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Unformatted text preview: ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 154* A 15—kg block of ice slides on a horizontal surface for 20 m before it stops. If the initial speed of the block was 15 m/s, determine The force of friction
between the block of 20m
ice and the surface. 1 Om/s The kinetic coefficient of
friction “k between the block of ice and the surface. Solution The free—body diagram includes the weight of
the block W = 15(9.81) = 147.15 N, the normal
force N and friction force F exerted on the
block by the surface. The equations of motion are +9 2F = ma : F
x x +T 2F ma N — 147.15 0
y y where ay = 0 since there is no motion in the vertical direction.
Therefore
N 147.15 N 15a —F Rewriting the xcomponent of acceleration using the chain—rule of differentiation and integrating 15V dv 2
15V /2 FR + C 1687.5  FR where the constant of integration has been chosen so that v = 15 m/s when x = 0. Then, if v  0 when x = 20 m,
F = 84.4 N and the kinetic coefficient of friction is 429 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 15—6 A lBOO—kg automobile is moving along a level road at a constant speed of 60 km/h. If the automobile accelerates at a constant rate and
reaches a speed of 80 km/h in 5 5, determine The force required to produce
this acceleration. The distance traveled by
the automobile during the
5—5 interval that it is accelerating. Solution The free—body diagram includes the weight
of the car an the normal force N and pushing
force F exerted on the tires by the road.
Only the x—component of the equation of motion gives useful information +9 2F = ma : F = 1500a = constant
x x Integrating the acceleration to get the velocity and position gives lSOOV Ft + C1 = Ft + 25,000 2 2
lSOOx Ft /2 + 25,000t + c2 = Ft /2 + 25,000t where the constants of integration have been chosen so that 60 km/h = 16.6657 m/s and X = 0 when t = 0. Then if V =
80 km/h = 22.2222 m/s when t = 5 s, 1500(22.2222) F(5) + 25,000 F 1667 N 833.33(5)2 + 25,000(5) 97.2 m 431 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 157* A force of 20 lb is applied to a 25lb block as shown in Fig. PIS—7. Let x = 0 and v = 0 when t = 0 and determine the velocity and displacement of the block at t = 5 3 if F The inclined plane supporting the
block is smooth. The kinetic coefficient of
friction between the inclined
plane and the block is “k = 0.25. Solution The freebody diagram includes the weight of
the block W, the normal force N and friction
force R exerted on the block by the surface, and
the applied force F. In terms of coordinates
along and normal to the surface, the equations
of motion are +£ 2F = ma : 20  25 sin 200  R
x x +5 2F = ma : N  25 cos 200 = O
Y Y where ay 0 since there is no motion in the direction normal to the surface. Therefore
N = 23.4923 lb
a. If R = 0, then the x—component of the equations of motion gives
a = 14.747 ft/s2 = constant
and integrating to get the velocity and position gives
V = 14.747t + Cl 14.747t
x = 7.37351:2 + C2 = 7.3735t2 where the constants of integration are both zero since the block starts
from rest at x = 0. Then, when t = 5 s V = 73.7 ft/s
x 184.3 ft R = 0.25N = 5.8731 lb, than
7.1824 ft/s2 constant
7.1824t + C 7.1824t 3
3.59121:2 + C4 = 3.5912t2 5 s,
35.9 ft/s
89.8 ft 432 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 15—10 A Saturn V rocket has a mass 6
of 2.75(10 ) kg and a thrust of 6
33(10 ) N. Determine The initial vertical acceleration
of the rocket. The rocket's velocity 10 s after
liftoff. The time required to reach an
altitude of 10,000 m. Solution A freebody diagram of the rocket
includes its weight W'and the thrust
T. Only the y—component of the
equations of motion gives useful information +T 2F = : T — mg = ma
Y Y
. 6
a. At liftoff, m = 2.75(10 ) kg,
6 2
33(10 ) N, g = 9.81 m/s , and the
initial vertical acceleration of the
rocket is 2
a = 2.19 m/s b. Assuming that the thrust and the mass both remain constant, the
acceleration is also constant. The first (T = constant) is probably a
good assumption but the second (m = constant) is probably not a good
assumption unless the time interval is very short. Then, integrating the acceleration to get the velocity of the rocket gives ‘ 2
a = 2.190 m/s = constant v = 2.190t m/S' (Problem 15—10 continues ...) 435 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges (Problem 1510 — cont.) where the constant of integration is zero since the rocket starts from rest when t = 0. And when t = 10 s, = (2.190)(10) = 21.9 m/s c. Integrating the velocity to get the position of the rocket gives 2
y = 1.095t m where the constant of integration is again zero since y = 0 when t = 0. Then the rocket will reach an altitude of 10,000 m when 2
10,000 = 1.095t
t = 95.6 5 (Over such a very long time interval, the rocket will have burned up
a large amount of fuel and its weight and mass will have decreased
significantly, Therefore, the constant mass assumption is very poor in this case and the answer is not likely to be very accurate.) 436 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 1521* Blocks A and B, which weigh 30 lb
and 50 lb, respectively, are connected by a
rope as shown. The kinetic coefficients of
friction “k are 0.35 for block A and 0.15
for block 3. During motion of the blocks down the inclined plane, determine a. The acceleration of block B. b. The tension in the rope. Solution So long as the rope remains taut,
the two blocks will move together and a
single free—body diagram may be drawn
of the two blocks or separate free—body
diagrams may be drawn for each of the
two‘blocks. Choosing to draw separate
freebody diagrams results in the equations of motion +§ ZFAX = 30 sin 350 + T — EA m a :
A Ax +3 ZFA . N — 30 cos 350 y A
+8 2F 50 s' 350 T F F6
BX 1“ B +& 2F NE  50 cos 350 O m a :
By B By
where the y—components of acceleration are both zero since there is no motion in the direction normal to the surface and the x—components of acceleration are the same since the blocks move together. Therefore N 4.575 lb F .3 .
A 2 A O SNA 8 6011 lb N . . .
B 40 958 lb F8 0 lSNb 6 1436 lb Finally, adding the xequations together gives ° 80
0 ‘ — , = ——————
8 Elm 35 14 7447 32.2 a 2
12.53 ft/s E .
3.07 lb 450 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 15—23 Blocks A and B, which weigh 200 lb and 120 lb, respectively, are connected by a rope as shown in Fig. a. During motion of the bodies, Determine the acceleration
of block A and the tension in the rope connecting the
bodies. Determine the acceleration
of block A if block B is
replaced with a constant
force of 120 lb as shown
in Fig. b. Solution a. Since the two bodies move in
different directions (A moves down
and B moves up), separate free—body
diagrams must be drawn for the two bodies. The equations of equilibrium give
200
+T 2F = m : — = ———
Ay AaAy 2T 200 32.2 aA (a)
120
+T Z = z  = —————
FB m a8 T 120 32.2 a (b) Measuring the position of both
blocks from the ceiling, the length of the rope can be expressed L=2 +
SA 53 Taking two time derivatives of this
relationship and noting that the length of the rope is constant gives 23 +2 =0
A a where s a =  u “
A A and 58 a8 (SA and 53 are positive downward while aA and a _ B
are positive upward). Therefore (Problem 1523 continues ...) 452 (Problem 15—23 — cont.) (a3) = 2<aA) and combining Eqs. a and b gives 200 2T = 200 + _§ETE_ a aA 1.894 ft/s2 T T = 105.9 lb b. If block 8 is replaced with a constant force of 120 lb, then the tension in the cable is just T = 120 1b and Eq. a gives the acceleration of block A 200
2(120) 200 _ 32.2 a aA = 6.44 ft/s2 T 453 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 1557 A circular disk rotates in a horizontal
plane. A 3—lb block rests on the disk 8 in. from
the axis of rotation. The static coefficient of
friction between the block and the disk is 0.50.
If the disk starts from rest with a constant
angular acceleration of 0.5 rad/52, determine the length of time required for the block to begin to slip. Solution The free—body diagram of the block
includes its weight W, a normal force
N exerted on the block by the surface,
and a friction force exerted on the
block by the surface. The friction
force is represented in terms of its
n— and t—components. The equations of
motion are 9+ 2F =
n +T 2F = ma : N  3 0
Z Z where the zcomponent of acceleration
is zero since the block has no
vertical motion,
2
a = 0.5 rad/s = constant w = 0.5t rad/s Therefore 3 lb , .
(Sde wew)
= 0.03106 lb 2
0.01553t lb 2 2 R 2 2
(0.03106) + (0.01553t ) s (0.50m) = (1.5 lb) and the block will begin to slip when
t = 9.83 s 506 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 1560 The car shown maintains a
constant speed of 100 km/h. At
both the bottom and top of the
hill, determine the force that
the car seat exerts on an 80—kg driver. solution The freebody diagram of the driver
includes his weight W and the normal
force N between the driver and the
seat. Using coordinates along and
normal to the road, the equations of motion are +9 ZFt mat: O = 80v +T 2F = ma : N 80(9.81) = 80.3
n n n . . 2 .
where the normal component of acceleration has magnitude v /p and pOints toward the center of curvature. When the car goes over the top of a hump in the road, then where v 100 km/h = 27.7778 m/s and p = 90 m. Therefore = 98.9 N 2
27.7778
90 N = 80 [9.81  which is only about 13 percent of the driver's weight! b. When the car goes over the bottom of a dip in the road, then where v = 100 km/h = 27.7778 m/s and p = 90 m. Now, 2
so [9.81+_27_;;78_]=1471N .......... ....ns.A which is about 1.87 times the driver's weight! 509 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 15—61 A highway is designed for traffic moving at 65 mi/h. Along a
certain portion of the highway, the radius of a curve is 900 ft. If the
curve is banked so that no friction is required to keep cars on the road, determine The required angle of banking
(angle 9) of the road. The minimum coefficient of
friction between the tires
and the road that would keep
traffic from skidding at this
speed if the curve were not
banked. Solution The free—body diagram of the car includes its weight W and the normal
force N and friction force F exerted on the tires by the road. Using
coordinates along and normal to the path of the car in the horizontal plane of motion, the equations of motion are e+ ZFn man F cos 6 + N sin 9 m(9s.33332/900) +T ZFZ ma : N cos 9 — F sin 9  mg 0 z
where the normal component of
acceleration is an = vz/p acting
toward the center of curvature
(horizontally to the left), V =
65 mi/h = 95.3333 ft/s, and the
z—component of acceleration is zero since the car has no vertical motion. a. If F = 0, then
N sin 6 = 10.0983m
N cos 9 32.2m
9 = tan—1(1O.O983/32.2) O b. If 6 = 0 , then
F = 10.0983m
N = 32.2m
u = F/N = l0.0983/32.2 510 ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges 1562 A curve of ZOO—m radius on a
level road is banked at the correct
angle for a speed of 65 km/h. If an automobile rounds this curve at
100 km/h, determine the minimum
coefficient of friction required
between the tires and the road so
that the automobile will not skid. Solution The free—body diagram of the car includes its weight W and the normal
force N and friction force F exerted on the tires by the road. Using
coordinates along and normal to the path of the car in the horizontal
plane of motion, the equations of motion are 9+ 2F; man F cos 9 + N sin 9 = mVZ/p +T 2F; maz: N cos 9 — F sin 9 — mg 0 where the normal component of
acceleration is an = VZ/p acting
toward the center of curvature
(horizontally to the left), and
the z—component of acceleration is zero since the car has no vertical motion. If F = 0 when v = 55 km/h =
18.0556 m/s, then N sin 9 1.6300m
N cos 6 9.81m 1 O
6 = tan (1.6300/9.81) = 9.434 Then, with 9 = 9.434° and v 100 km/h = 27.7778 m/s 0.98648F + 0.16391N 3.85802m
O.16391F + 0.98648N 9.81m F 2.19846m
N 10.3097m
F/N = 2.19846/10.3097 = 0.213 511 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 1564* A small sphere (m = 0.50 kg) is mounted
on a circular hoop. Friction between the sphere
and the hoop is negligible and the sphere is free
to slide when the hoop is rotated. Determine the
angle 6 and the force ﬁ exerted by the hoop on
the sphere when the hoop is rotating about a vertical diameter at a constant angular velocity of w = 120 rev/min. Solution The free body diagram of the sphere
includes its weight W and the normal
force P exerted on the sphere by the hoop. The equations of motion are e+ 2F = ma :
n n O.5(O.15 sin 6):»2 +T 2F ma :
z 2 P cos 9 — O.5(9.81) 0 where w = 120 rev/min = 4n rad/s, the
normal component of acceleration is
directed toward the center of the
circular path traveled by the sphere (in a horizontal plane, r = 0.15 sin 9),
and the zcomponent of acceleration is
zero since the sphere has no vertical motion. Therefore, P = 11.84 N 0
9 cos 0.41415 = 65.53 513 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 1566 The container is rotating about
a vertical axis with a constant angular
velocity. Determine the angular velocity
é for which the 2kg ball will maintain a
fixed position relative to the friction— less side of the container. Solution
The freebody diagram of the ball
includes its weight W and the normal force N. The equations of motion are 9+ 255 man N cos 300 = 2(o.1) 62 +T 2F ma :
Z Z N sin 30° — 2(9.81) where the normal component of
acceleration (an = 0.1 éz) is
directed toward the center of
the circular path traveled by
the ball, and the zcomponent of acceleration is zero since the ball has no vertical motion.
Therefore, N = 39.24 N 9 = 3.04 rad/s 515 ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges 1573* A S—lb sphere slides along a rod that is bent in a vertical . 2
plane into a shape that can be described by the equation x = By where x and y are both measured in feet. . . 12ft 12 ft
When the sphere is at the pOint x = 8 ft and y = 8 ft as shown,
it is moving along the rod at a
speed of 15 ft/s and is slowing
down at a rate of 3 ft/sz.
Determine the normal and
tangential components of the force being exerted on the sphere by the rod at this time. Solution Using components along and normal to the rod, the
equations of motion are +& EFL N  5 cos 9
+3 EFt 5 sin 9  T where the slope of the rod is given by
slope dy/dx = 0.25x At the position shown (x = 8 ft)
slope = —2 tan(—63.435°) e = 63.4350 The radius of curvature is given by 2 2
d dx 0.25 2 3/2 2 3/2
[1 + (dy/dx) ] [1 + (2) ] Therefore, with x = 8 ft, 6 = 63.4350, v = 15 ft/s, and 6 1_=
p 44.7214 ft
3.02 lb a 26.570
4.94 lb 5 63.430 525 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D..Sturges 1574 A lO—kg block A is held in place on a 20—kg cart 8 which rests against an obstruction C. If all surfaces are frictionless, determine The acceleration aA of the 10kg
block when it is released from rest.
The acceleration aA of the block
and the acceleration a3 of the cart if the obstruction C is
removed at the same time that the block A is released from rest. ~Solution a. If the cart does not move, only
a free—body diagram of the block A is
needed. In terms of coordinates along
and normal to the inclined surface, the equations of motion are +8 ZFAX = 10(9.81) sin 30° m a :
A Ax i& 25 [U a . I“ ‘ 10<9.81’ C05 30 where the y—component of acceleration is zero since there is no motion in the y—direction. Therefore, 2 0
BA = 4.91 m/s 3 30 b. Now the cart is also able to
move and a separate free—body diagram
must be drawn of it. In terms of
horizontal and vertical coordinates, the equations of motion are now +62 = ' 0:
th NA Sin 3O loaAX (a) m a :
A AX T2 . °— .
+ th NA cos 30 10(9 81) (b) 0
+6 2 .  '
FBX NA Bin 30 (c) +T Zpé  — NA cos 3o — 20(9.81) (d) (Problem 1574 continues ...) 526 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges (Problem 1574 — cont.) where the yhcomponent of acceleration of the cart 8 is zero since it does not move in the yhdirection. Since block A slides on the cart 8, the relative acceleration equation gives  = s +  3A as 311/3 0 o
= A + A — _ A
an 1 aA/B(cos 30 1 Sin 30 J) O
: = + 30
x aAX aBX aA/B cos (e) sin 30 Therefore, combining Eqs. a, c, and 9 gives NA sin 300 10(a + a cos 30°) = ZOa Bx A/B Bx 10a cos 30° (g) 3oan A/B and finally combining Eqs. a, e, and g and Eqs. b and f gives 0 O
N ' =
A Sln 30 (20/3)aA/B cos 30 10(9.81) — lOaA/B s1n 30 N 0
N cos 30
A Therefore D 5.773503A B
tan 30 = §§—I—:EEl——
' A/B
2
aA/B = 6.540 m/s 2
 1.888 m/s 75.52 N 7A 2 5A = 3.776 T — 3.27 J m/S
2 o
= 5.00 m/s E 40.89 ................. ... Ans.
E — 1 888 i 2
B — . m/s
2
= 1.888 m/s 6— ..... ..... ......... Ans. 527 ENGINEERING MECHANICS — Dynamics WF. Riley & L.D. Sturges 15104* A 5—kg block rests on a smooth conical surface that revolves about a vertical axis with a constant angular velocity u. The block is attached to the rotating shaft with a cable as Determine shown. a. The tension in the cable when the
system is rotating at 20 rev/min. b. The angular velocity in revolutions per
minute when the force between the‘
conical surface and the block is zero. Solution The free—body diagram of the block includes its weight W, the tension T and the normal force N in the cable, exerted by the surface on the block. Using xy coordinates along and normal to the surface, the equations of motion are +Z 2F = ma : ‘
x x ' ° 2
T — 5(9.81) sin 30 = 5(2)w cos 30o +5 2F = ma :
Y Y o 2 o N  S(9.81) cos 30 = —5(2)w sin 30 2 , _ 2
where the normal component of acceleration is an = v /r = rm and is directed toward the center of the circular path traveled by the block (in a horizontal plane); both the tangential
\ component and z— components of acceleration are zero. a. If m = 20 rpm = 2.09440 rad/s, then N = 20.546 N T = 62.5 N .......... ..... ......... ...... ............... Ans. b. [If N = O N, then w = 2.9147 rad/s = 27.8 rev/min ................. 582 ...
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