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```MCE 263
Final Exam
July 25, 2014
Please do all the work independently and do not write on back of this paper.
Problem 1 (50 points): The pitcher throws the baseball horizontally with a speed of 140 ft/s
from a height of 5 ft. The batter is 60 ft away. Determine:
(a) The initial horizontal and vertical velocity components of the ball.
9
8:28 AM
Page 71
(b) Equation of motion in the horizontal direction.
(c) Equation
of motion in the vertical direction.
6/8/09 8:28 AM Page 71
91962_01_s12-p0001-0176
(d) The time needed for the ball to arrive at the batter.
f this material may be reproduced,
any form
or by any
means,
without
permission
in writing
from the publisher.
(e)in The
height
h at
which
it passes
the
batter.
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
throws the baseball horizontally with
m a height of 5 ft. If the batter is 60 ft
h
me for the ball to arrive
at the The
batter
*12–88.
pitcher throws the baseball horizontally with
ich it passes the batter.
a speed of 140 ft>s from a height of 5 ft. If the batter is 60 ft
60 = 140t
5 ft
t = 0.4286 = 0.429 s +
A;B
1
t + ac t2
2
s = vt;
60 = 140t
5 ft
h
away, determine the time for the ball to arrive at the batter60 ft
and the height h at which it passes the batter.
60 ft
Ans.
Ans.
t = 0.4286 = 0.429 s
1
s = s0 + v0 t + ac t2
1
A + c 2B
2
+ 0 + (-32.2)(0.4286) = 2.04 ft
Ans.
2
1
h = 5 + 0 + (-32.2)(0.4286)2 = 2.04 ft
2
Ans.
vA
hrown off the top of the building. If it
uA
in 3 s, determine the initial velocity vA
le uA at which it was thrown.
Also,
find
A
•12–89. The ball is thrown off the top of the building. If it
all’s velocity when it strikes
the
ground.
strikes
the
ground at B in 3 s, determine the initial velocity v
vA
uA
A
and the inclination angle uA at which it was thrown. Also, find
the magnitude
of the
velocity
when it strikes the ground.
he x–y coordinate system
will be set so
thatball’s
its origin
coincides
A
Coordinate System: The x–y coordinate system will be set so that its origin coincides
75 ft
= vA cos u, xA = 0, and
B = 60
withxpoint
A.ft, and t = 3 s. Thus,
(vA)xt
A
cos u(3)
20
75 ft
x-Motion: Here, (vA)x = vA cos u, xA = 0, and xB = 60 ft, and t = 3 s. Thus,
+ B
A:
xB = xA + (vA)xt
60 = 0 + vA cos u(3)
(1)
B
60 ft
2
= vA sin u, ay = -g = -32.2 ft>s
, yAu == 020
, and yB = -75 ft,
vA cos
(1)
2
y-Motion: Here, (vA)y = vA sin u, ay = -g = -32.2 ft>s , yA = 0, and yB = -75 ft,
1 2
ay t
and t = 3 s. Thus,
2
1
1
vA sin u(3) + (-32.2)
A +Ac32B B yB = yA + (vA)y t + ay t2
2
2
1(2)
3.3
-75 = 0 + vA sin u(3) + (-32.2) A 32 B
2
) yields
v sin u = 23.3
(2)
(vA)y t +
B
60 ft
Problem 2 (50 points): At the instant shown car A is traveling with a velocity of 30 m/s and
has an acceleration of 2 m/s2 along the highway. At the same instant B is traveling on the
trumpet interchange curve with a speed of 15 m/s, which is decreasing at 0.8 m/s2 . At this
particular instant determine:
(a) The velocity vectors of A and B in your chosen coordinate system.
/09
9:33 AM
Page 174
(b) The relative velocity of B with respect to A at this instant, its magnitude and direction.
(c) The acceleration vector of A in your chosen coordinate system.
(d) The acceleration vector of B in your chosen coordinate system.
this material may be reproduced,
or by any
means, withoutof
permission
writing from
(e)in any
theform
relative
acceleration
B withinrespect
tothe
A,publisher.
its magnitude
and direction.
nt shown car A is traveling
with a
91962_01_s12-p0001-0176 6/8/09 9:33 AM Page 174
has an acceleration of 2 m>s2 along
ame instant B is traveling on the
rve with a speed of 15 m>s, which is
Determine the relative velocity and
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
B with respect to A at this instant.
*12–228. At the instant shown car A is traveling with a
velocity of 30 m>s and has an acceleration of 2 m>s2 along
the highway. At the same instant B is traveling on the
trumpet interchange curve with a speed of 15 m>s, which is
decreasing at 0.8 m>s2. Determine the relative velocity and
relative acceleration of B with respect to A at this instant.
A
B
60"
r ! 250 m
A
j = 30i + (vB>A)xi + (vB>A)y j
A)x
y
m>s ;
B
vB = vA + vB>A
60"
15 cos 60°i + 15 sin 60°j = 30i + (vB>A)xi + (vB>A)y j
15 cos 60° = 30 + (vB>A)x
15 sin 60° = 0 + (vB>A)y
(vB>A)x = -22.5 = 22.5 m>s ;
(vB>A)y = 12.99 m>s c
.99)2 = 26.0 m>s
0° b
26.0 m>s
vB>A = 2(22.5)2 + (12.99)2 =Ans.
Ans.
u = tan - 1 a
Ans.
(aB)n =
9 m>s2
12.99
b = 30° b
22.5
Ans.
v2
152
=
= 0.9 m>s2
r
250
aB = aA + aB>A
-0.8 cos 60°i - 0.8 sin 60°j + 0.9 sin 60°i - 0.9 cos 60°j = 2i + (aB>A)x i + (aB>A)y j
-0.8 cos 60° + 0.9 sin 60° = 2 + (aB>A)x
-0.8 sin 60° - 0.9 cos 60° = (a
)
B>A y
60°j + 0.9 sin 60°i - 0.9 cos 60°j = 2i + (aB>A)x i + (aB>A
)y j
(aB>A)x = -1.6206 ft>s = 1.6206 m>s2 ;
2
0° = 2 + (aB>A)x
0° = (aB>A)y
= 1.6206 m>s2 ;
(aB>A)y = -1.1428 ft>s2 = 1.1428 m>s2 T
aB>A = 2(1.6206)2 + (1.1428)2 = 1.98 m>s2
1.1428
f = tan - 1 ¢
≤ = 35.2° d
1.6206
= 1.1428 m>s2 T
1.1428)2 = 1.98 m>s2
Ans.
35.2° d
Ans.
Ans.
Ans.
r ! 250 m
Problem 3 (50 points): The man pushes on the 60-lb crate with a force F . The force is always
directed down at 30◦ from the horizontal as shown, and its magnitude is increased until the
crate begins to slide. The coefficient of static friction is µs = 0.6 and the coefficient of kinetic
friction is µk = 0.3. Determine:
AM
(a) Free-body and mass-acceleration diagrams with the appropriate xy axis.
Page 188
(b) The equation of motion in x direction.
(c) The equation of motion in y direction.
The This
minimal
force
neededunder
to produce
motion
and
corresponding
reserved.
material
is protected
laws as
theythe
currently
may be reproduced, in any form or by any means, without permission in writing from the publisher.
normal force.
(e) The crate’s initial acceleration right after it begins to slide.
91962_02_s13_p0177-0284
-lb crate with a force F.
own at 30° from the
de is increased until the
ne the crate’s initial
atic friction is ms = 0.6
n is mk = 0.3.
6/8/09
10:00 AM
Page 188
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
F
30⬚
*13–16. The man pushes on the 60-lb crate with a force F.
The force is always directed down at 30° from the
horizontal as shown, and its magnitude is increased until the
crate begins to slide. Determine the crate’s initial
acceleration if the coefficient of static friction is ms = 0.6
and the coefficient of kinetic friction is mk = 0.3.
F
30⬚
Force to produce motion:
:
x
Fcos 30° - 0.6N = 0
N - 60 - F sin 30° = 0
N = 91.80 lb
= 0
F = 63.60 lb
Since N = 91.80 lb,
:
x
x
0° = 0
63.60 cos 30° - 0.3(91.80) = a
60
ba
32.2
a = 14.8 ft>s2
Ans.
F = 63.60 lb
- 0.3(91.80) = a
60
ba
32.2
Ans.
188
Problem 4 (50 points): The ends of bar AB are confined to move along the paths shown. At
a given instant, A has a velocity of 8 ft/s and an acceleration of 3 ft/s2 . At this instant,
determine:
(a) The kinematic diagram of bar AB for the velocity analysis.
PM
Page 595
(b) The angular velocity of AB and the velocity of B.
(c) The kinematic diagram of bar AB for the acceleration analysis.
(d) The expressions for the acceleration vectors for A and B.
rial may be reproduced, in any (e)
form The
or by any
means, without
permissionof
in AB
writing
fromthe
the publisher.
angular
acceleration
and
acceleration
are confined to move along
ant, A has a velocity of 8 ft>s
s2. Determine the angular
on of AB at this instant.
4 ft
30⬚
B
30⬚
4 ft
Ans.
A
C a (4) S + C (2)2 (4 ) S
c 60°
g60°
30° = 0 + a(4) sin 60° + 16 cos 60°
30° = -3 + a(4) cos 60° - 16 sin 60°
Ans.
>A
+ (
(8)2
(8)2
) sin 30°i + (
) cos 30°j = -3j
4
4
30°j) - (2)2(-4 sin 30°i + 4 cos 30°j)
= -3.464a + 8
564 = -3 + 2a - 13.8564
Ans.
vA ⫽ 8 ft/s
aA ⫽ 3 ft/s2
of B.
the paths shown. At a given instant, A has a velocity of 8 ft>s
and an acceleration of 3 ft>s2. Determine the angular
velocity and angular acceleration of AB at this instant.
4 ft
30⬚
B
30⬚
8
4
4 ft
b
Ans.
vB = 4(2) = 8 ft>s
(aB)n =
A
vA ⫽ 8 ft/s
aA ⫽ 3 ft/s2
(8)2
= 16 ft>s2
4
aB = aA + aB>A
C 16 S + C (aB) t S = C 3 S + C a (4) S + C (2)2 (4 ) S
T
c 30°
c 60°
g60°
g30°
+ )
(:
16 sin 30° + (aB)t cos 30° = 0 + a(4) sin 60° + 16 cos 60°
(+ c )
16 cos 30° - (aB)t sin 30° = -3 + a(4) cos 60° - 16 sin 60°
b
Ans.
2
(aB)t = 30.7 ft>s
Also,
aB = aA + aAB * rB>A - v2rB>A
(aB)t cos 30°i - (aB)t sin 30°j + (
(8)2
(8)2
) sin 30°i + (
) cos 30°j = -3j
4
4
-(ak) * (-4 sin 30°i + 4 cos 30°j) - (2)2(-4 sin 30°i + 4 cos 30°j)
+ )
(:
A+cB
(aB)t cos 30° + 8 = -3.464a + 8
-(aB)t sin 30° + 13.8564 = -3 + 2a - 13.8564
b
Ans.
(aB)t = 30.7 ft>s2
595
Problem 5 (50 points): The uniform rod AB of mass m is released from rest when β = 70◦ .
Assuming that the friction force between end A and the surface is large enough to prevent
sliding, determine immediately after release:
(a) The free-body and mass-acceleration diagrams.
(b) The linear (force) equations of motion with respect to chosen coordinates.
(c) The angular (moment) equation of motion.
(d) The angular acceleration of the rod and the acceleration of its center of mass.
(e) The normal reaction at A and the friction force at A.
PROBLEM 16.157
The uniform rod AB of weight W is re
Assuming that the friction force betwee
enough to prevent sliding, determine im
angular acceleration of the rod, (b) th
friction force at A.
PROBLEM 16.157
The uniform rod AB of weight W is released from rest when β = 70°.
Assuming that the friction force between end A and the surface is large
enough to prevent sliding, determine immediately after release (a) the
angular acceleration of the rod, (b) the normal reaction at A, (c) the
friction force at A.
SOLUTION
SOLUTION
We note that rod rotates about A. ω = 0
We note that rod rotates about A. ω = 0
1
mL2
12
L
a= α
2
I =
1
I = mL2
12
L
a= α
2
ΣM A = Σ( M A )eff :
L
L

mg  cos β  = I α + (ma )
2
2

L
L

mg  cos β  = I α + (ma )
2
2

ΣM = Σ ( M ) :
1
1 A2
 L L
mgL cos β = mL
α +  m αA eff
2
12
 2 2
1 2
= mL α
3
1
1
 L L
mgL cos β = mL2α +  m α 
(1)
2
12
 2 2
1
FA = ma sin β
= mL2α
3
L
L  3 g cos β 
F = m α sin β = m
sin β
3 g cos β
α=
2 L
ΣFx = Σ( Fx )eff :
A
FA =
ΣFy = Σ( Fy )eff :
2  2
2
L
3
mg sin β cos β
4
L 
NA − mg = −ma cos β = − m  α  cos β
x 2 eff
x
ΣF = Σ ( F ) :
L  3 g cos β
2  2 L

 cos β

 3

2
NA = mg 1 − cos β 
 4

NA − mg = −m


α=
3 g cos β
(2)
2 L
FA = ma sin β
L
L  3 g cos β
FA = m(3) α sin β = m 
2
22 L
3
mg sin β cos β
4
FA =
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
ΣFy 6= Σ( Fy )eff :
1672
L 
NA − mg = −ma cos β = − m  α  cos β
2 
NA − mg = −m
L  3 g cos β
2  2 L

 cos β

```
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