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x(t  0)  x0
Newton’s Laws of Motion,
Friction and Inclined Planes
1. With no external force applied, a
body will move with a constant
dv
m

velocity v
dt
2. mass x acceleration = sum of
external forces*
3. If body A is in contact with body B
and exerts a force upon it, the force
upon A due to B is equal in
magnitude and opposite in direction.
Tension
y
x
g  9.81ms 2
T
a

Normal
R
contact force
m
Friction

F

Weight
mg
Consider a block being pulled
uphill via force T. The vector sum
of forces results in an acceleration a
directly up the hill. Surface contact is
maintained at all times.
Displacement
A body is said to be in
equilibrium if the vector sum
of external forces is zero
f
i
i
x(t )
The force due to gravity upon a
body of mass m is its weight
and has magnitude mg, where g
is the local gravitational field
strength.
dv d 2 x
a

dt dt 2
Velocity
Acceleration
y : 0  R  mg cos 
ma  T cos   mg sin   F
0  R  T sin   mg cos 
m 
mg sin 
mg cos 
Sliding friction model (friction always resists motion)
x:
y:
mg sin 
R
 R  mg cos 
x : 0  mg sin   m R  m 
Applying Newton Second Law in x and y directions
F  mR
F  mR
Displacement
Velocity and
Acceleration are
vector quantities
Example 1: Block is on the point of sliding, T
= 0. Note F will point in the opposite direction
as the block will slide downhill if  is increased.
Rough inclined plane
with coefficient of friction m
between block of mass m and
plane.
x:
y:
dx
v
dt
 m  tan 
y
In equilibrium
Sliding
g  9.81ms 2
x
ma  T cos   mg sin   m R
mg cos   T sin   R
R
F
m
0

ma  T cos   mg sin   m  mg cos   T sin  

T
a   cos   m sin    g  sin   m cos  
m
*Actually Newton #2 states the rate of change of momentum = sum of the external forces.
d  mv 
dt
mg
  fi If mass remains constant the LHS is mass x acceleration.
i
Mathematics topic handout: Forces, Friction and Newton’s Laws Dr Andrew French. www.eclecticon.info PAGE 1
Example 2: A block of 10kg is in equilibrium ‘at
the point of sliding’ uphill (this is called limiting
friction). If the plane is inclined at 30o and the
tension is at 45o to the plane, what is T given a
coefficient of friction of m = 1/5 ?
0
x:
T
 10  12 g  15 R
2
0 R
y:
Tension
y
x
g  9.81ms 2
T
0
Normal
R
contact force
 T   5 g  15 R  2
45
o
10
T
3
T
 10 g
 R  5g 3 
2
2
2
Friction
30o
F
30o

T 

 T   5 g  15  5 g 3 
 2
2 


Weight
10g
T  5 2 g  g 6  15 T
6
5

T g 5 2 6
T

5 5 2 6
6

g
Resolving forces
y
T  7.93 g
10
10
8
m
30o
x
45o
3
3
 max 


 ma y 
8
3 

5 3 

 max   10 cos 30  3cos 45 
2 




 

o
o
 ma y  10sin 30  3sin 45  8   3  3 


2 

o
o
The weight
of the puss
Mathematics topic handout: Forces, Friction and Newton’s Laws Dr Andrew French. www.eclecticon.info PAGE 2