Warmup
m
m
 x  2
s
s
m
 x  m
s
m
 x  s
s
m
m
x



2
s
s
m
x

m


2
s
m
x

s


2
s
Force
Affects on Velocity and Acceleration
Force Causes Change in Velocity
Oh!
Skeeter!
Ha,
ha,
good
one
This one time
at band
camp…
Force and Motion
Newton’s First Law





At rest – Stays at rest (until force is applied)
In motion – Stays in motion (until force is applied)
Force causes change in velocity
Force causes acceleration
Force causes a change in direction
Warm-up

A blue one with a nubbin was
moving at 5m/s straight down
when the problem started. The
difference between the bottom
and the top is 3 times the
height of a trans-atlantic 10m
building. This nubbin sporting
thing is earthbound. (include
units)
di 
df 
r
a
r
vi 
r
vf 
Types of Fundamental Force

Gravitational Force


Electromagnetic Force



Force we use in this section
Includes the contact forces we work with in this section
Nuclear Force
Weak Force
(summarize)
Electromagnetic Force

Contact Forces
Normal
 Friction



Static: friction when the object is not in motion
Sliding: friction when the object is in motion
Tension
 Spring

(Use article to organize with topic oval)
Force

Newton’s Third Law

Every action has an opposite
and equal reaction.
ur
ur
F A _ on _ B  F B _ on _ A
Force

Experiment with force sensors.

equal and opposite
Key Vocab

Net force (Fnet)
Vector addition/subtraction
5N
+
7N


The resultant is the net force
5N
+
7N
=
12N
kg gm
N 2
s
Newton’s Second Law

Two men pull a 50-kg box with forces 19.7 N
and 15.6 N in the directions shown below. Find
the net force of the box.
19.7 N
or
-19.7N
15.4 N
kg gm
N 2
s
Force

The pound-force or simply pound
(abbreviations: lb, lbf, or lbf) is a unit of force


1N=0.225lb; 1lb=4.45N
Normal plus lift (Fnet)


Weight from force gauge.
Upward force must be greater than gravity to have
upward acceleration.
Demo
HW:
16,17, (pg 97),
Example problem 2 pg 99 with elevator 3 times with a = 2.50, 3.00, 15.0m/s^2 and t=1.50, 3.25, 3.00s
19, 20, 22, 25 (pg 100 & 101)
Force

Newton’s Second Law………...........

Weight is a Force………………...........
1 lbf ≈ 4.448222 N

Defined with the universal constant ‘G’.
G  6.7 x10
11
N  m / kg
2
2
ur
r F net
a
m
ur
ur
F g  mg
ur
Gm1m2
Fg 
2
r
Force and Gravity
ur
GMm
Fg  2
r
G  6.7 x10
11
N  m / kg
2
2
ur
ur
F g  mg
GMm
mg  2
r
GM
g 2
r
kg gm
N 2
s
Key Vocab

Normal Force


Force due to gravity and mass is referred to as a
normal force or ‘the normal vector’.
Normal vector also refers to a vector that
intersects at 90 degrees.
(also stated as: A vector that is
perpendicular to the tangent
line at the interface)
Newton’s Second Law

Two men pull a 50-kg box with forces 19.7 N
and 15.6 N in the directions shown below. Find
the resultant acceleration of the box and the
direction in which the box moves.
19.7 N
or
-19.7N
15.4 N
kg gm
N 2
s
Practice 


A large helicopter is used to lift a heat pump to the
roof of a new building. The mass of the helicopter is
7.0x10^3 kg and the mass of the heat pump is 1700
kg.
a. How much force must the air exert on the
helicopter to lift the heat pump with an acceleration
of 1.2 m/s^2?
b. Two chains connected to the load each can
withstand 95,000 N. Can the load be safely lifted at
1.2 m/s^2?
Air


Drag Force: exerted by a fluid on an object
moving through the fluid.
Terminal Velocity: drag force is equal to force
of gravity
Friction


Static friction Ff,s
Friction
Coefficients:
Table 5.1 pg
129

Friction when there is no motion between the
objects.

Ff,s <= usFN
Sliding friction (or kinetic friction) Ff,k

Friction when surfaces are rubbing against each
other (in motion).

Ff,k = ukFN
Friction
Normal





A wood block on a wood
plank.
m= kg
us = 0.5
uk = 0.2
FN = N
Friction
Review Elevator Example



Fnet = FE + (-Fg)
Fnet = ma
(represents the force on the system as a whole)
FE = Fnet + Fg
Tension



Force exerted by pulling (usually a string or
rope).
For now, consider strings, ropes and pulleys
(also called sheaves or blocks) to be massless
and frictionless.
Provide a change in direction of force.
Tension Problem

The blocks shown
are placed on a
smooth horizontal
surface and
connected by a
piece of string. If
a 8.8-N force is
applied to the 8.8kg block, what is
the tension in the
string?
r
r
Fnet  mt a
mt  19.4kg  8.8kg  28.2kg
r
r Fnet
8.8 N
m
a

 0.312 2
mt
28.2kg
s
r
r
r
m
FT  Fa ,19.4  m19.4 a  19.4kg *0.312 2  6.1N
s
6.1N
Tension

Three blocks A, B,
and C are connected
by two massless
strings passing over
smooth pulleys as
shown below, with
the 3.4-kg block on a
smooth horizontal
surface. Calculate
the tension in the
strings connecting A
and B, and B and C.
54N
48N
Tension
r
r
Fnet  mt a
r
r
r
m
m
Fnet   FA  FB    mA g    mB g   (6.9kg *9.8 2 )  (4.1kg *9.8 2 )  68 N  40 N  28 N
s
s
mt  mA  mB  mC  6.9kg  3.4kg  4.1kg  14.4kg
r
r
r r F
28 N
m
Fnet  mt a  a  net 
 1.9 2
mt 14.4kg
s
r
r
Fa , BC  FB  FT , AB
r
r
r
FT , AB  FB  Fa , BC
r
r
m
Fa , BC  (mB  mC )a  (3.4  4.1)(1.9 2 )  14 N
s
r
FT , AB  40 N  (14 N )  54 N
Tension Practice
r
r
r
FT , BC  FA  Fa , AB
r
a
r
FB 
r
Fa , AB 
Sketch Problem
68N
68N
A
B
ABC
C
40N
40N
Sketch Problem
68N
68N
68N
A
AB
A
B
C
40N
C
BC
40N
40N
Fnet


Fnet = sum of all forces acting on a system = ma
Fnet = Ff,k + FT + Fpush + Fgravity = ma
Force

Newton’s Third Law

Every action has an opposite
and equal reaction.
ur
ur
F A _ on _ B  F B _ on _ A
Force

Newton’s Second Law………...........
ur
r F net
a
m
Force

Newton’s first law


When the net forces are zero, an object at rest remains
at rest and an object in motion remains in motion in the
same direction at the same speed.
For Fnet = 0


velocity is constant: v1 = v2
acceleration is zero
Friction


Static friction Ff,s
Friction
Coefficients:
Table 5.1 pg
129

Friction when there is no motion between the
objects.

Ff,s <= usFN
Sliding friction (or kinetic friction) Ff,k

Friction when surfaces are rubbing against each
other (in motion).

Ff,k = ukFN
Friction

Static friction Ff,s

Friction when there is no motion between the
objects.

Ff,s <= usFN

Ff,s <= usmg
5kg
10N
Review Elevator Example



Fnet = FE + (-Fg)
Fnet = ma
(represents the force on the system as a whole)
FE = Fnet + Fg
Setting up the Problem

Connected by a massless string, pulled along a
surface with a coefficient of friction k  0.75
100N
k  0.75
Steps

What forces are present?

What is the value of each force?
r
r
Fnet  mt a  100 N  k m3 g  k m2 g  k m1g
r
r
Fnet  mt a  100N  44N  29N 15N  12N

Draw the free body diagram
100N
Steps

List known variables, solve for unknown
r
r
Fnet  mt a  12 N
mt  2kg  4kg  6kg  12kg
r
r Fnet 12 N
m
a

1 2
mt 12kg
s
Break apart

Find the tension in each string

First, identify the forces of each part
r
for m2, Ff ,k ,m 2  29 N
r
for m1, Ff ,k ,m1  15 N
r
for m3, Ff ,k ,m3  44 N
100N
Consider each section as a system

Draw a free body diagram for this system
r
for m1, Ff ,k ,m1  15 N
r
for m2, Ff ,k ,m 2  29 N
r
r
Fnet  mt a 
Consider each section as a system

Draw a free body diagram for this system
r
for m1, Ff ,k ,m1  15 N
r
r
Fnet  mt a
Break apart

Draw a sketch (free body diagram) for each
string
Trig Identities




SOHCAHTOA
Adjacent Leg = Hypotenuse * cos
Opposite Leg = Hypotenuse * sin
Opposite Leg = Adjacent Leg * tan

Force components

The magnitude of the force 1 is 87 N, of
force 2 is 87 N, and of 3 is 87 N. The angles
θ 1 and θ 2 are 60° each. Use the
Pythagorean theorem and trig identities to
find the resultant of the forces 1, 2 , and 3 .
Drag Force


Terminal velocity is when an object in free fall
has reached equilibrium. That is, the drag
force is equal to the force of gravity.
The Fnet for a system in equilibrium is always
zero.
Friction with Force components

The system shown below is in equilibrium.
Calculate the force of friction acting on the
block A. The mass of block A is 7.10 kg and
that of block B is 7.30 kg. The angle θ is
48.0°.
Break Apart


The system shown below is in equilibrium: This
means Fnet=0, since Fnet=ma, then ma=0 and a=0
In order for the net force to be zero on block A (no
acceleration) the tension in the rope pulling block A
has to be matched by the friction between block A
and the surface.
The friction
here must be
equal to
the tension
pulling A
The tension
here must be
equal to
the friction
resisting
motion
Break Apart



There is a force from gravity acting on block
B.
Convert this force to its x and y components.
Using the concept of equilibrium, realize that
the sum of x components is 0
θ is 48.0°.
7.30kg
Two wire tension
Familiar Problem


Three blocks A, B, and C are connected by two
massless strings passing over smooth pulleys as
shown below, with the B block on a smooth
horizontal surface.
A=2kg, B=3kg, C=5kg
Redraw the problem
Force of gravity on A
Force of gravity on C
A

B
C
The problem may be redrawn to help visualize
which direction the forces are acting and
which direction is positive.
Breaking the problem down








Find the total mass of the system ABC.
Find the net force of the system ABC.
Find the acceleration of the system ABC.
Find the acceleration of the system C.
Find the total mass of the system C.
Calculate the force due to the acceleration of system C.
Calculate the force due to gravity on system C.
Calculate the tension in the strings connecting B and C.
Spring (restoring force)



F=-kx
k=constant
x=displacement
Force

Experiment with force sensors.



equal and opposite
normal plus lift (Fnet)
friction (static and sliding)