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Unformatted text preview: ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 154* A 15kg 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 r————————QOm——————————j
ice and the surface. lSnV Omk The kinetic coefficient of
friction Mk between the block of ice and the surface. Solution The freebody 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 +6 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 Fw + C = 1687.5  Ex 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 . . . . . . ... . . . . . ... . . . . . . . . . . . ...... . . . . . . . .... Ans. and the kinetic coefficient of friction is u F/N = 0.573 . . . . . . . . . . . . . . . . . . .... . . . . ... .. Ans. k 429 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 156 A 1500—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 s, determine a. 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 SE} = max: F = 1500a = constant Integrating the acceleration to get the velocity and position gives 1'! 1500V Ft + C1 = Ft + 25,000 H 2 2
Ft /2 + 25,000t + C = Ft /2 + 25,000t 1500
x 2 where the constants of integration have been chosen so that 60 km/h = 16.6667 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 . . . . . . . . . . . . . .. 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 25—lb block as shown in Fig.
P15—7. Let x = O and V = 0 when t = O and determine the velocity and displacement of the block at t = 5 5 if F The inclined plane supporting the
block is smooth. The kinetic coefficient of
friction between the inclined
plane and the block is uk = 0.25. Solution The free—body 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 +2 2F = ma : 2O — 25 sin 200 — R
x x +5 2F = ma : N — 25 cos 20° 0
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 + C 14.747t 1 2 2
x = 7.3735t + 02 7.3735t 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 If R = 0.25N = 5.8731 1b, then
2
a 7.1824 ft/s constant
v = 7.1824t + C 7.1824t 3
x 3.5912t2 + c; 3.5912t2 and at t = 5 s,
35.9 ft/s
89.8 ft ....... . . . . . . . . . . ....................... Ans. 432 ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges 1510 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 ybcomponent of the
equations of motion gives useful information +T 2F = : T — mg = ma
y y 6
a. At lift—off, m = 2.75(1o ) 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 1510 continues ...) 435 ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges (Problem 15—10  cont.) where the constant of integration is zero since the rocket starts from rest when t = 0. And when t = 10 s, V = (2.190)(10) = 21.9 m/s ......................... Ans. 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 s (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 15—21* Blocks A and B, which weigh 30 lb
and 50 lb, respectively, are connected by a
rope as shown. The kinetic coefficients of
friction pk 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 8. b. The tension in the rope. Solution So long as the rope remains taut,
the two blocks will move together and a
single freebody 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
free—body diagrams results in the equations of motion +2 EEAX 3o sin 350 + T  Eh m a :
A Ax + E .  o
l FAY V NA 30 cos 35 + E o ‘ 35o — 
E FBX 5 Elm T F3 Fe. 0
+1 2F m a : N  50 cos 35 0
By B. By B
where the yhcomponents 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 24. 75 . .
NA 5 lb Eh 0 35Nh 8 6011 lb N 40.95 b . .
B 8 1 FB 0 15Nﬁ 6 1436 lb Finally, adding the x—equations together gives 80 sin 35° — 14.7447 = 12.53 ft/s2 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 8 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 8 moves up), separate free—body
diagrams must be drawn for the two bodies. The equations of equilibrium give +T 2F
A Y +T EFBY Measuring the position of both
blocks from the ceiling, the length of the rope can be expressed = +
L 25A 53 Taking two time derivatives of this
relationship and noting that the length of the rope is constant gives +00 =
25A SB 0 h re = — a = — u
w e 3A aA nd 33 a3 (5A and 53 are positive downward while aA and a8 are positive upward). Therefore (Problem 1523 continues ...) 452 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges (Problem 1523  cont.) (a3) = 2(aA) and combining Eqs. 3 and b gives 200
= — = +
2T 200 + 32.2 3A 240 3A 1.894 ft/s2 T T = 105.9 lb b. If block 3 is replaced with a constant force of 120 lb, then the
tension in the cable is just T = 120 lb and Eq. a gives the acceleration
of block A 200
2(120) — 200 32.2 a = 6.44 ft/s2 T A 453 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 1557 A circular disk rotates in a horizontal
plane. A.3lb 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/sz, determine the length of time required for the block to begin to slip. Solution The freebody 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 e+ 2F
n +T 2F 0
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
0.03106 lb 0.01553tz lb 2 2 3 '
(0.03106) + (0.01553: ) _<_ (0.50M)2 = (1.5115)2 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 BOkg driver. Solution The free—body 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 +e 2F : = 80'
t mat 0 v +T EFL = ma : N  80(9.81) = 80a n n
2
where the normal component of acceleration has magnitude v /p and points toward the center of curvature. a. When the car goes over the top of a hump in the road, then = 100 km/h = 27.7778 m/s and p = 90 m. Therefore 2
27.7778 ] = 98.9 N N = 80 [9.81  90
is only about 13 percent of the driver's weight! When the car goes over the bottom of a dip in the road, then 2 2 = 100 km/h = 27.7778 m/s and p = 90 m. 2
N = 80 [9.81 + 1237—8] = 1471 N is about 1.87 times the driver's weight! 509 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 1561 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 a. The required angle of banking
(angle 9) of the road; b. 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 2
9+ ZFn man F'cos 6 + N sin 6 = m(95.3333 /900) +T SF; ma : N cos 6 — F sin 9  mg = O 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
zcomponent of acceleration is zero since the car has no vertical motion. a. If F = 0, then
N sin 9 = 10.0983m
N cos 6 = 32.2m
6 = tan—1(1O.0983/32.2) O b. If 9 = O , then
F = 10.0983m
N = 32.2m
u = F/N = 10.0983/32.2 510 ...
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 Spring '08
 McVay

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