Transcript Wednesday, March 4, 2009

PHYS 1441 – Section 002
Lecture #10
Wednesday, Mar. 4, 2009
Dr. Jaehoon Yu
•
•
Exam Solutions
Application of Newton’s Laws
–
•
Forces of Friction
–
Wednesday, Mar. 4, 2009
Motion without friction
Motion with friction
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
1
Announcements
• Changes of Exam Dates
– Mid-term exam scheduled on Mar. 11 now moved to Wed. Mar. 25
– Second non-comprehensive exam scheduled on Apr. 13 now
moved to Wed. Apr. 22
– Final comprehensive exam stays on the same date, May 11
• A fun colloquium today
– Physics of NASCAR at 4pm in SH101
Wednesday, Mar. 4, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
2
Wednesday, Mar. 4, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
3
Problems 10 – 14]
)
r
a  5i  m s 2 
• A car starts from rest and accelerates with the acceleration
for 15 s. It then travels at a constant speed for 50s and then slows down
at the rate of 1 m/s2 to a full stop. Ignore road friction and air
resistance. Be sure to express vector quantities in vector form.
10. What is the velocity after 40s from the start?
Solution: Since the acceleration is applied only in x-direction for the first 15s, the
velocity at 15 seconds is the same as that at 40s from the start. Using the
kinematic equation, we obtain
ax  5m s 2
ay  0 m s2
vx ,t  40  vx ,t 15  v  a t  0   5 15  75  m s 
xi
x
v y ,t  40  v y ,t 15  v yi  a y t  0   0 15  0  m s 
r
r
r
r
vt 40  vt 15
r
r
r
 vx ,t  40 i v y ,t  40 j  75i  0 j  75i  m s 
Wednesday, Mar. 4, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
4
Problems 10 – 14]
)
r
a  5i  m s 2 
• A car starts from rest and accelerates with the acceleration
for 15 s. It then travels at a constant speed for 50s and then slows down
at the rate of 1 m/s2 to a full stop. Ignore road friction and air
resistance. Be sure to express vector quantities in vector form.
11. What is the total displacement after 50s from the start?
Solution: We divide the motion in two segments: one from 0 – 15s where the car was
accelerated and 15 – 50, the 35 s period the car was running at the constant
velocity of . Using the kinematic equation, we obtain
1
1
x  x f  xi  vxi t  axt 2 ; y  y f  yi  v yi t  a y t 2
2
2
1
2
x015 s  0 15    5   15   563  m 
2
1
2
x1550 s   75   35   0   35   2625  m  ;
2
1
2
y050 s  y0s  0      0       0  m 
2 r
r
r
r
r
r
  r 050 s   x015 s  x1550 s  i  y050 s j   563  2625  i  0 j  3188im
Wednesday, Mar. 4, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
5
Problems 10 – 14]
)
r
a  5i  m s 2 
• A car starts from rest and accelerates with the acceleration
for 15 s. It then travels at a constant speed for 50s and then slows down
at the rate of 1 m/s2 to a full stop. Ignore road friction and air
resistance. Be sure to express vector quantities in vector form.
12. What is the velocity after 100s from the start??
Solution: Since the acceleration in opposite direction was applied on the car from 65s
after the start
ofrthe motion at which time the car was running at a constant
r
velocity, v  75 i m s . Using kinematic equation, we obtain
r
r
2
2
What is the acceleration? a  1i m s 2  ax  1 m s ; a y  0 m s
vx ,t 100  vx ,t 65  ax t  75   1  100  65  75   1   35  40  m s 
v y ,t 100  v y ,t 65  a y t  0   0  100  65  0   0   35  0  m s 
r
r
r
r
r
r
vt 100  vx ,t 100 i  v y ,t 100 j  40i 0 j  40i  m s 
Wednesday, Mar. 4, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
6
Problems 18 – 21]
• A projectile is launched with a speed of 100 m/s at an angle of 30 o with
respect to the horizontal axis. The magnitude of the gravitational
acceleration is 9.8m/s2. Ignore the air resistance.
18.What is the projectile’s maximum altitude?
Solution: From the information given in the problem, we obtain
r
o
Initial speed v0  100 m s
Launch Angle   30 ;
r
o
X-component of the initial velocity vx 0  v0 cos30  87 m s
r
o
y-component of the initial velocity vy 0  v0 sin 30  50 m s
Maximum altitude is determined only by the y component of the displacement!
2
2
v

v
y0
2  g  y  vyf2  vy20 Solve for y
2ax  v2f  vi2 In y
y  yf
2 g
0  502
Since at the maximum altitude is y
ymax 
 128m
component of the velocity is 0
2   9.8 
Wednesday, Mar. 4, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
7
Problems 18 – 21]
• A projectile is launched with a speed of 100 m/s at an angle of 30 o with
respect to the horizontal axis. The magnitude of the gravitational
acceleration is 9.8m/s2. Ignore the air resistance.
r
vx 0  v0 cos30o  87 m s
r
vy 0  v0 sin 30o  50 m s
19.What is the distance of the projectile from the starting point when it lands?
Solution: Since the time the projectile takes to reach back to land is twice the
time it takes to reach the maximum altitude.
Which component determines the flight time? y-component
v yf  v y 0  gt Solve for t
t
vyf  vy 0
g
; Time to reach
max altitude
Flight time is twice the time to reach the max height!!
ttop 
vyf  vy 0
g

0  50
 5.1s;
9.8
t flight  2  ttop  10.2s
The range is then determined only by the x component of the displacement!
x  x f  x0  vx 0  t flight  87 10.2  887  m
Wednesday, Mar. 4, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
8
Problems 18 – 21]
• A projectile is launched with a speed of 100 m/s at an angle of 30 o with
respect to the horizontal axis. The magnitude of the gravitational
acceleration is 9.8m/s2. Ignore the air resistance.
20.What is the projectile’s speed when it reaches the maximum altitude?
Solution: Since when the projectile reaches the maximum height the y component of
the velocity becomes 0, the speed at the maximum altitude is the same is the
magnitude of the x component of the initial velocity.
r
Initial speed
Launch Angle
  30o ;
v0  100 m s
X-component of the initial velocity
y-component of the initial velocity
x-component of
the final velocity
So the final
speed is
Wednesday, Mar. 4, 2009
r
vx 0  v0 cos30o  87 m s
r
vy 0  v0 sin 30o  50 m s
vxf  vx 0  87 m s
y-component of
the final velocity
v yf  0
r
Same as the
2
2
2
2
 v f  vxf  v yf  vxf  0  vxf  87 m s x-component
of the final velocity
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
9
Applications of Newton’s Laws
Suppose you are pulling a box on frictionless ice, using a rope.
M
What are the forces being
exerted on the box?
T
Gravitational force: Fg
n= -Fg
Free-body
diagram
n= -Fg
T
Fg=Mg
Normal force: n
T
Fg=Mg
Tension force: T
Total force:
F=Fg+n+T=T
If T is a constant
force, ax, is constant
Wednesday, Mar. 4, 2009
 Fx  T  Ma x
F
y
ax 
  Fg  n  Ma y  0
T
M
ay  0
v xf  vxi  axt  v xi  
T 
t
M 
1 T  2
x

x

v
t


t
x  f i xi
2M 
PHYS 1441-002, Spring 2009 Dr.
10
Jaehoon Yu What happened to the motion in y-direction?
Example for Using Newton’s Laws
A traffic light weighing 125 N hangs from a cable tied to two other cables
fastened to a support. The upper cables make angles of 37.0o and 53.0o with
the horizontal. Find the tension in the three cables.
37o
y
53o
T1
Free-body
Diagram
T2
37o
53o
T3
r r r
r
r
T

T

T

ma
0
F 1 2 3
i 3
x-comp. of
Tix  0
net force Fx  
i 1
y-comp. of
net force Fy 
i 3
T
i 1
iy
0
Wednesday, Mar. 4, 2009
 
x
Newton’s 2nd law
 
 T1 cos 37  T2 cos 53  0 T1 
 
 
T sin 53   0.754  sin 37   1.25T
 T
cos  37 
cos 53o
o
2

0.754T2
T1 sin 37o  T2 sin 53o  mg  0


2
2
 125N
0.754
T2  75.4 N
T2 PHYS
100
N ; T1 2009
1441-002, Spring
Dr.
Jaehoon Yu
11
Example w/o Friction
A crate of mass M is placed on a frictionless inclined plane of angle .
Determine the acceleration of the crate after it is released.
y
r
ur ur
r
F F g  n ma
n
Fx  Ma x  Fgx  Mg sin 
y
n
x


Fg
Free-body
Diagram
Supposed the crate was released at the
top of the incline, and the length of the
incline is d. How long does it take for the
crate to reach the bottom and what is its
speed at the bottom?
ax  g sin 
x
F= -Mg F  Ma  n  F  n  mg cos   0
y
y
gy
1 2 1
v
t

a x t  g sin  t 2  t 
d  ix
2
2
v xf  vix  axt g sin 
2d
g sin 
2d
 2dg sin 
g sin 
 vxf  2dg sin 
Wednesday, Mar. 4, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
12