© 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.
5–50.
SOLUTION
Set
, then
F
rom Eq. (2),
Ans.k=250 N/m
0.5 =k(0.002)
FC=FA=kyA
yB=4mm
yA=2mm
x=16.67
x
yA
=100 –x
10
x=50
3=16.67 mm
x
©Fy=0; FB=2F
©MB=0; FA=FC=F
The rigid metal strip of negligible weight is used as part
of an electromagnetic switch. Determine the maximum
stiffness kof the springs at Aand Bso that the contact at C
closes when the vertical force developed there is 0.5 N.
Originally the strip is horizontal as shown.
50 mm 50 mm
10 mm
A
B
C
k
k
5–51.
The cantilever footing is used to support a wall near its
edge Aso that it causes a uniform soil pressure under the
footing. Determine the uniform distribution loads and
, measured in lb ft at pads Aand B,necessary to support
the wall forces of 8 000 lb and 20 000 lb.
wB
w
A
w
A
AB
w
B
1.5 ft
8000 lb
20 000 lb
0.25 ft
SOLUTION
439
*5–52.
SOLUTION
The uniform beam has a weight Wand length land is
supported by a pin at Aand a cable BC.Determine the
horizontal and vertical components of reaction at Aand
the tension in the cable necessary to hold the beam in the
position shown.
C
B
A
l
5–53.
SOLUTION
a
Spring Formula: Applying , we have
Geometry: The angle of tilt is
Ans.a=tan–1a0.10464 +0.07848
1b=10.4°
a
¢A=1177.2
15(103)=0.07848 m ¢B=1569.6
15(103)=0.10464 m
¢=F
k
+©MA=0; F
B(1) –392.4(4) =0F
B=1569.6 N
B
A
1m 3 m
A boy stands out at the end of the diving board, which is
supported by two springs A and B, each having a stiffness of
k = 15kN>m. In the position shown the board is horizontal.
If the boy has a mass of 40 kg, determine the angle of tilt
which the board makes with the horizontal after he jumps
off. Neglect the weight of the board and assume it is rigid.
5–54.
h
C
B
l
The uniform rod has a length of .If,
determine the distance hof placement at the end Aalong the
smooth wall for equilibrium.
SOLUTION
Using this result to write the force equation of equilibrium along yaxis,
(1)
Geometry: Applying the sine law with by referring to Fig. b,
Substituting Eq. (2) into (1) yields
since , then
(3)
Again, applying law of cosine by referring to Fig. b,
(4)
Equating Eqs. (3) and (4) yields
Ans.h=0.645 m
3h2=1.25
4
3h=h2+1.25
3h
cos (u–f)=h2+1.25
3h
l2=h2+1.52–2(h)(1.5) cos (u–f)
cos (u–f)=(4>3)hcos (u–f)–(4>3)h
sin uZ0
sin u[cos (u–f)–4
3 h]=0
sin (180° –u)=sin u
sin u cos (u–f)–2 sin f=0
a15 sin u
sin fb cos (u–f)–3=0+c©F
y=0;
T=1.5 sin u
sin f
s=1.5 ml=1 m30-N
5–55.
SOLUTION
Using the result ,
(1)
Geometry: Applying the sine law with , we have
(2)
Substituting Eq. (2) into (1) yields
(3)
Using the cosine law,
(4)
Equating Eqs. (3) and (4) yields
Ans.h=
A
s2–l2
3
2h
s=h2+s2–l2
2hs
cos (u–f)=h2+s2–l2
2hs
l2=h2+s2–2hs cos (u–f)
cos (u–f)=2h
s
sin f
h=sin u
ssin f=h
ssin u
sin (180° –u)=sin u
sin ucos (u–f)–2 sin f=0
+c©Fy=0; Wsin u
2 sin fcos (u–f)–W=0
T=Wsin u
2 sin f
T=Wsin u
2 sin f
h
s
C
A
The uniform rod has a length l and weight W. It is supported
at one end A by a smooth wall and the other end by a cord
of length s which is attached to the wall as shown. Determine
the placement h for equilibrium.
*5–56.
SOLUTION
a
(1)
(2)
(3)
Substituting Eqs
. (1) and (3) into Eq. (2):
Ans.
u
=tan–1a1
2cot c–1
2cot fb
tan
u=sin f–cos ftan c
2 sin ftan c
sin
u(2 sin ftan c)–cos u(sin f–cos ftan c)=0
2
cos (f–u) tan c–cos utan ccos f–cos usin f=0
aW–Wcos ucos f
2 cos (f–u)btan c–Wcos usin f
2 cos (f–u)=0
NB=W–NAcos f
cos c
+c©Fy=0; NBcos c+NAcos f–W=0
:
+©Fx=0; NBsin c–NAsin f=0
NA=Wcos u
2 cos (f–u)
+©MB=0; –WaL
2cos ub+NAcos f(Lcos u)+NAsin f(Lsin u)=0
T
he uniform rod of length Land weight Wis supported on
the
smooth planes. Determine its position for equilibrium.
N
eglect the thickness of the rod.
u
L
u
f
c
5–57.
The beam is subjected to the two concentrated loads.
Assuming that the foundation exerts a linearly varying load
distribution on its bottom, determine the load intensities
and for equilibrium if and .L=12 ftP=500 lbw2
w1
P2P
w2
w1
L
––
3
L
––
3
L
––
3
© 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.
5–58.
directly by summing moments about point A.
a
Ans.
Ans.w2=4P
L
1
2aw2–2P
LbL+2P
L1L2–3P=0+c©Fy=0;
w1=2P
L
PaL
3b–w1LaL
6b=0+©MA=0;
P2P
L
––
3
L
––
3
L
––
3
The beam is subjected to the two concentrated loads.
Assuming that the foundation exerts a linearly varying load
distribution on its bottom, determine the load intensities w1
and w2 for equilibrium in terms of the parameters shown.
© 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.
5–59.
SOLUTION
Solving by trial and error,
Ans.u=23.2° and u=85.2°
cos u–1.5 sin 2u+0.1667 =0
100 +600 cos u–1800 sin ucos u=0
Fs=kx;Fs=50 (6 sin u)
3 ft
3 ft
2ft
100 lb ft k50 lb/ft
B
Au
The rod supports a weight of 200 lb and is pinned at its
end A. If it is also subjected to a couple moment of
100 lb # ft, determine the angle ufor equilibrium. The
spring has an unstretched length of 2 ft and a stiffness
of k = 50 lb/ft.
*5–60.
Determine the distance
d
for placement of the load Pfor
equilibrium
of the smooth bar in the position as shown.
Neglect the weight of the bar.
u
SOLUTION
a
Ans.
Also;
Require forces to be concurrent at point O.
T
hus,
Ans.d=a
cos3
u
AO =dcos u=a>cos u
cos u
d=a
cos3u
Rd cos2u=Raa
cos ub
+©MA=0; –P(dcos u)+Raa
cos ub=0
+c©Fy=0; Rcos u–P=0
d
a
u
P
5–61.
If ,and ,determine the normal reaction at
the smooth supports and the required distance afor the
placement of the roller if .Neglect the weight of
the bar.
SOLUTION
(1)
(2)
Ans.
Substitute this result into Eq (2),
Ans.
Substitute this result into Eq (1),
Ans.
=0.650 maa=0.6495 m
692.82 =450
a
N=693 NN
B=692.82
N
B–0.5(346.41) =600 cos 30°
N
A=346.41 N =346 N
N
A cos 30° –600 sin 30° =0
+
Q ©F
x¿=0;
N
B–0.5N
A=600 cos 30°
N
B–N
A sin 30° –600 cos 30° =0
a
+©F
y¿=0;
N
B =450
a
N
B =aa
cos 30° b–600 cos 30°(1) =0
a
+©MA=0;
P=600 N
u=30°d=1
m
P
d
u
5–62.
The uniform load has a mass of 600 kg and is lifted using a
uniform 30-kg strongback beam BAC and four wire ropes
as shown. Determine the tension in each segment of rope
and the force that must be applied to the sling at A.
SOLUTION
Ans.
The force Fapplied to the sling Amust support the weight of the load and
strongback beam. Hence
Ans.F=6180.3 N =6.18 kN
F–60019.812–3019.812=0©Fz=0;
T=1839.375 N =1.84 kN
2 m
1.25 m
1.25 m
F
A
BC
5–63.
A
B
D
E
6ft
z
x
y
C
SOLUTION
Solving,
Ans.
Ans.
Ans.RF=13. 7 k i p
RE=22. 6 k i p
RD=22. 6 k i p
©F
z=0; RD+RE+RF–8000 –6000 –45 000 =0
©My=0; 8000( 4) +45 000( 7) +6000( 4) –RF( 27) =0
Due to an unequal distribution of fuel in the wing tanks, the
c en t er s o f gr avi t y f o r t h e ai r p l an e f u s el age Aand wings B
an d Car e l o c at ed as s h o w n . I f t h es e c o m p o n en t s h ave
weights and
determine the normal reactions of the wheels D, E, an d F
on the ground.
W
C=6000 l b ,W
B=8000 l b ,W
A=45 000 l b ,
5–65.
The
5
0-lb mulching machine has a center of gravity at G.
Determine the vertical reactions at the wheels Cand Band
the smooth contact point A.
z
G
4 ft
453
SOLUTION
11.25
11.25
11.25
11.25
5–66.
The smooth uniform rod AB is supported by a ball-and-socket
joint at A, the wall at B, and cable BC. Determine the
components of reaction at A, the tension in the cable, and the
normal reaction at B if the rod has a mass of 20 kg.
z
A
B
2 m
0.5 m
C
*5–68.
The 100-lb door has its center of gravity at G. Determine the
components of reaction at hinges Aand Bif hinge Bresists
only forces in the xand ydirections and Aresists forces in
the x,y,zdirections.
SOLUTION
B
z
18 in.
24 in.