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Ballistic Cart Demo
Discuss law of cosines for planeinwindb problem
Other HW problems?
Chapter 4
Forces and Mass
Classical Mechanics
does not apply for
 very tiny objects (< atomic sizes)
 objects moving near the speed of light
Newton’s First Law

If the net force SF exerted on an object is
zero the object continues in its original state of
motion. That is, if SF = 0, an object at rest
remains at rest and an object moving with some
velocity continues with the same velocity.
 Contrast with Aristotle!
Forces



Usually a push or pull
Vector
Either contact or field force
Contact and Field Forces
Fundamental (Field) Forces

Types
 Strong nuclear force
 Electromagnetic force
 Weak nuclear force
 Gravity
Strong Nuclear Force


QCD (Quantum chromodynamics) confines quarks
by exchaning gluons
Nuclear force: binds protons and neutrons
by exchanging pions
Electromagnetic Forces



Opposites attract, like-signs repel
Electric forces bind electrons in atoms
Magnetic forces arise from moving charges
Weak Nuclear Force


Involves exchange of heavy W or Z particle
Responsible for decay of neutrons
Gravity



Attractive force between any two bodies
Proportional to both masses
Inversely proportional to square of distance
F G
m1 m2
r
2
Inertia (Newton’s First Law)

Tendency of an object to continue in its original
motion
Mass



A measure of the resistance of an object to
changes in its motion due to a force
Scalar
SI units are kg
Newton’s Second Law

Acceleration is proportional to net force and
inversely proportional to mass.


 F  ma
Units of Force

SI unit is Newton (N)
F  ma
kg  m
1N 1 2
s

US Customary unit is pound (lb)
 1 N = 0.225 lb
Weight
Weight is magnitude of gravitational force
mass
weight
w  mg
M earth m
wG
2
r
GM earth
g
2
Rearth
Weight vs. Mass


Mass is an inherent property
Weight depends on location
Newton’s Third Law


F12   F21
Force on “1” due to “2”


Single isolated force cannot exist
For every action there is an equal and opposite
reaction
Newton’s Third Law cont.


F12 is action force F21
is reaction force
 You can switch
action <-> reaction
Action & reaction
forces act on
different objects
Action-Reaction Pairs


n  n
'
Fg   Fg
Define the OBJECT (free body)



Newton’s Law uses the
forces acting ON object
n and Fg act on object
n’ and Fg’ act on other
objects
Assumptions for F=ma


Objects behave as particles
 ignore rotational motion (for now)
Consider only forces acting ON object
 neglect reaction forces
Problem Solving Strategy
Identify object (free body)
 Label all forces acting on object
 Resolve forces into x- and y-components, using
convenient coordinate system
 Apply equations, keep track of signs!
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Mechanical Forces



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Strings, ropes and Pulleys
Gravity
Normal forces
Friction
Springs (later in the book)
Some Rules for Ropes and Pulleys



Force from rope points AWAY from object
Magnitude of the force is called tension
Tension does not change when going over a
pulley (if frictionless)
Equilibrium
F  0
Cable Pull Demo
Example 4.1
Given that Mlight = 25 kg, find all three tensions
T3 = 245.3 N, T1 = 147.4 N, T2 = 195.7 N
Example 4.2
2) Which statements are correct?
Assume the objects are static.
A)
B)
C)
D)
T1 must = T2
T2 must = T3
T1 must be < Mg
T1+T2 must be > Mg
A) T
B) T
C) T
D) T
cos(10o)=0.985
sin(10o)=0.173
Example 4.3
a) Find acceleration
b) Find T, the tension above
the bowling ball
c) Find T3, the tension in the
rope between the pails
d) Find force ceiling must exert
on pulley
a) a = g/6 = 1.635 m/s2
b) T = 57.2 N
c) T3=24.5 N
d) Fpulley=2T = 114.5 N
Inclined Planes


Choose x along the
incline and y
perpendicular to incline
Replace force of gravity
with its components
Fg , x  mg sin 
Fg , y  mg cos 
Example 4.4
Find the acceleration and the tension
a = 4.43 m/s2, T= 53.7 N
Example 4.5
M
Find M such that the box slides at constant v
M=15.6 kg
Forces of Friction


Resistive force between object and neighbors or
the medium
Examples:
 Sliding a box
 Air resistance
 Rolling resistance
Sliding Friction
f  s N
f  k N
s  k



Direction parallel to
surface, opposite to
other forces
Nearly independent of
the area of contact
The coefficient of friction (µ) depends on the
surfaces in contact
Coefficients
of Friction
f  s N
f  k N
s  k
Static Friction, ƒs


If F   s n, f s   F


s is coefficient of
static friction
n is the normal force
f
F
Kinetic
Friction, ƒk
If F   s n,
f  k n
k is coefficient of
kinetic friction
 Friction force opposes F
 n is the normal force

f
F
Friction Demo
Example 4.6
The man pushes/pulls with a force of 200 N. The
child and sled combo has a mass of 30 kg and the
coefficient of kinetic friction is 0.15. For each case:
What is the frictional force opposing his efforts?
What is the acceleration of the child?
f=59 N, a=3.80 m/s2
/
f=29.1 N, a=4.8 m/s2
Example 4.7
Given m1 = 10 kg and m2 = 5 kg:
a) What value of s would stop the block from sliding?
b) If the box is sliding and k = 0.2, what is the
acceleration?
c) What is the tension of the rope?
a) s = 0.5
b) a=1.96 m/s2
c) 39.25 N
Example 4.8
What is the minimum s required to
prevent a sled from slipping down a
hill of slope 30 degrees?
s = 0.577
Other kinds of friction
 Air resistance, F ~ Area  v2
 Rolling resistance, F ~ v
Terminal velocity:
Fresistance  CAv
 mg at termina l velocity
2
Coffee Filter Demo
Accelerating Reference Frames

Equivalent to “Fictitious” gravitational force
g fictitious  a frame
Fictitious Force: Derivation
1 2
x  v0t  at
2
1 F 2 Eq. of motion in fixed frame
 v0t 
t
2m
1 2
x0 (t )  a f t
2
1 ( F  ma f ) 2
x  x0 (t )  v0t 
t
2
m
F-maf looks like force in new frame,
maf acts like fake gravitational force!
Example 4.9
An elevator falls with acceleration a = 8.0 m/s2.
If a 200-lb person stood on a bathroom scale
during the fall, what would the scale read?
36.9 lbs
Example 4.10
You are calibrating an accelerometer so that you
can measure the steady horizontal acceleration of a
car by measuring the angle a ball swings backwards.
If M = 2.5 kg and the acceleration, a = 3.0 m/s2:
a) At what angle does the ball swing backwards?
b) What is the tension in the string?

 = 17 deg
T= 25.6 N