17–1.
z
Determine the moment of inertia Iy for the slender rod. The
rod’s density r and cross-sectional area A are constant.
Express the result in terms of the rod’s total mass m.
l
SOLUTION
A
x
Iy =
LM
x 2 dm
l
=
=
L0
x 2 (r A dx)
1
r A l3
3
m = rAl
Thus,
Iy =
1
m l2
3
Ans.
y
17–14.
If the large ring, small ring and each of the spokes weigh
100 lb, 15 lb, and 20 lb, respectively, determine the mass
moment of inertia of the wheel about an axis perpendicular
to the page and passing through point A.
4 ft
1 ft
O
SOLUTION
Composite Parts: The wheel can be subdivided into the segments shown in Fig. a.
The spokes which have a length of (4 - 1) = 3 ft and a center of mass located at a
3
distance of a1 + b ft = 2.5 ft from point O can be grouped as segment (2).
2
Mass Moment of Inertia: First, we will compute the mass moment of inertia of the
wheel about an axis perpendicular to the page and passing through point O.
IO = a
100
1 20
20
15
b (4 2) + 8c a
b (32) + a
b(2.52) d + a
b(12)
32.2
12 32.2
32.2
32.2
= 84.94 slug # ft2
The mass moment of inertia of the wheel about an axis perpendicular to the page
and passing through point A can be found using the parallel-axis theorem
100
20
15
IA = IO + md 2, where m =
+ 8a
b +
= 8.5404 slug and d = 4 ft.
32.2
32.2
32.2
Thus,
IA = 84.94 + 8.5404(42) = 221.58 slug # ft2 = 222 slug # ft2
Ans.
A
17–25.
The door has a weight of 200 lb and a center of gravity at G.
Determine the constant force F that 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 A and B.
6 ft
3 ft
+ )s = s + v t + 1 a t2
(:
0
0
G
2
12 = 0 + 0 +
1
aG(5)2
2
ac = 0.960 ft>s2
F =
200
(0.960)
32.2
F = 5.9627 lb = 5.96 lb
NB(12) - 200(6) + 5.9627(9) =
NB = 99.0 lb
+ c ©Fy = m(aG)y ;
12 ft
F
SOLUTION
a + ©MA = ©(Mk)A ;
B
G
C
+ ©F = m(a ) ;
:
x
G x
6 ft
A
Ans.
200
(0.960)(7)
32.2
Ans.
NA + 99.0 - 200 = 0
NA = 101 lb
Ans.
5 ft
17–29.
1 ft
If the strut AC can withstand a maximum compression
force of 150 lb before it fails, determine the cart’s maximum
permissible acceleration. The crate has a weight of 150 lb
with center of gravity at G, and it is secured on the platform,
so that it does not slide. Neglect the platform’s weight.
G
1 ft
C
30⬚
SOLUTION
Equations of Motion: FAC in terms of a can be obtained directly by writing the
moment equation of motion about B.
A
30⬚
+ ©MB = ©(Mk)B;
150(2) - FAC sin 60°(3) = - a
150
150
ba cos 30°(1) - a
ba sin 30°(2)
32.2
32.2
FAC = (3.346a + 115.47) lb
Assuming AC is about to fail,
FAC = 150 = 3.346a + 115.47
a = 10.3 ft>s2
Ans.
2 ft
a
B
P
17–35.
At the start of a race, the rear drive wheels B of the 1550-lb
car slip on the track. Determine the car’s acceleration and
the normal reaction the track exerts on the front pair of
wheels A and rear pair of wheels B. The coefficient of kinetic
friction is mk=0.7, and the mass center of the car is at G.
The front wheels are free to roll. Neglect the mass of all
the wheels.
0.75 ft
G
A
SOLUTION
Equations of Motion: Since the rear wheels B are required to slip, the frictional
force developed is FB = msNB = 0.7NB.
1550
a
32.2
+ ©F = m(a ) ;
;
x
G x
0.7NB =
+ c ©Fy = m(aG)y;
NA + NB - 1550 = 0
(2)
a + ©MG = 0;
NB(4.75) - 0.7NB(0.75) - NA(6) = 0
(3)
(1)
Solving Eqs. (1), (2), and (3) yields
NA = 640.46 lb = 640 lb
NB = 909.54 lb = 910 lb
a = 13.2 ft>s2
B
6 ft
Ans.
4.75 ft
17–38.
If the 4500-lb van has rear-wheel drive, and the coefficient
of static friction between the front wheels B and the road is
ms = 0.8, determine the normal reactions on the pairs of
front and rear wheels when the van has maximum
acceleration. The front wheels are free to roll. Neglect the
mass of the wheels.
G
2.5 ft
B
SOLUTION
Equations of Motion: The maximum acceleration occurs when the rear wheels are
about to slip. Thus, FB = msNB = 0.8 NB. Referring to Fig. a,
+ ©F = m(a ) ;
:
x
G x
0.8NB = a
4500
ba
32.2 max
+ c ©Fy = m(aG)y;
NA + NB - 4500 = a
+ ©MG = 0;
NA(3.5) + 0.8NB(2.5) - NB(6) = 0
(1)
4500
b (0)
32.2
(2)
(3)
Solving Eqs. (1), (2), and (3) yields
NA = 2.40 kip
NB = 2.10 kip
amax = 12.02 ft>s2 = 12.0 ft>s2
Ans.
6 ft
A
3.5 ft
17–45.
The 2-Mg truck achieves a speed of 15 m>s with a constant
acceleration after it has traveled a distance of 100 m,
starting from rest. Determine the normal force exerted on
each pair of front wheels B and rear driving wheels A. Also,
find the traction force on the pair of wheels at A. The front
wheels are free to roll. Neglect the mass of the wheels.
G 0.75 m
A
2m
SOLUTION
Kinematics: The acceleration of the truck can be determined from
v2 = v02 + 2ac(s - s0)
152 = 0 + 2a(100 - 0)
a = 1.125 m>s2
Equations of Motion: NB can be obtained directly by writing the moment equation
of motion about point A.
+ ©MA = (Mk)A;
NB(3.5) - 2000(9.81)(2) = -2000(1.125)(0.75)
NB = 10 729.29 N = 10.7 kN
Ans.
Using this result and writing the force equations of motion along the x and y axes,
+ ©F = m(a ) ;
:
x
G x
FA = 2000(1.125) = 2250 N = 2.25 kN
+ c ©Fy = m(aG)y;
NA + 10 729.29 - 2000(9.81) = 0
NA = 8890.71 N = 8.89 kN
B
Ans.
Ans.
1.5 m
17–49.
A
If the cart’s mass is 30 kg and it is subjected to a horizontal
force of P = 90 N, determine the tension in cord AB and
the horizontal and vertical components of reaction on end
C of the uniform 15-kg rod BC.
30⬚
1m
C
SOLUTION
Equations of Motion: The acceleration a of the cart and the rod can be determined
by considering the free-body diagram of the cart and rod system shown in Fig. a.
+ ©F = m(a ) ;
:
x
G x
90 = (15 + 30)a
a = 2 m>s2
The force in the cord can be obtained directly by writing the moment equation of
motion about point C by referring to Fig. b.
+ ©MC = (Mk)C;
FAB sin 30°(1) - 15(9.81) cos 30°(0.5) = -15(2) sin 30°(0.5)
FAB = 112.44 N = 112 N
Ans.
Using this result and applying the force equations of motion along the x and y axes,
+ ©F = m(a ) ;
:
x
G x
-Cx + 112.44 sin 30° = 15(2)
Cx = 26.22 N = 26.2 N
+ c ©Fy = m(aG)y;
Ans.
Cy + 112.44 cos 30° - 15(9.81) = 0
Cy = 49.78 N = 49.8 N
Ans.
30⬚
B
P
17–53.
The arched pipe has a mass of 80 kg and rests on the surface
of the platform. As it is hoisted from one level to the next,
a = 0.25 rad>s2 and v = 0.5 rad>s at the instant u = 30°. If
it does not slip, determine the normal reactions of the arch
on the platform at this instant.
500 mm
G
A
1m
θ
B
200 mm
ω, α
1m
SOLUTION
+ c ©Fy = m(aG)y ;
NA + NB - 80(9.81) = 20 sin 60° - 20 cos 60°
NA + NB = 792.12
a + ©MA = ©(Mk)A ;
NB(1) - 80(9.81)(0.5) = 20 cos 60°(0.2) + 20 sin 60°(0.5)
- 20 cos 60°(0.5) + 20 sin 60°(0.2)
NB = 402 N
Ans.
NA = 391 N
Ans.
17–55.
1 ft
At the instant shown, link CD rotates with an angular
velocity of vCD = 8 rad>s. If link CD is subjected to a
couple moment of M = 650 lb # ft, determine the force
developed in link AB and the angular acceleration of the
links at this instant. Neglect the weight of the links and the
platform. The crate weighs 150 lb and is fully secured on
the platform.
D
M ⫽ 650 lb⭈ft
Equilibrium: Since the mass of link CD can be neglected, Dt can be obtained
directly by writing the moment equation of equilibrium about point C using the
free-body diagram of link CD, Fig. a,
Dt(4) - 650 = 0
Dt = 162.5 lb
Equations of Motion: Since the crate undergoes curvilinear translation,
(aG)n = v2rG = 82(4) = 256 ft>s2 and (aG)t = arG = a(4). Referring to the freebody diagram of the crate, Fig. b, we have
©Ft = m(aG)t;
162.5 =
150
C a(4) D
32.2
150
(256)
32.2
©Fn = m(aG)n;
Dn + FAB + 150 =
a ©MG = 0;
Dn(1) - FAB(2) + 162.5(1) = 0
a = 8.72 rad>s2
Ans.
(1)
(2)
Solving Eqs. (1) and (2), we obtain
FAB = 402 lb
Dn = 641 lb
B
4 ft
vCD ⫽ 8 rad/s
SOLUTION
a + ©MC = 0;
G
1 ft
Ans.
A
C
3 ft
17–59.
The uniform slender rod has a mass m. If it is released from
rest when u = 0°, determine the magnitude of the reactive
force exerted on it by pin B when u = 90°.
L
3
A
B
u
2 L
3
SOLUTION
Equations of Motion: Since the rod rotates about a fixed axis passing through point
L
L
B, (aG)t = a rG = a a b and (aG)n = v2rG = v2 a b . The mass moment of inertia
6
6
1
2
of the rod about its G is IG =
mL . Writing the moment equation of motion about
12
point B,
-mg cos u a
+ ©MB = ©(Mk)B;
a =
L
L
L
1
b = -mc aa b d a b - a mL2 ba
6
6
6
12
3g
cos u
2L
1
This equation can also be obtained by applying ©MB = IBa, whereIB =
mL2 +
12
2
1
L
ma b = mL2. Thus,
6
9
-mg cos u a
+ ©MB = IBa;
L
1
b = - a mL2 b a
6
9
3g
cos u
2L
a =
Using this result and writing the force equation of motion along the n and t axes,
mg cos u - Bt = mc a
©Ft = m(aG)t;
Bt =
3g
L
cos u b a b d
2L
6
3
mg cos u
4
Bn - mg sin u = m cv2 a
©Fn = m(aG)n;
(1)
L
bd
6
1
mv2L + mg sin u
6
Bn =
(2)
Kinematics: The angular velocity of the rod can be determined by integrating
L
vdv =
v
L0
adu
u
vdv =
v =
L
3g
cos u du
2L
L0
3g
sin u
BL
When u = 90°, v =
3g
. Substituting this result and u = 90° into Eqs. (1) and (2),
AL
3
mg cos 90° = 0
4
3g
1
3
Bn = m a b (L) + mg sin 90° = mg
6
L
2
Bt =
FA = 3At2 + An2 =
2
3
3
02 + a mgb = mg
C
2
2
Ans.
C
17–62.
The 10-lb bar is pinned at its center O and connected to a
torsional spring. The spring has a stiffness k = 5 lb # ft>rad, so
that the torque developed is M = 15u2 lb # ft, where u is in
radians. If the bar is released from rest when it is vertical at
u = 90°, determine its angular velocity at the instant u = 0°.
1 ft
u
1 ft
O
SOLUTION
c + ©MO = IOa; - 5u = [
1 10
(
)(2)2]a
12 32.2
- 48.3 u = a
a du = v dv
v
o
-
Lp2
48.3 u du =
L0
v dv
1
48.3 p 2
( ) = v2
2 2
2
v = 10.9 rad/s
Ans.
17–67.
Determine the position rP of the center of percussion P of
the 10-lb slender bar. (See Prob. 17–66.) What is the
horizontal component of force that the pin at A exerts on
the bar when it is struck at P with a force of F=20 lb?
A
rP
4 ft
P
SOLUTION
Using the result of Prob 17–66,
rGP =
k2G
rAG
B
=
1 ml2 2
a
bR
B 12 m
l
2
=
1
l
6
Thus,
rP =
c + ©MA = IA a;
1
2
2
1
l + l = l = (4) = 2.67 ft
6
2
3
3
20(2.667) = c
Ans.
1 10
a
b (4)2 d a
3 32.2
a = 32.2 rad>s2
(a G)t = 2(32.2) = 64.4 ft>s2
+ ©F = m(a ) ;
;
x
G x
- Ax + 20 = a
Ax = 0
10
b (64.4)
32.2
Ans.
F
17–70.
The door will close automatically using torsional springs
mounted on the hinges. If the torque on each hinge is
M = ku, where u is measured in radians, determine the
required torsional stiffness k so that the door will close
1u = 0°2 with an angular velocity v = 2 rad>s when it is
released from rest at u = 90°. For the calculation, treat the
door as a thin plate having a mass of 70 kg.
M
1.5 m
A
SOLUTION
©MA = IAa;
2M = - B
0.4 m
1
(70)(1.2)2 + 70(0.6)2 R (a)
12
ku = - 16.8a
a du = v dv
0
Lp2
udu = 16.8
M
1.2 m
M = - 16.8a
-k
θ
B
0.4 m
2
L0
v dv
16.8
k p 2
( ) =
(2)2
2 2
2
k = 27.2 N # m>rad
Ans.
17–74.
The 5-kg cylinder is initially at rest when it is placed in
contact with the wall B and the rotor at A. If the rotor
always maintains a constant clockwise angular velocity
v = 6 rad>s, determine the initial angular acceleration of
the cylinder. The coefficient of kinetic friction at the
contacting surfaces B and C is mk = 0.2.
B
125 mm
SOLUTION
v
45
Equations of Motion: The mass moment of inertia of the cylinder about point O is
1
1
given by IO = mr 2 = (5)(0.1252) = 0.0390625 kg # m2. Applying Eq. 17–16,
2
2
we have
+ ©F = m(a ) ;
:
x
G x
NB + 0.2NA cos 45° - NA sin 45° = 0
(1)
+ c ©Fy = m(a G)y ;
0.2NB + 0.2NA sin 45° + NA cos 45° - 5(9.81) = 0
(2)
a + ©MO = IO a;
0.2NA (0.125) - 0.2NB (0.125) = 0.0390625a
(3)
Solving Eqs. (1), (2), and (3) yields;
NA = 51.01 N
NB = 28.85 N
a = 14.2 rad>s2
Ans.
C
A
17–83.
The two-bar assembly is released from rest in the position
shown. Determine the initial bending moment at the fixed
joint B. Each bar has a mass m and length l.
A
l
B
l
SOLUTION
C
Assembly:
IA =
1 2
1
l
(m)(l)2 + m(l2 + ( )2)
ml +
3
12
2
= 1.667 ml2
c + ©MA = IA a;
l
mg( ) + mg(l) = (1.667ml2)a
2
a =
0.9 g
l
Segment BC:
c + ©MB = ©(Mk)B;
M = c
M =
l>2
1
l
l
ml2 d a + m(l 2 + ( )2)1>2 a(
)( )
l 2 2
2
12
2
l + (2)
0.9g
1 2
1
ml a = ml2 (
)
3
3
l
M = 0.3gml
Ans.
17–85.
The bar has a weight per length of w. If it is rotating in the
vertical plane at a constant rate V about point O, determine
the internal normal force, shear force, and moment as a
function of x and u.
O
V
U
L
SOLUTION
x
a = v2 aL -
x u
bh
z
Forces:
wx 2
x
v aL - b u = N u + S au + wx T
h
g
z h
(1)
Moments:
x
Ia = M - S a b
2
O = M -
1
Sx
2
(2)
Solving (1) and (2),
N = wx B
v2
x
A L - B + cos u R
g
2
S = wx sin u
M =
1
wx2 sin u
2
Ans.
Ans.
Ans.
17–94.
The wheel has a weight of 30 lb and a radius of gyration of
kG = 0.6 ft. If the coefficients of static and kinetic friction
between the wheel and the plane are ms = 0.2 and
mk = 0.15, determine the wheel’s angular acceleration as it
rolls down the incline. Set u = 12°.
G
1.25 ft
SOLUTION
u
+b©Fx = m(aG)x ;
+a©Fy
= m(aG)y ;
a + ©MG = IG a;
30
ba
30 sin 12° - F = a
32.2 G
N - 30 cos 12° = 0
F(1.25) = c a
30
b (0.6)2 da
32.2
Assume the wheel does not slip.
aG = (1.25)a
Solving:
F = 1.17 lb
N = 29.34 lb
aG = 5.44 ft>s2
a = 4.35 rad>s2
Fmax = 0.2(29.34) = 5.87 lb 7 1.17 lb
Ans.
OK
17–102.
The 2-kg slender bar is supported by cord BC and then
released from rest at A. Determine the initial angular
acceleration of the bar and the tension in the cord.
C
30°
B
A
300 mm
SOLUTION
+ ©F = m(a ) ;
:
x
G x
T cos 30° = 2(aG)x
+ c ©Fy = m(aG)y ;
T sin 30° - 19.62 = 2(aG)y
a + ©MG = IGa ;
T sin 30°(0.15) = [
1
(2)(0.3)2]a
12
aB = aG + aB>G
aB sin 30°i - aB cos 30°j = (aG)xi + (aG)y j + a (0.15)j
+ )
(:
(aB) sin 30° = (aG)x
(+ c)
(aB) cos 30° = - (aG)y - a (0.15)
Thus,
1.7321(aG)x = - (aG)y - 0.15a
T = 5.61 N
Ans.
(aG)x = 2.43 m s2
(aG)y = - 8.41 m s2
a = 28.0 rad s2
Ans.
17–106.
The spool has a mass of 500 kg and a radius of gyration
kG = 1.30 m. It rests on the surface of a conveyor belt for
which the coefficient of static friction is ms = 0.5 and the
coefficient of kinetic friction is mk = 0.4. If the conveyor
accelerates at a C = 1 m>s2, determine the initial tension in
the wire and the angular acceleration of the spool. The
spool is originally at rest.
0.8 m
G
aC
SOLUTION
+
:
a Fx = m(aG)x;
+ c a Fy = m(aG)y;
c + a MG = IGa;
-Fs + T = 500aG
Ns - 500(9.81) = 0
Fs(1.6) - T(0.8) = 500(1.30)2a
ap = aG + ap>G
(ap)yj = aG i - 0.8ai
aG = 0.8a
Ns = 4905 N
Assume no slipping
a =
ac
1
=
= 1.25 rad>s
0.8
0.8
Ans.
aG = 0.8(1.25) = 1 m>s2
T = 2.32 kN
Fs = 1.82 kN
Since
(Fs)max = 0.5(4.905) = 2.45 7 1.82
(No slipping occurs)
1.6 m
Ans.
17–109.
The 500-kg concrete culvert has a mean radius of 0.5 m. If
the truck has an acceleration of 3 m>s2, determine the
culvert’s angular acceleration. Assume that the culvert does
not slip on the truck bed, and neglect its thickness.
0.5m
SOLUTION
Equations of Motion: The mass moment of inertia of the culvert about its mass
center is IG = mr 2 = 500 A 0.52 B = 125 kg # m2. Writing the moment equation of
motion about point A using Fig. a,
a + ©MA = ©(Mk)A ;
0 = 125a - 500a G(0.5)
(1)
Kinematics: Since the culvert does not slip at A, (aA)t = 3 m>s2. Applying the
relative acceleration equation and referring to Fig. b,
a G = a A + a * rG>A - v2rG>A
a Gi - 3i + (aA)n j + (ak * 0.5j) - v2(0.5j)
aGi = (3 - 0.5a)i + C (aA)n - 0.5v2 D j
Equating the i components,
a G = 3 - 0.5a
(2)
Solving Eqs. (1) and (2) yields
aG = 1.5 m>s2 :
a = 3 rad>s2
3 m/s2
4m
Ans.
17–110.
v0
The 10-lb hoop or thin ring is given an initial angular
velocity of 6 rad>s when it is placed on the surface. If the
coefficient of kinetic friction between the hoop and the
surface is mk = 0.3, determine the distance the hoop moves
before it stops slipping.
6 in.
O
SOLUTION
+ c ©Fy = m(aG)y ;
+ ©F = m(a ) ;
;
x
G x
c + ©MG = IGa;
N- 10 = 0
0.3(10) =
N = 10 lb
A
6
0.3(10) A 12
B =
10
32.2
B aG
aG = 9.66 ft>s2
10
A 32.2
B A 126 B 2 a
a = 19.32 rad>s2
When slipping ceases, vG = v r = 0.5v
(1)
v = v0 + at
(a + )
v = 6 + (- 19.32)t
+ B
A;
(2)
vG = (vG)0 + aGt
vG = 0 + 9.66t
(3)
Solving Eqs. (1) to (3) yields:
t = 0.1553 s
+ B
A;
vG = 1.5 ft>s
v = 3 rad>s
s = s0 + (vG)0 t + 12 aG t2
= 0 + 0 + 12 (9.66)(0.1553)2
= 0.116 ft = 1.40 in.
6 rad/s
Ans.
17–113.
v0
The uniform disk of mass m is rotating with an angular
velocity of v0 when it is placed on the floor. Determine the
initial angular acceleration of the disk and the acceleration
of its mass center. The coefficient of kinetic friction between
the disk and the floor is mk.
r
SOLUTION
Equations of Motion. Since the disk slips, the frictional force is Ff = mkN. The mass
1
moment of inertia of the disk about its mass center is IG = m r2. We have
2
+ c ©Fy = m(aG)y;
N - mg = 0
N = mg
+ ©F = m(a ) ;
;
x
G x
mk(mg) = maG
aG = mkg ;
+ ©MG = IGa;
1
-mk(mg)r = a mr2 b a
2
a =
2mkg
r
Ans.
Ans.
17–115.
ω
The 16-lb bowling ball is cast horizontally onto a lane such
that initially v = 0 and its mass center has a velocity
v = 8 ft>s. If the coefficient of kinetic friction between the
lane and the ball is mk = 0.12, determine the distance
the ball travels before it rolls without slipping. For the
calculation, neglect the finger holes in the ball and assume
the ball has a uniform density.
8 ft/s G
0.375 ft
SOLUTION
16
a
32.2 G
+ ©F = m(a ) ;
:
x
G x
0.12NA =
+ c ©Fy = m(aG)y ;
NA - 16 = 0
a + ©MG = IG a;
2 16
0.12NA(0.375) = c a
b (0.375)2 da
5 32.2
Solving,
NA = 16 lb;
aG = 3.864 ft>s2;
a = 25.76 rad>s2
When the ball rolls without slipping v = v(0.375),
(a+)
v = v0 + ac t
v
= 0 + 25.76t
0.375
v = 9.660t
+ B
A;
v = v0 + ac t
9.660t = 8 - 3.864t
t = 0.592 s
+ B
A;
s = s0 + v0 t +
1 2
a t
2 c
s = 0 + 8(0.592) s = 4.06 ft
1
(3.864)(0.592)2
2
Ans.