum friction force:
0.40(120.436 lb)
48.174 lb
ms
FN
m
=
=
=
We note that
.
m
FF>
Thus, block moves up
Actual friction force:
0.30(120.436 lb) 36.131 lb,
kk
FF N
m
= = = =
36.1 lb=F
30.0°
consent of McGraw–Hill Education.
PROBLEM 4.72
Solve Prob. 4.69, assuming that the hinge at A has been removed and
that the hinge at B can exert couples about axes parallel to the x and y
axes.
PROBLEM 4.69 A 10–kg storm window measuring 900 × 1500 mm is
held by hinges at A and B. In the position shown, it is held away from
the side of the house by a 600-mm stick CD. Assuming that the hinge at
A does not exert any axial thrust, determine the magnitude of the force
exerted by the stick and the components of the reactions at A and B.
SOLUTION
Free–Body Diagram: Since CD is a two–force member,
CD
F
2
(10 kg)(9.81 m/s ) (98.10 N)=−=−Wj
(0.75 m)sin 23.074 (0.75 m)cos23.074 (0.45 m)
(0.29394 m) (0.690 m) (0.45 m)
(cos11.5370 sin11.5370 )
(0.97980 0.20 )
G
G
CD CD
CD CD
F
F
=−+
= −+
= +
= +
r i jk
r i jk
F ij
F ij
0:() () 0
B B x B y G C CD
M MMΣ = + +×+× =i jr Wr F
( ) ( ) (0.29394 0.690 0.45 ) ( 98.10)
(1.5 0.9 ) (0.97980 0.20 ) 0
Bx By
CD
MM
F
+ + − + ×−
++ × + =
i j i jk j
jk i j
( ) ( ) 28.836 44.145 1.46970 0.88182 0.180 0
B x B y CD CD CD
MM F F F+− ++ + − =ijki k j i
consent of McGraw–Hill Education.
SOLUTION Continued
Equating the coefficients of the unit vectors to zero,
28.836 1.46970 0; 19.6203 N
CD CD
FF−+ = =k:
( ) 0.88182(19.6203) 0; ( ) 17.3016 N m
By By
MM+ = =−⋅j:
( ) 44.145 0.180(19.6203) 0; ( ) 40.613 N m
Bx Bx
MM+− = =− ⋅i:
19.62 N
CD
F=
(40.6 N m) (17.30 N m)
B
=− ⋅− ⋅
M ij
0: cos11.5370 0
19.6203cos11.5370 0
19.22 N
x x CD
x
x
F BF
B
B
Σ = + °=
+=
= −
0: sin11.5370 0
19.6230sin11.5370 98.1 0
94.2 N
y y CD
y
y
F BF W
B
B
Σ= + −=
+ −=
=
0: 0
zz
FB
Σ= =
(19.22 N) (94.2 N)=−+B ij
PROBLEM 4.73
The assembly shown is welded to collar A that fits on the
vertical pin shown. The pin can exert couples about the x
and z axes but does not prevent motion about or along the y–
axis. For the loading shown, determine the tension in each
cable and the reaction at A.
SOLUTION
Free–Body Diagram:
First note:
22
22
(0.08 m) (0.06 m)
(0.08) (0.06) m
( 0.8 0.6 )
(0.12 m) (0.09 m)
(0.12) (0.09) m
(0.8 0.6 )
CF CF CF CF
CF
DE DE DE DE
DE
TT
T
TT
T
−+
= = +
= −+
−
= = +
= −
ij
Tλ
ij
jk
Tλ
jk
From F.B.D. of assembly:
0: 0.6 0.8 480 N 0
y CF DE
F TTΣ= + − =
or
0.6 0.8 480 N
CF DE
TT
+=
(1)
0: (0.8 )(0.135 m) (0.6 )(0.08 m) 0
y CF DE
MT TΣ= − + =
or
2.25
DE CF
TT=
(2)
Substituting Equation (2) into Equation (1),
0.6 0.8[(2.25) ] 480 N
CF CF
TT+=
200.00 N
CF
T=
or
200 N
CF
T=
and from Equation (2):
2.25(200.00 N) 450.00
DE
T= =
or
450 N
DE
T=
SOLUTION Continued
From F.B.D. of assembly:
0: (0.6)(450.00 N) 0 270.00 N
zz z
FA AΣ= − = =
0: (0.8)(200.00 N) 0 160.000 N
xx x
FA AΣ= − = =
or
(160.0 N) (270 N)= +A ik
0: (480 N)(0.135 m) [(200.00 N)(0.6)](0.135 m)
[(450 N)(0.8)](0.09 m) 0
x
xA
MMΣ= + −
−=
16.2000 N m
x
A
M=−⋅
0: (480 N)(0.08 m) [(200.00 N)(0.6)](0.08 m)
[(450 N)(0.8)](0.08 m) 0
z
zA
MMΣ= − +
+=
0
z
A
M=
or
(16.20 N m)
A
=−⋅
Mi
PROBLEM 4.74
Three rods are welded together to form a “corner” that is supported
by three eyebolts. Neglecting friction, determine the reactions at
A, B, and C when
240P=
lb,
12 in., 8 in.,ab= =
and
10 in.c=
SOLUTION
From F.B.D. of weldment:
///
0: 0
O AO BO CO
Σ = ×+ ×+ ×=M r Ar Br C
12 0 0 0 8 0 0 0 10 0
000
yz x z xy
AA B B CC
++ =
i j k i jk i j k
( 12 12 ) (8 8 ) ( 10 10 ) 0
z y zx y x
A A BB C C− + + − +− + =j k ik i j
From i–coefficient:
8 10 0
zy
BC−=
or
1.25
zy
BC=
(1)
j–coefficient:
12 10 0
zx
AC−+ =
or
1.2
xz
CA=
(2)
k–coefficient:
12 8 0
yx
AB−=
or
1.5
xy
BA=
(3)
0: 0Σ= + + − =F ABCP
or
( ) ( 240 lb) ( ) 0
xx yy zz
BC AC AB+ + +− + + =i jk
From i–coefficient:
0
xx
BC+=
or
xx
CB= −
(4)
j–coefficient:
240 lb 0
yy
AC+− =
or
240 lb
yy
AC+=
(5)
consent of McGraw–Hill Education.
SOLUTION Continued
PROBLEM 4.75
Determine whether the block shown is in equilibrium and find the magnitude and
direction of the friction force when
θ
= 25° and P = 750 N.
SOLUTION
Assume equilibrium:
0: (1200 N)sin 25° (750 N)cos 25° 0
x
FFΣ= + − =
172.6 NF= +
172.6 N=F
0: (1200 N)cos 25° (750 N)sin 25° 0
y
FNΣ= − − =
1404.5 NN=
Maximum friction force:
0.35(1404.5 N) 491.6 N
ms
FN
m
= = =
Since
,
m
FF<
block is in equilibrium
Friction force:
172.6 N=F
25.0°
consent of McGraw–Hill Education.
PROBLEM 4.76
Determine whether the block shown is in equilibrium and find the magnitude and
direction of the friction force when
θ
= 30° and P = 150 N.
SOLUTION
Assume equilibrium:
0: (1200 N)sin 30° (150 N)cos 30° 0
x
FFΣ= + − =
470.1 NF= −
470.1 N=F
0: (1200 N)cos 30 (150 N)sin 30 0
y
FNΣ = − °− °=
1114.2 NN=
(a) Maximum friction force:
0.35(1114.2 N)
390.0 N
ms
FN
m
=
=
=
Since F is and
,
m
FF>
block moves down
(b) Actual friction force:
0.25(1114.2 N) 279 N
kk
FF N
m
= = = =
279 N=F
30.0°
consent of McGraw–Hill Education.
PROBLEM 4.77
Determine whether the block shown is in equilibrium and find the
magnitude and direction of the friction force when P = 120 lb.
SOLUTION
Assume equilibrium:
0: (50 lb)sin 30 (120 lb)cos 40 0
x
FFΣ = + °− °=
66.925 lbF= +
0: (50 lb)cos 30 (120 lb)sin 40 0
y
FNΣ = − °− °=
120.436 lbN= +
Maximum friction force:
0.40(120.436 lb)
48.174 lb
ms
FN
m
=
=
=
We note that
.
m
FF>
Thus, block moves up
Actual friction force:
0.30(120.436 lb) 36.131 lb,
kk
FF N
m
= = = =
36.1 lb=F
30.0°
consent of McGraw–Hill Education.