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PROBLEM 3.107
A 160-lb force P is applied at Point A of a structural member.
Replace P with (a) an equivalent force-couple system at C,
(b) an equivalent system consisting of a vertical force at B
and a second force at D.
SOLUTION
(a) Based on
: 160 lb
C
FPPΣ==
or
160 lb
C=P
60°
:
C C x cy y Cx
M M Pd PdΣ =−+
where
(160 lb)cos60
80 lb
(160 lb)sin 60
138.564 lb
4 ft
2.75 ft
x
y
Cx
Cy
P
P
d
d
= °
=
= °
=
=
=
(80 lb)(2.75 ft) (138.564 lb)(4 ft)
220 lb ft 554.26 lb ft
334.26 lb ft
C
M= +
= ⋅+ ⋅
= ⋅
or
334 lb ft
C
= ⋅M
(b) Based on
: cos60
x Dx
FP PΣ=°
(160 lb)cos60
80 lb
= °
=
: ( cos60 )( ) ( )
D DA B DB
M P d PdΣ °=
[(160 lb)cos60 ](1.5 ft) (6 ft)
20.0 lb
B
B
P
P
°=
=
or
20.0 lb
B=P
consent of McGraw-Hill Education.
SOLUTION Continued
: sin60
y B Dy
F P PPΣ °= +
22
22
(160 lb)sin 60 20.0 lb
118.564 lb
()()
(80) (118.564)
143.029 lb
Dy
Dy
D Dx Dy
P
P
PP P
°= +
=
= +
= +
=
1
1
tan
118.564
tan 80
55.991
Dy
Dx
P
P
θ
−
−
=
=
= °
or
143.0 lb
D
P=
56.0°
consent of McGraw-Hill Education.
PROBLEM 3.108
A regular tetrahedron has six edges of length a. A force P is
directed as shown along edge BC. Determine the moment of P
about edge OA.
SOLUTION
We have
/
()
OA OA C O
M=⋅×rP
λ
From triangle OBC:
() 2
1
() ()tan30 23 23
x
zx
a
OA
aa
OA OA
=
= °= =
Since
222 2
() ()() ( )
x yz
OA OA OA OA
=++
or
2
2
22
22
2
()
223
2
() 4 12 3
y
y
aa
a OA
aa
OA a a
=++
= −−=
Then
/
2
23
23
AO
aa
a=++
ri jk
and
121
23
23
OA =++ij k
λ
/
( sin30 ) ( cos30 ) () ( 3)
2
BC
CO
aa P
PP
a
a
°− °
= = = −
=
ik
P ik
ri
λ
121
23
23
()
10 0 2
10 3
2(1)( 3)
23 2
OA
P
Ma
aP aP
=
−
= − −=
2
OA aP
M=
consent of McGraw-Hill Education.
PROBLEM 3.107
A 160-lb force P is applied at Point A of a structural member.
Replace P with (a) an equivalent force-couple system at C,
(b) an equivalent system consisting of a vertical force at B
and a second force at D.
SOLUTION
(a) Based on
: 160 lb
C
FPPΣ==
or
160 lb
C=P
60°
:
C C x cy y Cx
M M Pd PdΣ =−+
where
(160 lb)cos60
80 lb
(160 lb)sin 60
138.564 lb
4 ft
2.75 ft
x
y
Cx
Cy
P
P
d
d
= °
=
= °
=
=
=
(80 lb)(2.75 ft) (138.564 lb)(4 ft)
220 lb ft 554.26 lb ft
334.26 lb ft
C
M= +
= ⋅+ ⋅
= ⋅
or
334 lb ft
C
= ⋅M
(b) Based on
: cos60
x Dx
FP PΣ=°
(160 lb)cos60
80 lb
= °
=
: ( cos60 )( ) ( )
D DA B DB
M P d PdΣ °=
[(160 lb)cos60 ](1.5 ft) (6 ft)
20.0 lb
B
B
P
P
°=
=
or
20.0 lb
B=P
consent of McGraw-Hill Education.
SOLUTION Continued
: sin60
y B Dy
F P PPΣ °= +
22
22
(160 lb)sin 60 20.0 lb
118.564 lb
()()
(80) (118.564)
143.029 lb
Dy
Dy
D Dx Dy
P
P
PP P
°= +
=
= +
= +
=
1
1
tan
118.564
tan 80
55.991
Dy
Dx
P
P
θ
−
−
=
=
= °
or
143.0 lb
D
P=
56.0°
consent of McGraw-Hill Education.
PROBLEM 3.108
A regular tetrahedron has six edges of length a. A force P is
directed as shown along edge BC. Determine the moment of P
about edge OA.
SOLUTION
We have
/
()
OA OA C O
M=⋅×rP
λ
From triangle OBC:
() 2
1
() ()tan30 23 23
x
zx
a
OA
aa
OA OA
=
= °= =
Since
222 2
() ()() ( )
x yz
OA OA OA OA
=++
or
2
2
22
22
2
()
223
2
() 4 12 3
y
y
aa
a OA
aa
OA a a
=++
= −−=
Then
/
2
23
23
AO
aa
a=++
ri jk
and
121
23
23
OA =++ij k
λ
/
( sin30 ) ( cos30 ) () ( 3)
2
BC
CO
aa P
PP
a
a
°− °
= = = −
=
ik
P ik
ri
λ
121
23
23
()
10 0 2
10 3
2(1)( 3)
23 2
OA
P
Ma
aP aP
=
−
= − −=
2
OA aP
M=
consent of McGraw-Hill Education.
