6: Applications of Newton's Laws
Friction
opposes motion
due to surfaces sticking together
Kinetic Friction: surfaces are moving relative to each other
a.k.a. Sliding Friction
Static Friction: surfaces are not moving relative to each other.
Static Friction prevents stationary objects from moving until sufficient force has been
applied.
Friction
Applied Force
Coefficient of Friction
Frictional forces depend upon
how hard the surfaces are being pressed together
-> force perpendicular to the surface = normal force
the types of surfaces that are in contact
-> coefficient of friction
Kinetic Friction
Static Friction
f k = k N
f s , max = s N
f s ≤ s N
p250c6:1
Typical coefficients of friction
Materials
kinetic, mk
static, mS
Rubber on concrete (dry)
0.80
0.90
Rubber on concrete (wet)
0.25
0.30
Steel on Steel
0.57
0.74
Glass on glass
0.40
0.94
Wood on leather
0.40
0.50
Copper on steel
0.36
0.53
Steel on ice
0.06
0.10
Waxed ski on snow
0.05
0.10
Teflon on Teflon
0.04
0.04
Example: A 50.0 g saltshaker is slid down a counter top with an initial speed of 1.15 m/s whereupon it
comes to rest in a distance of 0.840 m. What is the coefficient of kinetic friction between the shaker and
the table?
p250c6:2
Example: A sea lion slides from rest down an 3.00 m long ramp into a pool of water which is inclined at an
angle of 23°. If the coefficient of kinetic friction is 0.26, how long does the sea lion's ride last?
p250c6:3
Example: A flatbed truck slowly tilts back. At an angle of 23.2° a crate on the flatbed begins to slide. What
is the coefficient of static friction between the bed and the truck?
p250c6:4
From the previous example, what is the force of static friction when the truck bed is level? When it is at
20°?
Thought Question: What type of friction occurs between the tires of a car and the road when operating normally?
p250c6:5
Strings and springs
Tension: the force necessary to hold a string (rope, cable, etc) taut.
(frictionless, low mass) pulleys do not reduce tension.
T
T
both sides of rope contribute to tension on pulley
T
add forces as vectors!
T
T
What are the readings?
1.00 kg
1.00 kg
1.00 kg
1.00 kg
p250c6:6
Force: Hooke’s Law for elastic deformations
stretch or compress
unstretched
F = kx
F x =−k x
half stretched
stretched
x
F=0
F = k x/2
F=kx
Example: A 1.50 kg object is hung from a spring with spring constant k = 250 N/m. How far is the spring
stretched from its equilibrium length?
p250c6:7
Equilibrium
Static Equilibrium: no acceleration with bodies at rest.
Dynamic Equilibrium: no acceleration with moving bodies
Equilibrium implies Fi = 0
so that Fxi = 0 and Fyi = 0
Connected Objects
use action-reaction
objects will have related changes in motion
chose coordinates so that velocity and acceleration are the same
p250c6:8
Example: Find expressions for the acceleration of the system and
the tension in the string. Find values if m1 is 1.50 kg and m2 is
0.750 kg.
T=F 1
m1
1
(+)
(+)
m2 2
T
w 1 -T=F 2
=m1 g-T
w
p250c6:9
T
(+)
(+)
w1
w2 -T=F 2
1
m1 = 3.1kg
2
m2 = 4.4kg
p250c6:10
A curved path requires an “inward” force
“Center seeking” = Centripetal
Centripetal force is the force perpendicular to the
velocity of an object moving along a curved path.
v
v
v
The centripetal force is directed toward the center of
curvature of the path.
v = at
= (F/m)dt
examples: ball on a string, car rounding a corner.
Centrifugal Effect: the “fictitious force” felt by an object when the frame of reference
moves along (and therefore accelerates) along a curved path. This effect is simply
inertia. Stop the force and the object will undergo straight line motion.
p250c6:11
Uniform Circular Motion
motion in a circle at constant speed
centripetal force Fc and centripetal acceleration ac is always directed towards the center
centripetal force and acceleration have constant magnitudes
v2
acp 
r
mv 2
Fcp  macp 
r
p250c6:12
A 1200kg car rounds a corner with radius r = 45m. If the static friction between the tires and the road is
ms = 0.82, what is the greatest speed the car can have in the corner without skidding?
phys
p250c6:13
A roadway is banked at the proper angle so that a car, without any assistance from friction, can round the
corner safely. Find the appropriate banking for a 900 kg car traveling at 20.5 m/s in a turn of radius 85.0
m/s.
ph y
s
p250c6:14
While driving along at constant speed of 17.0 m/s you encounter a dip in the road. The dip can be
approximated as a circular arc, 65.0 m in radius. What is the normal force exerted by a car seat on a 80.0
kg passenger at the bottom of the dip? How does this compare to the weight of the passenger?
p250c6:15