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PROBLEM 3.11
Rod AB is held in place by the cord AC. Knowing that the tension in
the cord is 1350 N and that c = 360 mm, determine the moment about
B of the force exerted by the cord at point A by resolving that force
into horizontal and vertical components applied (a) at point A, (b) at
point C.
PROBLEM 3.12
Rod AB is held in place by the cord AC. Knowing that c = 840 mm
and that the moment about B of the force exerted by the cord at point
A is 756 N∙m, determine the tension in the cord.
SOLUTION
Free-Body Diagram of Rod AB:
22
(756 N m ) (450) (1080) 1170 mm
BAC=−⋅ = + =Mk
450 1080
cos sin
1170 1170
αα
= =
cos sin
450 1080
1170 1170
FF
FF
αα
=
= +
F i+ j
ij
F
/(0.45 m) (0.24 m)
AB =−−r ij
/
( 0.45 0.24 ) (450 1080 )1170
B AB
F
= ×=− − × +Mr F i j i j
F F F
( 486 108 )1170
F
=−+kk
378
1170 F
= −
k
Substituting for
:
B
M
378
756 1170 F−=−
2340 NF=
consent of McGraw-Hill Education.
PROBLEM 3.13
Determine the moment about the origin O of the force F = 4i − 3j + 5k that acts at a Point A. Assume that the
position vector of A is (a) r = 2i + 3j − 4k, (b) r = −8i + 6j − 10k, (c) r = 8i − 6j + 5k.
SOLUTION
O
= ×M rF
(a)
23 4
4 35
O
= −
−
ijk
M
(15 12) ( 16 10) ( 6 12)= − +− − +−−i jk
3 26 18
O
=−−
M i jk
(b)
8 6 10
4 35
O
=−−
−
ijk
M
(30 30) ( 40 40) (24 24)= − +− + + −i jk
0
O
=M
(c)
8 65
4 35
O= −
−
i jk
M
( 30 15) (20 40) ( 24 24)=−+ + − +−+ij k
15 20
O=−−M ij
determinant are proportional.
consent of McGraw-Hill Education.
PROBLEM 3.14
Determine the moment about the origin O of the force F = 2i + 3j − 4k that acts at a Point A. Assume that the
position vector of A is (a) r = 3i − 6j + 5k, (b) r = i − 4j − 2k, (c) r = 4i + 6j − 8k.
SOLUTION
O
= ×M rF
(a)
3 65
23 4
O
= −
−
ijk
M
(24 15) (10 12) (9 12)= − + + ++i jk
9 22 21
O=++M i jk
(b)
142
23 4
O
= −−
−
ijk
M
(16 6) ( 4 4) (3 8)= + +−+ + +i jk
22 11
O
= +
M ik
(c)
46 8
23 4
O
= −
−
ijk
M
( 24 24) ( 16 16) (12 12)=−+ +−+ + −i jk
0
O=M
determinant are proportional.
consent of McGraw-Hill Education.
PROBLEM 3.15
A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the
resulting force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B.
SOLUTION
We have
(6 lb)cos 8 5.9416 lb
xz
T= °=
Then
sin 30 2.9708 lb
sin 8 0.83504 lb
cos 30 5.1456 lb
x xz
y BC
z xz
TT
TT
TT
= °=
= °=−
= °=−
Now
/A B A BC
= ×Mr T
where
/
(6sin 45 ) (6cos 45 )
6 ft ()
2
BA
= °− °
= −
r jk
jk
Then
or (28.5 N) (106.3 N)=−+F jk
or
(25.4 lb ft) (12.60 lb ft) (12.60 lb ft)
A
=−⋅−⋅−⋅M ijk
consent of McGraw-Hill Education.
PROBLEM 3.16
The wire AE is stretched between the corners A and E of a
bent plate. Knowing that the tension in the wire is 435 N,
determine the moment about O of the force exerted by the
wire (a) on corner A, (b) on corner E.
SOLUTION
2 22
(0.21 m) (0.16 m) (0.12 m)
(0.21 m) ( 0.16 m) (0.12 m) 0.29 m
AE
AE
=−+
= +− + =
ijk
(a)
0.21 0.16 0.12
(435 N) 0.29
(315 N) (240 N) (180 N)
A A AE
AE
FF
AE
= =
−+
=
=−+
F
ijk
i jk
λ
/
(0.09 m) (0.16 m)
AO
=−+r ij
0.09 0.16 0
315 240 180
O
= −
−
i jk
M
28.8 16.20 (21.6 50.4)=+ +−ij k
(28.8 N m) (16.20 N m) (28.8 N m)
O
= ⋅+ ⋅− ⋅M i jk
(b)
(315 N) (240 N) (180 N)
EA
=−=− + −FF i j k
/
(0.12 m) (0.12 m)
EO
= +
r ik
0.12 0 0.12
315 240 180
O=
−−
i jk
M
28.8 ( 37.8 21.6) 28.8=− +− + +i jk
(28.8 N m) (16.20 N m) (28.8 N m)
O
=− ⋅− ⋅+ ⋅M i jk
consent of McGraw-Hill Education.
PROBLEM 3.17
The 12-ft boom AB has a fixed end A. A steel cable is stretched from
the free end B of the boom to a point C located on the vertical wall. If
the tension in the cable is 380 lb, determine the moment about A of the
force exerted by the cable at B.
SOLUTION
First note
2 22
( 12) (4.8) ( 8)
15.2 ft
BC
d= − + +−
=
Then
380 lb ( 12 4.8 8 )
15.2
BC = −+ −T i jk
We have
/
A B A BC
= ×
Mr T
where
/
(12 ft)
BA
=ri
Then
380 lb
(12 ft) ( 12 4.8 8 )
15.2
A
= × −+ −M i i jk
or
(2400 lb ft) (1440 lb ft)
A
= ⋅+ ⋅
M jk
consent of McGraw-Hill Education.
PROBLEM 3.18
A wooden board AB, which is used as a temporary prop to support
a small roof, exerts at Point A of the roof a 57-lb force directed
along BA. Determine the moment about C of that force.
SOLUTION
We have
/C A C BA
= ×Mr F
where
/(48 in.) (6 in.) (36 in.)
AC = −+r ij k
and
222
(5 in.) (90 in.) (30 in.) (57 lb)
(5) (90) (30) in.
(3 lb) (54 lb) (18 lb)
BA BA BA
F=
−+ −
=
++
=−+ −
Fλ
ijk
i jk
48 6 36 lb in.
3 54 18
(1836 lb in.) (756 lb in.) (2574 lb in.)
C
= ⋅
=− ⋅+ ⋅+ ⋅
i jk
M
ij
or
(153.0 lb ft) (63.0 lb ft) (215 lb ft)
C
=− ⋅+ ⋅+ ⋅M i jk
consent of McGraw-Hill Education.
PROBLEM 3.19
A 200-N force is applied as shown to the bracket ABC.
Determine the moment of the force about A.
SOLUTION
We have
/A CA C
= ×Mr F
where
/
(0.06 m) (0.075 m)
(200 N)cos 30 (200 N)sin 30
CA
C
= +
=− °+ °
r ij
F jk
Then
200 0.06 0.075 0
0 cos30 sin 30
200[(0.075sin 30 ) (0.06sin 30 ) (0.06cos 30 ) ]
A
=
−° °
= °− °− °
ij k
M
ijk
or
(7.50 N m) (6.00 N m) (10.39 N m)
A
= ⋅− ⋅− ⋅M ij k
consent of McGraw-Hill Education.
PROBLEM 3.20
A small boat hangs from two davits, one of which is shown in the
figure. The tension in line ABAD is 82 lb. Determine the moment
about C of the resultant force RA exerted on the davit at A.
SOLUTION
We have
2
A AB AD
= +R FF
where
(82 lb)
AB
= −Fj
and
6 7.75 3
(82 lb) 10.25
(48 lb) (62 lb) (24 lb)
AD AD
AD
AD
AD
−−
= =
=−−
i jk
FF
F ijk
FFF
Thus
2 (48 lb) (226 lb) (24 lb)
A AB AD
= += − −R FF i j k
Also
/
(7.75 ft) (3 ft)
AC
= +r jk
Using Eq. (3.21):
0 7.75 3
48 226 24
(492 lb ft) (144.0 lb ft) (372 lb ft)
C
=
−−
= ⋅+ ⋅− ⋅
ijk
M
i jk
(492 lb ft) (144.0 lb ft) (372 lb ft)
C
= ⋅+ ⋅− ⋅
M i jk
consent of McGraw-Hill Education.
PROBLEM 3.12
Rod AB is held in place by the cord AC. Knowing that c = 840 mm
and that the moment about B of the force exerted by the cord at point
A is 756 N∙m, determine the tension in the cord.
SOLUTION
Free-Body Diagram of Rod AB:
22
(756 N m ) (450) (1080) 1170 mm
BAC=−⋅ = + =Mk
450 1080
cos sin
1170 1170
αα
= =
cos sin
450 1080
1170 1170
FF
FF
αα
=
= +
F i+ j
ij
F
/(0.45 m) (0.24 m)
AB =−−r ij
/
( 0.45 0.24 ) (450 1080 )1170
B AB
F
= ×=− − × +Mr F i j i j
F F F
( 486 108 )1170
F
=−+kk
378
1170 F
= −
k
Substituting for
:
B
M
378
756 1170 F−=−
2340 NF=
consent of McGraw-Hill Education.
PROBLEM 3.13
Determine the moment about the origin O of the force F = 4i − 3j + 5k that acts at a Point A. Assume that the
position vector of A is (a) r = 2i + 3j − 4k, (b) r = −8i + 6j − 10k, (c) r = 8i − 6j + 5k.
SOLUTION
O
= ×M rF
(a)
23 4
4 35
O
= −
−
ijk
M
(15 12) ( 16 10) ( 6 12)= − +− − +−−i jk
3 26 18
O
=−−
M i jk
(b)
8 6 10
4 35
O
=−−
−
ijk
M
(30 30) ( 40 40) (24 24)= − +− + + −i jk
0
O
=M
(c)
8 65
4 35
O= −
−
i jk
M
( 30 15) (20 40) ( 24 24)=−+ + − +−+ij k
15 20
O=−−M ij
determinant are proportional.
consent of McGraw-Hill Education.
PROBLEM 3.14
Determine the moment about the origin O of the force F = 2i + 3j − 4k that acts at a Point A. Assume that the
position vector of A is (a) r = 3i − 6j + 5k, (b) r = i − 4j − 2k, (c) r = 4i + 6j − 8k.
SOLUTION
O
= ×M rF
(a)
3 65
23 4
O
= −
−
ijk
M
(24 15) (10 12) (9 12)= − + + ++i jk
9 22 21
O=++M i jk
(b)
142
23 4
O
= −−
−
ijk
M
(16 6) ( 4 4) (3 8)= + +−+ + +i jk
22 11
O
= +
M ik
(c)
46 8
23 4
O
= −
−
ijk
M
( 24 24) ( 16 16) (12 12)=−+ +−+ + −i jk
0
O=M
determinant are proportional.
consent of McGraw-Hill Education.
PROBLEM 3.15
A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the
resulting force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B.
SOLUTION
We have
(6 lb)cos 8 5.9416 lb
xz
T= °=
Then
sin 30 2.9708 lb
sin 8 0.83504 lb
cos 30 5.1456 lb
x xz
y BC
z xz
TT
TT
TT
= °=
= °=−
= °=−
Now
/A B A BC
= ×Mr T
where
/
(6sin 45 ) (6cos 45 )
6 ft ()
2
BA
= °− °
= −
r jk
jk
Then
or (28.5 N) (106.3 N)=−+F jk
or
(25.4 lb ft) (12.60 lb ft) (12.60 lb ft)
A
=−⋅−⋅−⋅M ijk
consent of McGraw-Hill Education.
PROBLEM 3.16
The wire AE is stretched between the corners A and E of a
bent plate. Knowing that the tension in the wire is 435 N,
determine the moment about O of the force exerted by the
wire (a) on corner A, (b) on corner E.
SOLUTION
2 22
(0.21 m) (0.16 m) (0.12 m)
(0.21 m) ( 0.16 m) (0.12 m) 0.29 m
AE
AE
=−+
= +− + =
ijk
(a)
0.21 0.16 0.12
(435 N) 0.29
(315 N) (240 N) (180 N)
A A AE
AE
FF
AE
= =
−+
=
=−+
F
ijk
i jk
λ
/
(0.09 m) (0.16 m)
AO
=−+r ij
0.09 0.16 0
315 240 180
O
= −
−
i jk
M
28.8 16.20 (21.6 50.4)=+ +−ij k
(28.8 N m) (16.20 N m) (28.8 N m)
O
= ⋅+ ⋅− ⋅M i jk
(b)
(315 N) (240 N) (180 N)
EA
=−=− + −FF i j k
/
(0.12 m) (0.12 m)
EO
= +
r ik
0.12 0 0.12
315 240 180
O=
−−
i jk
M
28.8 ( 37.8 21.6) 28.8=− +− + +i jk
(28.8 N m) (16.20 N m) (28.8 N m)
O
=− ⋅− ⋅+ ⋅M i jk
consent of McGraw-Hill Education.
PROBLEM 3.17
The 12-ft boom AB has a fixed end A. A steel cable is stretched from
the free end B of the boom to a point C located on the vertical wall. If
the tension in the cable is 380 lb, determine the moment about A of the
force exerted by the cable at B.
SOLUTION
First note
2 22
( 12) (4.8) ( 8)
15.2 ft
BC
d= − + +−
=
Then
380 lb ( 12 4.8 8 )
15.2
BC = −+ −T i jk
We have
/
A B A BC
= ×
Mr T
where
/
(12 ft)
BA
=ri
Then
380 lb
(12 ft) ( 12 4.8 8 )
15.2
A
= × −+ −M i i jk
or
(2400 lb ft) (1440 lb ft)
A
= ⋅+ ⋅
M jk
consent of McGraw-Hill Education.
PROBLEM 3.18
A wooden board AB, which is used as a temporary prop to support
a small roof, exerts at Point A of the roof a 57-lb force directed
along BA. Determine the moment about C of that force.
SOLUTION
We have
/C A C BA
= ×Mr F
where
/(48 in.) (6 in.) (36 in.)
AC = −+r ij k
and
222
(5 in.) (90 in.) (30 in.) (57 lb)
(5) (90) (30) in.
(3 lb) (54 lb) (18 lb)
BA BA BA
F=
−+ −
=
++
=−+ −
Fλ
ijk
i jk
48 6 36 lb in.
3 54 18
(1836 lb in.) (756 lb in.) (2574 lb in.)
C
= ⋅
=− ⋅+ ⋅+ ⋅
i jk
M
ij
or
(153.0 lb ft) (63.0 lb ft) (215 lb ft)
C
=− ⋅+ ⋅+ ⋅M i jk
consent of McGraw-Hill Education.
PROBLEM 3.19
A 200-N force is applied as shown to the bracket ABC.
Determine the moment of the force about A.
SOLUTION
We have
/A CA C
= ×Mr F
where
/
(0.06 m) (0.075 m)
(200 N)cos 30 (200 N)sin 30
CA
C
= +
=− °+ °
r ij
F jk
Then
200 0.06 0.075 0
0 cos30 sin 30
200[(0.075sin 30 ) (0.06sin 30 ) (0.06cos 30 ) ]
A
=
−° °
= °− °− °
ij k
M
ijk
or
(7.50 N m) (6.00 N m) (10.39 N m)
A
= ⋅− ⋅− ⋅M ij k
consent of McGraw-Hill Education.
PROBLEM 3.20
A small boat hangs from two davits, one of which is shown in the
figure. The tension in line ABAD is 82 lb. Determine the moment
about C of the resultant force RA exerted on the davit at A.
SOLUTION
We have
2
A AB AD
= +R FF
where
(82 lb)
AB
= −Fj
and
6 7.75 3
(82 lb) 10.25
(48 lb) (62 lb) (24 lb)
AD AD
AD
AD
AD
−−
= =
=−−
i jk
FF
F ijk
FFF
Thus
2 (48 lb) (226 lb) (24 lb)
A AB AD
= += − −R FF i j k
Also
/
(7.75 ft) (3 ft)
AC
= +r jk
Using Eq. (3.21):
0 7.75 3
48 226 24
(492 lb ft) (144.0 lb ft) (372 lb ft)
C
=
−−
= ⋅+ ⋅− ⋅
ijk
M
i jk
(492 lb ft) (144.0 lb ft) (372 lb ft)
C
= ⋅+ ⋅− ⋅
M i jk
consent of McGraw-Hill Education.
