physics 207: lecture 8, pg 1 lecture 8 l goals: solve 1d & 2d motion with friction utilize...
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Physics 207: Lecture 8, Pg 1
Lecture 8
Goals:Goals:
Solve 1D & 2D motion with friction
Utilize Newton’s 2nd Law
Differentiate between Newton’s 1st, 2nd and 3rd Laws
Begin to use Newton’s 3rd Law in problem solving
Physics 207: Lecture 8, Pg 2
Friction with no acceleration
No net force So frictional force just cancels applied force
FFAPPLIED
ffFRICTION mgg
NN
ii
j j
Physics 207: Lecture 8, Pg 3
Friction...
Friction is caused by the “microscopic” interactions between the two surfaces:
Physics 207: Lecture 8, Pg 4
Friction: Static frictionStatic equilibrium: A block with a horizontal force F applied,
As F increases so does fs
Fm
FBD
fs
N
mg
Physics 207: Lecture 8, Pg 5
Friction: Static frictionStatic equilibrium: A block with a horizontal force F applied,
Fm
FBD
fs
N
mg
Fx = 0 = -F + fs
fs = F
Fy = 0 = - N + mg
N = mg
Force applied
Fri
ctio
nF
orce
Physics 207: Lecture 8, Pg 6
Static friction, at maximum (just before slipping)
fS is proportional to the magnitude of N
fs = s N
Fm fs
N
mg
Force applied
Fri
ctio
nF
orce
Physics 207: Lecture 8, Pg 7
Model of Static Friction
Magnitude:
f is proportional to the applied forces such that
fs ≤ s N s called the “coefficient of static friction”
Direction: If just a single “applied” force, friction is in opposite direction
Physics 207: Lecture 8, Pg 8
Kinetic (fk < fs)
Dynamic equilibrium, moving but acceleration is still zero
F
m
FBD
fk
N
mg
Fx = 0 = -F + fk fk = F
Fy = 0 = - N + mg N = mg v
fk = k N
Physics 207: Lecture 8, Pg 9
Model of Sliding Friction
Direction: to the normal force vector N N and
opposite to the velocity.
Magnitude: ffk is proportional to the magnitude of N N
ffk = k NN
The constant k is called the “coefficient of kinetic friction”
Logic dictates that S > K for any system
Physics 207: Lecture 8, Pg 10
Coefficients of Friction
Material on Material s = static friction k = kinetic friction
steel / steel 0.6 0.4
add grease to steel 0.1 0.05
metal / ice 0.022 0.02
brake lining / iron 0.4 0.3
tire / dry pavement 0.9 0.8
tire / wet pavement 0.8 0.7
Physics 207: Lecture 8, Pg 11
Sliding friction (fk < fs) but now |a| > 0
A change in velocity
As F increases fk remains nearly constant
(but now there is acceleration)
Fm
FBD
fk
N
mg
Fx = -F + fk = net Force
Fy = 0 = - N + mg N = mg v
fk = k N
Physics 207: Lecture 8, Pg 12
Acceleration, Inertia and Mass The tendency of an object to resist any attempt to
change its velocity is called Inertia Mass is that property of an object that specifies how
much resistance an object exhibits to changes in its velocity (acceleration)
If mass is constant then
If force constant
Mass is an inherent property of an object Mass is independent of the method used to measure it Mass is a scalar quantity The SI unit of mass is kg
netFa
ma 1||
|a|
m
Physics 207: Lecture 8, Pg 14
ExerciseNewton’s 2nd Law
A. increasingB. decreasingC. constant in timeD. Not enough information to decide
An object is moving to the right, and experiencing a net force that is directed to the right. The magnitude of the force is decreasing with time (read this text carefully).
The speed of the object is
Physics 207: Lecture 8, Pg 15
1st: Frictionless experiment (with a ≠ 0)
Two blocks are connected on the table as shown. The
table is frictionless. Find the acceleration of mass 2.
Requires two FBDsT
Mass 2
Fx = m2ax = -T
Fy = 0 = N – m2g
m1
m2
m2g
N
m1g
T
Mass 1
Fy = m1ay = T – m1g
Notice ay = ax = a
Eliminate T
m1a + m2a = m1
a = m1 / (m2+m1)g
Physics 207: Lecture 8, Pg 16
Experiment with friction (with a ≠ 0)
Two blocks, of m1 & m2 , are connected on the table as shown. The table has unknown static and kinetic friction coefficients.
Given an a, find K.
Similar but now with friction.
T
Mass 2
Fx = m2a = -T + fk = -T + k N
Fy = 0 = N – m2g
m1
m2
m2g
N
m1g
T
fk
Mass 1
Fy = m1a = T – m1g
T = m1g + m1a = k m2g – m2a k = (m1(g+a)+m2a)/m2g
Physics 207: Lecture 8, Pg 18
To repeat, net force acceleration
In physics: A force is an action which causes an object to
accelerate (translational & rotational)
This is Newton’s Second Law
zz
yy
xx
maF
maF
maF
amFF
0net
Physics 207: Lecture 8, Pg 19
Home Exercise Newton’s 2nd Law
A. A
B. B
C. D
D. F
E. G
A mass undergoes motion along a line with velocities as given in the figure below. In regards to the stated letters for each region, in which is the magnitude of the force on the mass at its greatest?
Physics 207: Lecture 8, Pg 20
Remember: Forces are Conditional
Notice what happens if we change the direction of the applied force
The normal force can increase or decrease Here the normal force exceeds mg
Let a=0
F
fF mg
N F sin +mg
ii
j j F sin
Physics 207: Lecture 8, Pg 27
Forces at different angles
Case 1 Case 2
F
mg
N
Case1: Downward angled force with friction
Case 2: Upwards angled force with friction
Cases 3,4: Up against the wall
Questions: Does it slide?
What happens to the normal force?
What happens to the frictional force?
mg
Cases 3, 4
mg
NN
F
F
ff
ff
ff
Physics 207: Lecture 8, Pg 28
Example (non-contact)
Consider the forces on an object undergoing projectile motion
FB,E = - mB g
EARTH
FE,B = mB g
FB,E = - mB g
FE,B = mB g
Physics 207: Lecture 8, Pg 29
Gravity
Newton also recognized that gravity is an attractive, long-range force between any two objects.
When two objects with masses m1 and m2 are separated by distance r, each object “pulls” on the other with a force given by Newton’s law of gravity, as follows:
Physics 207: Lecture 8, Pg 30
Cavendish’s Experiment
F = m1 g = G m1 m2 / r2
g = G m2 / r2
If we know big G, little g and r then will can find m2 the mass of the Earth!!!
Physics 207: Lecture 8, Pg 31
Example (non-contact)
FB,E = - mB g
EARTH
FE,B = mB g
FB,E = - mB g
FE,B = mB g
Compare: g = G m2 / 40002 g’ = G m2 / (4000+40)2
g / g’ = / (4000+40)2 / 40002 = 0.98
Question: By how much does g change at an altitude of 40 miles? (Radius of the Earth ~4000 mi)
Physics 207: Lecture 8, Pg 32
Recap
Assignment: HW4, (Chapter 6 & 7 due 10/4)
For Monday finish Chapter 7