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17–21.
SOLUTION
Ans.=4.45 kg #m2
=1
12 (3)(2)2+3(1.781 -1)2+1
12 (5)(0.52+12)+5(2.25 -1.781)2
IG=©IG+md2
G
2m
1m
0.5 m
y
O
T
he
pendulum
consists
of
the
3-kg
slender
rod
and
the
5-kg thin plate. Determine the location y of the center of
mass G
of
the
pendulum;
then
calculate
the
moment
of
inertia
of
the
pendulum
about
an
axis
perpendicular
to
the
page and passing through G.
17–22.
90 mm
50 mm
20 mm
20 mm
x
50 mm 30 mm
30 mm
180 mm
Determine the moment of inertia of the overhung crank
about the xaxis.The material is steel having a destiny of
.r=7.85 Mg>m3
SOLUTION
+c1
12 (0.8478)
A
(0.03)2+(0.180)2
B
d
Ix=2c1
2 (0.1233)(0.01)2+(0.1233)(0.06)2d
mr=7.85(103)((0.03)(0.180)(0.02)) =0.8478 kg
mc=7.85(103)
A
(0.05)p(0.01)2
B
=0.1233 kg
17–23.
90 mm
50 mm
20 mm
20 mm
x
50 mm 30 mm
30 mm
180 mm
Determine the moment of inertia of the overhung crank
about the axis.The material is steel having a destiny of
r=7.85 Mg>m3.
x¿
SOLUTION
+c1
12 (0.8478)
A
(0.03)2+(0.180)2
B
+(0.8478)(0.06)2d
Ix¿=c1
2 (0.1233)(0.01)2d+c1
2 (0.1233)(0.02)2+(0.1233)(0.120)2d
mp=7.85
A
103
BA
(0.03)(0.180)(0.02)
B
=0.8478 kg
mc=7.85
A
103
BA
(0.05)p(0.01)2
B
=0.1233 kg
*17–24.
SOLUTION
a
Ans.
Ans.
(:
+)s=s0+
v
0t+1
2aGt2
NA=105 lb
+c©Fy=m(aG)y;NA+95.0 -200 =0
NB=95.0 lb
+©MA=©(Mk)A;NB(12) -200(6) +30(9) =(200
32.2)(4.83)(7)
aG=4.83 ft>s2
:
+©Fx=m(aG)x;30=(200
32.2)aG
The door has a weight of 200 lb and a center of gravity at G.
Determine how far the door moves in 2 s, starting from rest,
if a man pushes on it at Cwith a horizontal force
Also, find the vertical reactions at the rollers Aand B.
F=30 lb.
6ft6ft
AB
CG12 ft
F
17–25.
The door has a weight of 200 lb and a center of gravity at G.
Determine the constant force Fthat must be applied to the
door to push it open 12 ft to the right in 5 s, starting from
rest. Also, find the vertical reactions at the rollers Aand B.
Ans.
a
Ans.
Ans.N
A
=101 lb
+c©Fy=m(aG)y;NA+99.0 -200 =0
NB=99.0 lb
+©MA=©(Mk)A;NB(12) -200(6) +5.9627(9) =200
32.2(0.960)(7)
F=5.9627 lb =5.96 lb
:
+©Fx=m(aG)x;F=200
32.2(0.960)
ac=0.960 ft>s2
12 =0+0+1
2aG(5)2
6ft6ft
AB
CG12 ft
F
17–26.
The jet aircraft has a total mass of 22 Mg and a center of
mass at G. Initially at take-off the engines provide a thrust
and Determine the acceleration of
the plane and the normal reactions on the nose wheel and
each of the two wing wheels located at B. Neglect the mass
of the wheels and, due to low velocity, neglect any lift
caused by the wings.
T¿=1.5 kN.2T=4kN
a
Ans.
Ans.
Ans.aG=0.250 m>s2
By=71.6 kN
Ay=72.6 kN
+©MB=©(MK)B; 4(2.3) -1.5(2.5) -22(9.81)(3) +Ay(9) =-22aG(1.2)
+c©Fy=0; 2By+Ay-22(9.81) =0
:
T¿2TG
2.5m 2.3m B
1.2m
A
6m3m
817
17–27.
SOLUTION
Solving,
Ans.
Ans.
Ans.t=2.72 s
10 =0+3.68 t
(:
+)v=v0+act
aG=3.68 ft>s2
NB=857 lb
NA=1393 lb
+c©Fy=m(aG)y;2NB+2NA-4500 =0
:
+©Fx=m(aG)x; 0.3(2NB)=4500
32.2 aG
The sports car has a weight of 4500 lb and center of gravity
at G. If it starts from rest it causes the rear wheels to slip
as it accelerates. Determine how long it takes for it to reach
a speed of 10 ft/s. Also, what are the normal reactions at
each of the four wheels on the road? The coefficients of
static and kinetic friction at the road are and
respectively. Neglect the mass of the wheels.mk=0.3,
ms=0.5 4ftAB
G
2ft
2.5 ft
Ans:
NA=1393
lb
NB=857
lb
t=2.72 s
818
*17–28.
The assembly has a mass of 8 Mg and is hoisted using the
boom and pulley system. If the winch at B draws in the cable
with an acceleration of 2 m
>
s2, determine the compressive
force in the hydraulic cylinder needed to support the boom.
The boom has a mass of 2 Mg and mass center at G.
G
C
4 m
2 m
6 m
SOLUTION
sB+2sL=l
a
L=-1 m>s2
Assembly:
+
c
Σ
Fy
=
may; 2T
-
8
(
10
3
)
(9.81)
=
8
(
10
3
)
(1)
T=43.24 kN
Boom:
a
+ΣMA=0;
F
CD
(2)
-
2
(
10
3
)
(9.81)(6 cos 60°)
-
2(43.24)
(
10
3
)
(12 cos 60°)
=
0
FCD =289 kN
Ans.
Ans:
FCD =289 kN
819
17–29.
The assembly has a mass of 4 Mg and is hoisted using the
winch at B. Determine the greatest acceleration of the
assembly so that the compressive force in the hydraulic
cylinder supporting the boom does not exceed 180 kN. What
is the tension in the supporting cable? The boom has a mass
of 2 Mg and mass center at G.
Ans:
a
=
2.74 m
>
s
2
T=25.1 kN
G
C
4 m
2 m
6 m
SOLUTION
17–30.
The uniform girder AB has a mass of 8 Mg. Determine the
internal axial, shear, and bending-moment loadings at the
center of the girder if a crane gives it an upward acceleration
of 3 m
>
s2.
SOLUTION
C
3 m/s2
821
17–31.
A car having a weight of 4000 lb begins to skid and turn with
the brakes applied to all four wheels. If the coefficient of
kinetic friction between the wheels and the road is m
k=0.8,
determine the maximum critical height h of the center of
gravity G such that the car does not overturn. Tipping will
begin to occur after the car rotates 90° from its original
direction of motion and, as shown in the figure, undergoes
translation while skidding. Hint: Draw a free-body diagram
of the car viewed from the front. When tipping occurs, the
normal reactions of the wheels on the right side (or passenger
side) are zero.
Ans:
h=3.12 ft
SOLUTION
NA
represents the reaction for both the front and rear wheels on the left side.
d
+
ΣF
x=
m(a
G
)
x
; 0.8NA=
4000
32.2
aG
+
c
Σ
Fy=m(aG)y ;
NA-4000 =0
a
+ΣMA=Σ(Mk)A;
4000(2.5) =
4000
32.2
(aG)(h)
Solving,
NA=4000 lb
aG=25.76 ft>s2
h=3.12 ft
Ans.
y
x
2.5 ft
2.5 ft h
z
G
*17–32.
A force of
P=300 N
is applied to the 60-kg cart. Determine
the reactions at both the wheels at A and both the wheels
at B. Also, what is the acceleration of the cart? The mass
center of the cart is at G.
0.3 m
0.4 m
30
AB
G
P
823
17–33.
Determine the largest force P that can be applied to the
60-kg cart, without causing one of the wheel reactions,
either at A or at B, to be zero. Also, what is the acceleration
of the cart? The mass center of the cart is at G.
Ans:
P=579 N
SOLUTION
Equations of Motions. Since
(0.38 m) tan 30°=0.22 m 70.1 m
, the line of action
of P passes below G. Therefore, P tends to rotate the cart clockwise. The wheels at A
will leave the ground before those at B. Then, it is required that
NA=0
. Referring,
to the FBD of the cart, Fig.a
+
c
ΣF
y
=m(a
G
)
y;
N
B
+P sin 30°-60(9.81) =60(0)
(1)
a+
ΣMG=0
;
P cos 30°(0.1) -P sin 30°(0.38) +NB(0.2) =0
(2)
Solving Eqs. (1) and (2)
P=578.77 N =579 N
Ans.
NB=299.22 N
0.3 m
0.08 m
0.2 m
0.3 m
0.4 m
30
AB
G
P
824
17–34.
SOLUTION
Using this result to write the moment equation about point A,
a
Ans.
Using this result to write the force equation of motion along the yaxis,
Ans.N
A=326.81 N =327 N
+c©F
y=m(aG)y;N
A+1144.69-150(9.81) =150(0)
N
B=1144.69N=1.14 kN
+©MA=(Mk)A; 150(9.81)(1.25) -600(0.5) -N
B(2) =-150(4)(1.25)
The trailer with its load has a mass of 150 kg and a center of
mass at G. If it is subjected to a horizontal force of
, determine the trailer’s acceleration and the
normal force on the pair of wheels at Aand at B.The
wheels are free to roll and have negligible mass.
P=600 N
1.25 m
0.25 m
0.25 m 0.5 m
G
BA
P600
N
Ans:
a
=
4
m
>
s
2
S
NB=1.14
kN
NA=327
N
17–35.
The desk has a weight of 75 lb and a center of gravity at G.
Determine its initial acceleration if a man pushes on it with
a force
F=60 lb.
The coefficient of kinetic friction at A
and B is m
k=0.2.
SOLUTION
S
+Σ
F
x=
ma
x
; 60 cos 30°-0.2NA-0.2NB=
32.2
aG
+
c
ΣF
y
=ma
y;
NA+NB-75 -60 sin 30°=0
a+
ΣMG=0
;
60 sin 30°(2) -60 cos 30°(1) -NA(2) +NB(2) -0.2NA(2)
-0.2NB(2) =0
Solving,
a
G=
13.3 ft
>
s
2
Ans.
NA=44.0 lb
NB=61.0 lb
G
F
A
30
B
1 ft
2 ft
826
*17–36.
The desk has a weight of 75 lb and a center of gravity at G.
Determine the initial acceleration of a desk when the man
applies enough force F to overcome the static friction at A
and B. Also, find the vertical reactions on each of the
two legs at A and at B. The coefficients of static and kinetic
friction at A and B are m
s=0.5
and m
k=0.2,
respectively.
G
F
A
30
B
1 ft
2 ft
SOLUTION
2
Ans:
a
G=
13.6 ft
>
s
2
N
A
′=22.1 lb
N
B
′=30.6 lb
827
17–37.
The 150-kg uniform crate rests on the 10-kg cart. Determine
the maximum force P that can be applied to the handle
without causing the crate to tip on the cart. Slipping does
not occur.
Ans:
P=785 N
1 m
0.5 m
P
SOLUTION
828
17–38.
The 150-kg uniform crate rests on the 10-kg cart. Determine
the maximum force P that can be applied to the handle
without causing the crate to slip or tip on the cart. The
coefficient of static friction between the crate and cart is
m
s=0.2.
Ans:
P=314 N
1 m
0.5 m
P
SOLUTION
17–39.
The bar has a weight per length w and is supported by the
smooth collar. If it is released from rest, determine the
internal normal force, shear force, and bending moment in
the bar as a function of x.
30
x
SOLUTION
*17–40.
The smooth 180-lb pipe has a length of 20 ft and a negligible
diameter. It is carried on a truck as shown. Determine the
maximum acceleration which the truck can have without
causing the normal reaction at A to be zero. Also determine
the horizontal and vertical components of force which the
truck exerts on the pipe at B.
SOLUTION
B
A
20 ft
5 ft
