B 105 n a setup similar to the one shown at right is

(b) 105 N
A setup similar to the one
shown at right is often used
in hospitals to support and
apply a traction force to an
injured leg. (a) Determine
the force of tension in the
rope supporting the leg. (b)
What is the traction force
exerted on the leg? Assume
the traction force is
horizontal.

Example Problem #4.30
13 N
An object of mass 2.0 kg starts from rest and slides down an
inclined plane 80 cm long in 0.50 s. What net force is acting on
the object along the incline?
Geometry:


 ° 


Solution (details given in class):


Example Problem #4.36
Find the acceleration of each block and the tension in the
cable for the following frictionless system:
m
1
m
2
+ y
+ x
Coupled system : mass m
2
moves
same distance in same time as
mass m
1
 v
1
= v
2
 a
1
= a
2
Free–body diagrams: m
1
m
2
N
m
1
g
m
2
g
T
T
Apply Newton’s 2
nd
Law to block m
1
 F
x
= m
1
a
x
 T = m
1
a
x
= m
1
a
 F
y
= 0  m
1
g – N = 0  m
1
g = N ( a = 0 in y –direction)
5.00 kg
10.0 kg
= a
:
(1)

Example Problem (continued)
Apply Newton’s 2
nd
Law to block m
2
:
 F
y
= m
2
a
y
= m
2
a
 m
2
g – T = m
2
a (2)
Combining equations (1) and (2):
 m
2
g – m
1
a = m
2
a  a = [ m
2
/ ( m
1
+ m
2
)] g = 6.53 m/s
2
Using equation (1) and plugging in for a :
 T = m
1
a = 32.7 N
(Note that this analysis will be useful for Experiment 5 in lab!)

CQ5: Interactive Example Problem:
Rocket Blasts Off
What is the maximum height reached by the
rocket?
ActivPhysics Problem #2.4, Pearson/Addison Wesley (1995–
A) 0 m
B) 240 m
C) 960 m
D) 2880 m
E) 3840 m
2007)

Friction
•
Friction is a contact force between two surfaces
that always opposes motion
•
The kind of friction that acts when a body slides
over a surface is called the kinetic friction force
(“kinetic” for motion)
–
The magnitude of is proportional to the magnitude of
the normal force N
– Constant

k
= coefficient of kinetic friction (depends on
the two surfaces in contact)
W
N
F
f
is always  to
f

N
k
f

k
f

N f
k k
 
:


Friction
•
When pushing a car, have you ever noticed that it’s
harder to start car moving than to keep it moving?
•
The magnitude of the frictional force varies!
•
There is a static frictional force , with variable
magnitude, that is almost always larger (at it’s
maximum value) than the kinetic frictional force
•
For the case of pushing a piano across the floor:
s
f

W
N
(1) (no pushing)
W
N
(pushing but no sliding)
(2)
F
f
s

Friction
•
f
s
has a variable magnitude:
– Constant

s
is the coefficient of static friction ( 
s
> 
k
)
•
In situation (1) above:
•
In situation (2) above:
•
In situation (3) above:
•
In situation (4) above:
N f
s s
 
N f f
s s
  
N f f
s s
  
N f f
k k
  
0  f
W
N
(just about to slide)
(3)
F
f
s, max
f
s ,max
= f
s
max. value
W
N
(piano sliding)
(4)
F
f
k

Friction
•
Variable magnitude of the
friction force as a function
of pushing force F
summarized in the graph
at right
•
Friction responsible for motion of wheeled vehicles
(Note that f is in the same direction
as the motion of the car, but
opposite to the motion of the tires !)