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3–1.
SOLUTION
Solving:
Ans.
Ans.F1=1.83 kN
F2=9.60 kN
0.3420F2-0.8660F1=1.7
+c©Fy=0; F2cos 70° +5 sin 30° -F1sin 60° -3
5(7) =0
0.9397F2+0.5F1=9.930
5(7) =0
The members of a truss are pin connected at joint O.
Determine the magnitudes of and for equilibrium.
Set .u=60°
F2
F1
u
F2
70fi
30fi
5 kN
y
x
O
3
5
3–2.
SOLUTION
Solving:
Ans.
Ans.F1=4.31 kN
u=4.69°
F1sin u=0.3521
+c©Fy=0; 6 cos 70° +5 sin 30° -F1sin u-3
5(7) =0
F1cos u=4.2920
5(7) =0
The members of a truss are pin connected at joint O.
Determine the magnitude of and its angle for
equilibrium. Set .F
2=6kN
uF1
u
F2
70fi
30fi
5 kN
y
x
O
3
5
*3–4.
The bearing consists of rollers, symmetrically confined
within the housing.The bottom one is subjected to a 125-N
force at its contact Adue to the load on the shaft.
Determine the normal reactions NBand NCon the bearing
at its contact points Band Cfor equilibrium.
Ans.
Ans.NB=105 N
:
+©Fx=0; NB-163.176 sin 40° =0
NC=163.176 =163 N
125 N
A
40°
N
C
3–5.
SOLUTION
Solving,
Ans.
Ans.F=5.40 kN
T=7.20 kN
+c©F
y=0; 9 -Tsin 53.13° -Fa3
5b=0
:
+©F
x=0; Tcos 53.13° -Fa4
5b=0
f=90° -tan-1a3
4b=53.13°
The members of a truss are connected to the gusset plate.If
the forces are concurrent at point O, determine the
magnitudes of Fand Tfor equilibrium. Take .u=90°
x
y
A
O
F
T
B
9kN
C
4
53
u
3–6.
SOLUTION
(2)
Rearrange then divide Eq. (1) into Eq. (2):
Ans.
Ans.u=f+tan-1a3
4b=70.1°
T=7.66 kN
tan f=0.656,
f=33.27°
+c©F
y=0; 9 -8a3
5b-Tsin f=0
+©F
The gusset plate is subjected to the forces of three members.
Determine the tension force in member Cand its angle for
equilibrium. The forces are concurrent at point O.Take
.F=8kN
u
y
A
F
T
B
9kN
C
4
53
u
167
3–7.
SOLUTION
Thus,
(1)
(2)
Divide Eq.(2) by Eq.(1)
Ans.
Ans.T30.5 lb
u=20°
tan u
2=tan 10°
2Tsin
u
2cos
u
2=60 sin 10°
T(2cos2u
2)=60 cos 10°
T(1 +cos u)=60 cos 10°
+Q©Fy¿=0; Tsin u-60 sin 10° =0
+R©Fx¿=0; 60 cos 10° -T-Tcos u=0
The man attempts to pull down the tree using the cable and
small pulley arrangement shown. If the tension in AB is
60 lb, determine the tension in cable CAD and the angle
which the cable makes at the pulley.
u
20°
θ
B
A
C
D
30°
SOLUTION
*3–8.
The cords ABC and BD can each support a maximum load
of 100 lb. Determine the maximum weight of the crate, and
the angle
u
for equilibrium.
12
5
13
B
A
D
u
3–9.
Determine the maximum force F that can be supported in
the position shown if each chain can support a maximum
tension of 600 lb before it fails.
C
A
B
45
3
30fi
F
SOLUTION
*3–12.
The lift sling is used to hoist a container having a mass of
500 kg.Determine the force in each of the cables AB and
AC as a function of If the maximum tension allowed in
each cable is 5 kN,determine the shortest lengths of cables
AB and AC that can be used for the lift. The center of
gravity of the container is located at G.
u.A
F
173
© 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.
3–13.
A nuclear-reactor vessel has a weight of 500
(
10
3
)
lb.
Determine the horizontal compressive force that the
spreader bar AB exerts on point A and the force that each
cable segment CA and AD exert on this point while the
vessel is hoisted upward at constant velocity.
SOLUTION
AB
C
DE
30fi30fi
3–14.
Determine the stretch in each spring for equlibrium of the
2-kg block. The springs are shown in the equilibrium
position.
SOLUTION
3 m
3
m4
m
kAB fi 30 N/m
kAC fi 20 N/m
CB
3–15.
The unstretched length of spring AB is 3 m. If the block is
held in the equilibrium position shown, determine the mass
of the block at D.
3m
3m 4 m
k
AC
20 N/m
k
AB
30 N/m
CB
SOLUTION
W=84 N
+c©Fy=0; -W+67.88 sin 45° +60a3
5b=0
*3–16.
Determine the mass of each of the two cylinders if they
cause a sag of when suspended from the rings at
Aand B.Note that when the cylinders are removed.s=0
s=0.5 m
1m 2 m2m
1.5m
s
BA
CD
k100 N/mk100 N/m
SOLUTION
+c©Fy=0; 32.84sin 45° - m(9.81) = 0
TAC =100 N>m (2.828 - 2.5) = 32.84 N
© 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.
3–17.
Determine the stiffness kT of the single spring such that the
force F will stretch it by the same amount s as the force F
stretches the two springs. Express kT in terms of stiffness k1
and k2 of the two springs.
SOLUTION
F=ks
s=s1+s2
s=
F
k
T
=
F
k
1
+
F
k
2
1
k
T
=
1
k
1
+
1
k
2
Ans.
s
Unstretched
position
k1
s
k2
kT
F
F
© 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:
k=176 N>m
SOLUTION
3–18.
If the spring DB has an unstretched length of 2 m, determine
the stiffness of the spring to hold the 40-kg crate in the
position shown.
2 m
2
m3
m
k
CB
D
SOLUTION
3–19.
Determine the unstretched length of DB to hold the 40-kg
crate in the position shown. Take k = 180 N
>
m.
2 m
2
m3
m
k
CB
D
*3–20.
Avertical force is applied to the ends of the 2-ft
cord AB and spring AC.If the spring has an unstretched
length of 2 ft,determine the angle for equilibrium. Take
k=15 lb>ft.
u
P=10 lb
SOLUTION
(1)
(2)
From Eq. (1):
From Eq.(2):
Set
Solving for by trial and error,
Ans.u=35.0°
u
k=15 lb>ft
tan ua25-4 cos u-1b
25-4 cos u
=10
4k
a25-4 cos u-1b
25-4 cos u
(2 tan u-sin u+sin u)=10
2k
2ka25-4 cos -1b(2 -cos u)
25-4 cos u
tan u+
2ka25-4 cos u-1b2sin u
225-4 cos u
=10
T=2k
A
25-4 cos u-1
B
¢
2-cos u
25-4 cos u
≤
a1
cos ub
T=F
sacos f
cos ub
F
s=ks =2k(25-4 cos u-1)
s=2(4)2+(2)2-2(4)(2) cos u-2=225-4 cos u-2
+c©F
y=0; Tsin u+F
ssin f-10 =0
:
+©Fx=0; Fscos f-Tcos u=0
2 ft
k
2 ft
A
BC
u
