PHYS 1441 – Section 501
Lecture #19
Monday, Aug. 9, 2004
Dr. Jaehoon Yu
•
•
Problem #35
And many others
Monday, Aug. 9, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
1
Problem #35
Water is filled to a height H (=32m) behind a dam of width w (=1200m).
Determine the resultant force exerted by the water on the dam.
Since the water pressure varies as a function of depth, we
will have to do some calculus to figure out the total force.
h
H
dy
y
The pressure at the depth h is
P   gh   g  H  y 
The infinitesimal force dF exerting on a small strip of dam dy is
dF  PdA   g  H  y  wdy
Therefore the total force exerted by the water on the dam is
yH
F

y 0
 g  H  y  wdy   gw  Hy 
Monday, Aug. 9, 2004
yH
1
1 2
y 
  gwH 2 
2  y 0 2

1
2
3
 10  9.8 1200   32   6.0  109 N
2
PHYS 1441-501, Summer 2004
2
Dr. Jaehoon Yu
Problems 5, 25 and 29
Problem 5: Rotational kinematics
 f   i 47  47

 47 rad / s 2

2.0
t
Problem 25: Elastic properties of matter
Weight of the person W  100  9.8  980N
Total force to be supported is Fsup port  4W  3920 N
Thus, from the definition of tensile strength
3920 N
Fsup port
5
2
2

1.96

10
m

0.2
cm

A
8
2.00

10
TensileStength
Problem 29: Conditions for equilibrium
 Fx  T sin   500N  0
F
y
 N  T cos  0
Where  is the angle between the support cable and the pole
500 N
500

 2.5 103
Thus, solving Fx equation for T, we obtain
sin 
25  1
Monday, Aug.
2004
PHYSnot
1441-501,
Summer
3
Bad9, problem
since it does
provide
all2004
necessary information!!
T
Dr. Jaehoon Yu
Problems 36 and 42
Problem 36:Pressure measurements
P  PAir  Pgauge  1.0atm  16 psi  14.7 psi  16 psi
30.7
 30.7 psi 
1.01105  2.1105 N
14.7
Problem 42: Flow rate and equation of continuity problem
Flow rate through the vent
M
2
  Air  A  v   Air     0.05  1.4m / s  0.011  Air kg / s
t
Total mass of air in the room
M air room   AirV   Air  4.0  3.5  2.5  35   Air kg
Thus the time needed to replace the entire air in the room is
M air  room
35   Air kg

 3182s  53min
Flow  rate 0.011   Air kg / s
Monday, Aug. 9, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
4
Problems 56 and 57
Problem 56: Simple Harmonic Motion
1
k
2
2
2
2
f 
; Thus, k  4    m  f  4   10  1.68  1114 N / m
2 m
fA 
331
k
1114
 0.662 Hz; Thus, mA 

 64kg
2
2
500
 2    f A   2    0.662 
Problem 57: Equation of Motion of a SHM
x   A sin  t  ;
a   A 2 sin  t     3.33   4.22  sin  4.22 1.15   58.7m / s 2
2
Monday, Aug. 9, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
5