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Unformatted text preview: ENGINEERING MECHANICS  Dynamics W.F. Riley & LJD. Sturges 173 The 4lb pendulum bob is released from 0
rest when 9 = 75 . Determine the work done on the pendulum bob By the cord tension T as the bob swings D D
from 6 = 75 to 6 = o .
By the gravitational force W as the bob swings from 9 = 750 to 9 = 0°. Solution
a. The cord tension T always
acts normal to the motion and hence does no work. UT = 0 lb'ft .......... ................................... Ans.
b. The gravitational force W
is constant in both magnitude and
direction. As the 4lb pendulum
bob swings from 9 = O0 to 9 = 750 it drops (moves in the direction of the force an a distance of d = 25(cos Oo — cos 75°)
= 18.52952 in. = 1.54413 ft and does work Fd = 4(1.54413)
6.18 lb'ft 833 3.7.7. 5 :m» w..." «Mme We.. 7., .,i_ __ "mp ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 175* A fully loaded Boeing 747 has a takeoff weight of 660,000 lb and
its engines develop a combined thrust of 200,000 lb. If the runway is
flat and level, determine the work done on the aircraft
a. By the engines as the aircraft moves 1000 ft down the runway.
b. By the gravitational force W as the aircraft moves 1000 ft down the runway.
Solution 3. The engine thrust is constant and acts directly in line with the
motion. Therefore, the work done
by the engine thrust is U: = Fd = 200,000(1000) = 200(106) lb'ft b. The weight force always acts in a vertical
direction while the motion of the plane is always
horizontal. Therefore, the weight force is always
perpendicular to the motion and does no work UW = O lb'ft ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 177 A ZOOlb block is pushed along a horizontal surface by a constant force F as shown. If the kinetic coefficient of friction between the block and the surface is ”k = 0.25, determine the work done on the block By the frictional force on the bottom
of the block as the block moves 20 ft
along the surface. By the gravitational force W as the
block moves 20 ft along the surface. By the loolb force F as the block moves 20 ft along the surface.
Solution Since the block has no motion in
the vertical direction, it has no
acceleration in that direction and the equations of motion give +T 2F = ma : N  100 sin 34°  200 =
y y Therefore N 255.919 lb Eb 0.25N = 63.9798 lb
0
a. friction force is constant and acts 180 to the direction of
motion of the block. Therefore
U = Fd = 63.9798(20) u
= 1280 lb‘ft ... ....... . ........ ............ ....... .. Ans. b. The weight force (vertical) is always perpendicular to the motion (horizontal), and therefore the weight force does no work
”W = 0 lb'ft ............................................. Ans. c. The force F is also constant. The component of F normal to the
surface does no work since there is no motion in that direction. The work
done by the component of F which acts along the surface (which is also
constant) is 0 U = Fd = (100 cos 34 )(20) F
= 1658 lb'ft ................................... 837 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 17—9* The pair of blocks shown are connected by a light inextensible
cord and are released from rest when the spring is unstretched. The
static and kinetic coefficients of friction are “5 = 0.2 and pk = 0.1,
respectively. If the 10lb block moves 9 in. to the right, determine
the work done On the 5—lb block by the gravitational  mnmﬁ force. {if ff: On the 10block by the frictional force
acting on the bottom of the block. On the 10lb block by the spring
force F5 = 5(4  to) where the spring constant is 5 = 20 lb/ft
and the unstretched length of
the spring is to = 12 in. Solution Since the 10lb block has no motion
in the direction normal to the surface,
it has no acceleration in that direction and the equations of motion give +TEF =ma: N—1o=o
y y Therefore
N 10 lb
F 0.1N = 1 lb a. The weight of the 5lb block is constant
and always acts vertically downward. As the
10—lb block moves 9 in. = 0.75 ft to the right,
the 5lb block will fall 0.75 ft and the weight force will do positive work on the block U5 Fd = 5(o.75) 3.75 lb'ft ...... ............ ...... ................ Ans. The friction force is constant and acts 180° to the direction of of the 10lb block. Therefore (Problem 179 continues ...) 839 ENGINEERINGMECHANICS  Dynamics W.F. Riley & L.D. Sturges 1710* The pair of blocks shown are connected by a light inextensible
cord and are released from rest when the spring is stretched 600 mm. The
static and kinetic coefficients of friction are NS = 0.3 and “k = 0.2, respectively. If the Skg block moves 300 mm to the left, determine the work done a. 0n the Skg block by the
frictional force acting on the
bottom of the block. 0n the 5kg block by the
spring force F; = £(£  £0) where the spring constant is
5 = 1 kN/m and the unstretched
length of the spring is to = 250 mm. 0n the lO—kg block by the gravitational force. 0n the lO—kg block by the frictional force acting on the
bottom of the block. Solution Since the blocks have no motion in
the direction normal to the incline, they
have no acceleration in that direction and the equations of motion give +1‘ 2F = ma : N  5(9.81)
y y 5 + 2 = : — . o
A Eh man N10 10(9 81) cos 50 Therefore
N5 49.05 N F3 0.2N = 9.81 N NlO 63.0575 N F10 0.2N = 12.6115 N a. The friction force IE is constant and acts 180° to the direction of motion
of the block. Therefore
U = Fd = —9.81(0.3) F5
= 2.94 N'm = 2.94 J ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges 1712 The pair of blocks shown are connected by a light inextensible
cord. The blocks are released from rest from the position shown with the
spring stretched 150 mm. If the 2—kg block rises 100 mm, determine the work done On the 2—kg block by the
gravitational force. On the 10kg.block by the
spring force F; = £(C  (b) where the spring constant is
i = 1.2 kN/m. Solution a. The weight force W5 = 19.62 N is constant and acts vertically downward.
As the 2kg block rises 0.1 m, the work done by the weight force W5 is Uh = FE = 19.62(O.1) = —l.962 N°m = ‘1.962 J b. The length of the cord L = d + 52 + c is constant. Initially, 510 = 400 mm, d = 500 mm, and 52 = 300 mm. After the 2—kg block rises 100 mm, 52 = 200 mm, d = 600 mm, 510 = 519.615 mm, and the stretch in the
spring has been reduces to 30.385 mm. The
spring exerts a force F; 1200x (where x = 0
when the spring is unstretched) on the 10kg
block. If x increases to the right, then the
spring force acts to the left (1800 to the
motion), d0; = —1200x dx, and 0.030385
0‘ f 1200x dx = 1200[
0.150 2 0.030385
X
2 ] 0.150 = 600[(0.030385)2 — (0.150)2] = 12.95 N'm = 12.95 J ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 1714* A 2kg package slides on a frictionless
floor and strikes the bumpers shown. The two
linear springs are identical with spring constants
of 5 = 1.5 kN/m. Determine the work done on the
package by the springs as spring 1 is compressed 120 mm (and spring 2 is compressed 20 mm). Solution Each spring exerts a force F; = lSOOx
(where x = 0 when the spring is unstretched)
on the 2kg block. If x increases to the
right, then the spring forces act to the
left (180° to the motion), dUa = 1500x dx, 0.120 0.020
I 1500x dx  I 1500x dx
0 0 x
2 x
2 2 ]O.120 2 0.020
1500[  1500[ ]
0 0 750[(0.120)2  O]  750[(0.020)2  0] 11.10 N‘m = 11.10 J ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 1723 A 7500—1b truck is traveling on the freeway at 65 mi/h when the
driver suddenly notices a moose standing on the road 200 ft straight
ahead. If it takes the driver 0.4 s to apply the brakes and the
kinetic coefficient of friction between the tires and the road is “k = 0.5, Can the driver avoid hitting
the moose without steering to
one side? Where will the truck come to rest relative to the moose? If the driver must steer to one side, determine the speed of the
truck as it passes the moose. Solution
Initially, the truck is traveling at 65 mi/h = 95.33 ft/s = constant. Therefore, the truck will travel a distance of A51 = v At = (95.33)(o.4) = 38.13 ft before the driver applies the brakes. The normal force is +T 2F = ma : N  W = O
Y Y N = W = 7500 lb
and the friction force is F uN 0.5(7500) = 3750 lb and it is constant. Only the friction force does work and the workenergy equation gives 1 7500 2 '
—_ .— . _ A =
2 [ 32_2 ](95 33) 3750 52 0 A52 = 282.2 ft
The total distance to stop is As1 + As 320.4 ft. Therefore
The driver must swerve. Stop 120.4 ft past the moose Again use the work—energy equation to find the speed of the truck when it passes the moose at As = 200  A51 = 161.9 ft of braking 1 7500 2 l 7500 2
2 [ 32.2 ](95.33)  (3750)(161.9)  —2_[—§2T2_]V 'v = 62.3 ft/s s 42.4 mi/h ......... ....... ...... Ans. 855 ember». .. "w, r“, 37...e w __..i,. 1,. , ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 1743 A 1.5—lb weight slides on a frictionless vertical rod as shown. The undeformed length of the spring is to = 8 in., the spring modulus is i = so lb/ft, and the distance d = 12 in. If
the slider is released from rest when b = 9 in., determine the speed of the slider when b = 0. Solution
The normal force N'does no work.
As the slider rises 9 in. 0.75 ft, the constant weight force does work = =—.5o.
UW Fd 1 ( 75) = 1.125 lb'ft The component of the spring force that is
normal to the rod does no work; the work done by the component along the rod is
U; = I —(£6 cos 9) db where 5 = ﬂ — £0 = t  8/12, cos 9 = b/ﬂ, and b is positive downward. Then Ué = J0 80[/ 1 + b2  ——8—]————
75 O. 12 2 O
80 b  *—§— 1 + b2 = 9.16667 lb'ft
12
0.75 Therefore, the workenergy equation Ti + U = T f gives 0 — 1.125 + 9.16667 = 18.58 ft/s ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 1757 The two blocks A and 3 shown weigh NA = 25 lb and WE = 50 1b, respectively. The blocks are at rest and the spring (5 = 25 lb/ft) is unstretched when the blocks are in
the position shown. Determine
The speed of block 3 when it is 1 ft below its initial
position. The maximum distance that block 3 will drop.
So1ution The tension is an internal force — its
work will cancel out when the workenergy
equations for the two blocks are added
together. When block block B falls a
distance d, block A rolls to the right
the same distance. The spring force
F5 = ﬁx acts 180° to the motion of block
A and the work done by the spring
force and the weight of block 3 is d
U = I (25x) dx + 50d = 50d  12.5d2 lb'ft
0 Therefore, the workenergy equation 11 + U = Tf 2 l 25 2 1 50 2
+ _ . = __ .___—_ __ —
(0 O) + (50d 12 5d ) 2 [ 32.2 ]v + 2 [ 32.2 ]V gives v2 = 42.9333d  10.7333az2 a. When d = 1 ft, VB = v = 5.67 ft/s ¢ ................................... Ans. b. The maximum distance that B will fall corresponds to v = 0
0 42.9333d  10.7333d2 d d = 4.00 ft ........................................ Ans.
max ...
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