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228
4–1.
If
A
,
B
,an
d
D
are g
i
ven vectors, prove t
h
e
distributive law for the vector cross product, i.e.,
.A:(B+D)=(A:B)+(A:D)
SOLUTION
Note also,
(QED)=(A *B) +(A *D)
=3ijk
AxAyAz
BxByBz3+3ijk
AxAyAz
DxDyDz3
+[(AyDz-AzDy)i-(AxDz-AzDx)j+(AxDy-AyDx)k
=[(AyBz-AzBy)i-(AxBz-AzBx)]j+(AxBy-AyBx)k
+[Ax(By+Dy)-Ay(Bx+Dx)]k
-[Ax(Bz+Dz)-Az(Bx+Dx)]j
=[Ay(Bz+Dz)-Az(By+Dy)]i
A*(B+D)=3ijk
AxAyAz
Bx+DxBy+DyBz+Dz3
D=Dxi+Dyj+Dzk
B=Bxi+Byj+Bzk
A=Axi+Ayj+Azk
229
4–2.
Prove t
h
e tr
i
p
l
e sca
l
ar pro
d
uct
id
ent
i
ty
A # (B : C) = (A : B) # C.
SOLUTION
Thus,
But,
Thus,
Since |(A * B) # C| represents this same volume then
(QED)
Also,
Thus,
(QED)A # (B : C) = (A : B) # C
LHS =RHS
=AxByCz-AxBzCy-AyBxCz+AyBzCx+AzBxCy-AzByCx
=Cx(AyBz-AzBy)-Cy(AxBz-AzBx)+Cz(AxBy-AyBx)
=3ijk
AxAyAz
BxByBz3#(Cxi+Cyj+Czk)
RHS =(A : B) # C
=AxByCz-AxBzCy-AyBxCz+AyBzCx+AzBxCy-AzByCx
=Ax(ByCz-BzCy)-Ay(BxCz-BzCx)+Az(BxCy-ByCx)
=(Axi+Ayj+Azk)#3ijk
BxByBz
CxCyCz3
LHS =A # (B : C)
A # (B : C) = (A : B) # C
Volume =|A#(B * C)|
|h|=|A#u(B*C)|=
`
A#aB*C
|B*C|b
`
Volume of parallelepiped is |B*C||h|
230
4–3.
Given the three nonzero vectors
A
,
B
,and
C
,show that if
, the three vectors must lie in the same
plane.
A#(B:C)=0
SOLUTION
If , then the volume equals zero, so that A, B, and C are coplanar.A#(B *C) =0
=volume of parallelepiped.
=BC |h| sin f
=|h||B*C|
=(|A| cos u)|B*C|
231
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Ans:
(
MF
1
)
A=3.00 kip
#
ft (Clockwise)
(
MF
2
)
A=5.60 kip
#
ft (Clockwise)
(
MF
3
)
A=2.59 kip
#
ft (Clockwise)
*4–4.
Determine the moment about point Aof each of the three
forces acting on the beam.
SOLUTION
(Clockwise) Ans.
a
(Clockwise) Ans.
a
(Clockwise) Ans.=-2593 lb #ft =2.59 kip #ft
+1MF32A=-1601cos 30°21192+160 sin 30°10.52
=-5600 lb #ft =5.60 kip #ft
+1MF22A=-500a4
5b1142
=-3000 lb #ft =3.00 kip #ft
= 500 lbF2
4
3
5
= 375 lbF
1
8ft6ft
0.5 ft
30˚
5ft
B
A
232
3
233
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Ans:
MP=341 in. #lb
d
MF=403 in. #lb
b
Not sufficient
4–6.
The crowbar is subjected to a vertical force of P
=
25 lb at the
grip, whereas it takes a force of F
=
155 lb at the claw to pull
the nail out. Find the moment of each force about point A and
determine if P is sufcient to pull out the nail. The crowbar
contacts the board at point A.
SOLUTION
20
3 in.
1.5 in.
60
O
A
F
P
14 in.
234
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Ans:
MB=150 N #m
d
MB=600 N #m
d
MB=0
*4–8.
Determine the moment of each of the three forces about point B.
SOLUTION
For F2,
a
dAns.
Since the line of action of F3 passes through B,its moment arm about point Bis
zero. Thus
Ans. MB=0
=600 N #m
+MB=300 sin 60° (0) +300 cos 60° (4)
2 m3 m
60
30
F1 250 N
F2 300 N
A
4
3
236
237
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Ans:
MA=195
lb #ft
d
4–10.
20
2.5 ft
A
FB
FC
0.75 ft
30
B
C
25
If and determine the resultant
moment about the bolt located at A.
FC=45 lb,FB=30 lb
SOLUTION
d=195 lb #ft
SOLUTION
4–13.
The 20-N horizontal force acts on the handle of the socket
wrench. What is the moment of this force about point B.
Specify the coordinate direction angles
a,
b
,
g
of the
moment axis.
O
x
z
BA
y
50 mm
200 mm
10 mm
20 N
60
SOLUTION
M
Od
3.5096
4–14.
The 20-N horizontal force acts on the handle of the socket
wrench. Determine the moment of this force about point O.
Specify the coordinate direction angles
a,
b
,
g
of the
moment axis.
O
x
z
BA
y
50 mm
200 mm
10 mm
20 N
60
242
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Ans:
(MA)C=768
lb #ft
b
(MA)B=636
lb #ft
d
Clockwise
4–15.
Two men exert forces of and on the
ropes.Determine the moment of each force about A.Which
way will the pole rotate,clockwise or counterclockwise?
P
=
50 lbF
=
80 lb
A
P
F
B
C
6ft
45
12 ft
3
4
5
SOLUTION
adAns.
Since
Clockwise Ans.
(MA)C7(MA)B
+(MA)B=50 (cos 45°)(18) =636 lb #ft
243
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Ans:
F=39.8 lb
*4–16.
If the man at Bexerts a force of on his rope,
determine the magnitude of the force Fthe man at Cmust
exert to prevent the pole from rotating, i.e., so the resultant
moment about Aof both forces is zero.
P
=
30 lb
P
F
B
6ft
45
12 ft
3
4
5
SOLUTION
Ans.F=39.8 lb
244
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Ans:
m=
al
d
+
l
b
M
4–17.
SOLUTION
d+lb
The torque wrench ABC is used to measure the moment or
torque applied to a bolt when the bolt is located at A and a
force is applied to the handle at C. The mechanic reads the
torque on the scale at B. If an extension AO of lengthd is
used on the wrench, determine the required scale reading if
the desired torque on the bolt at O is to be M.
A
F
B
C
dl
O
M
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Ans:
M
P
=(537.5
cos
u
+75 sin
u
) lb #ft
4–18.
The tongs are used to grip the ends of the drilling pipe P.
Determine the torque (moment) that the applied force
exerts on the pipe about point Pas a function of
. Plot this moment versus for .0…u…90°uMP
u
F=150 lb
MP
43 in.
6 in.
F
P
M
P
u
SOLUTION
Ans.
At , is maximum.
Also (MP)max =150 lb
¢
a43
12 b2
+a6
12 b2
≤
1
2
=543 lb #ft
(MP)max =538 cos 7.943° +75 sin 7.943° =543 lb #ft
MP
u=7.943°
dMP
du
=-537.5 sin u+75 cos u=0 tan u=75
537.5 u=7.943°
=(537.5 cos u+75 sin u)lb#ft
=(6450 cos u+900 sin u)lb#in.
MP=150 cos u(43) +150 sin u(6)
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Ans:
F=239 lb
4–19.
The tongs are used to grip the ends of the drilling pipe P.If
a torque (moment) of is needed at Pto
turn the pipe, determine the cable force Fthat must be
applied to the tongs. Set .u=30°
MP=800 lb #ft
43 in.
6 in.
F
P
u
Ans.F=239 lb
800(12) =Fcos 30°(43) +Fsin 30°(6)
