Transcript Part 2a: Newton and His Laws

Part 2a: Newton and His Laws
(1642-1727)
This topic will be taught largely by example problems.
Initially we must define the concept of “force”:
A force is that which causes an object to accelerate.
NEWTON I: (Newton’s First Law of Motion)
In the absence of external forces, an object at rest remains at rest
and an object in motion continues in motion at constant velocity (in
a straight line).
NEWTON II:
The acceleration of an object is directly proportional to the net force
acting on it and inversely proportional to its mass (F=ma).
NEWTON III:
If two objects interact, the force F12 exerted by object 1 on object 2
is equal in magnitude and opposite in direction to the force F21
exerted by object 2 on object 1 (F12 = - F21)
Applications of Newton’s Laws
EXAMPLE 1
An American traffic-light weighing 125 N hangs from a cable tied to two other
cables fastened to a support. The upper cables make angles of 37o and 53o
with the horizontal. Find the tension in the three cables.
T1
37o
T1
53o
T2
T2
y
T3
37o
T3
53o
x
T3
Fg
Solution: We need to consider the forces
acting on the two bodies involved: the traffic
light and the intersection of the cables (knot).
T3 = Fg since the light is held up by the tension
in the rope above it.
Equilibrium !!!
Resolving we get:
SFx=0= -T1cos37 + T2cos53
SFy=0=T1sin37 + T2sin53 – T3
Solving yields T1=75.1N, T2=99.9N
and T3=125N.
Crate on a frictionless incline
A crate of mass m is placed on a frictionless inclined plane of angle q.
Determine (a) the acceleration of the crate after it is released from rest
and (b) how long does it take the front edge of the crate to travel a distance
d down the plane?
y
a
n
mg sin θ
d
θ
θ
mg cos θ
mg
Solution: It is convenient to choose the surface of the plane as the x-axis.
The forces acting on the crate can be seen in the figure:
SFx  mg sin q  max
SFy  n  mg cos q  ma y
Thus we can see that the acceleration ax is independent of the mass:
ax  g sin q
We can take this further by:
d  vxi t  12 axt 2
The crate starts from rest so:
d  12 a x t 2
2d
t

ax
Finally we can find the final velocity:
vxf  vxi  2a x d
2
2
vxf  2a x d  2dg sin q
2
2d
g sin q
vxf  2dg sin q
Friction: Static and Dynamic
When a body is moving either on a surface or in a viscous medium
(eg. a liquid), there is resistance to its motion due to the interaction
of the body with its surroundings. This is the force of friction.
If a book is at rest on a desk and we apply a force, F, then if the book
does not move, then another force of equal magnitude must be acting
against the applied force. This is called the force of static friction, fs.
f, frictional force
If the force is gradually increased, eventually
the book will move and then the retarding force
against the motion is called the force of kinetic
friction, fk. If the applied force is then removed,
then fk acts on the book to decelerate it; if the
applied force is increased above fk, then the book
will accelerate. Experimentally, these frictional
forces are approximately proportional to the normal
force:
fs  sn
F, applied force
Static
Kinetic
fk  k n
Experimental determination of coefficients of friction
We gradually increase the angle of incline
until the block moves:
STATIC CASE (just before the point where the block moves):
n
 Fx  mg sin q  f s  ma x  0
 Fy  n  mg cosq  may  0
y
x
f
mg sin θ
θ
mg cos θ
From (2),
(1)
(2)
n
so, substituting into (1) gives:
cosq
n sin q
(3)
f s  mg sin q 
 n tan q
cosq
mg 
When the incline is at the critical angle θc,
mg
θ
f s  f s max  s n  n tan qc
  s  tan q c
KINETIC CASE: When the block is accelerating, it is possible to reduce θc’ (< θc) so
that it moves down the incline at constant velocity. In this case:
 k  tan q c '
Coefficients of Friction
Steel on steel
Aluminium on steel
Rubber on concrete
Wood on wood
Glass on glass
Waxed wood on wet snow
Ice on ice
Teflon on teflon
Synovial joints in humans
s
k
0.74
0.61
1.00
0.25-0.50
0.94
0.14
0.10
0.04
0.01
0.57
0.47
0.80
0.20
0.40
0.10
0.03
0.04
0.003