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Unit is the NEWTON(N)
Is by definition a push or a pull
Can exist during physical
contact(Tension, Friction, Applied Force)
Can exist with NO physical contact,
called FIELD FORCES ( gravitational,
electric, etc)
Types of Forces
•
•
•
•
Weight(W or Fg) – Always drawn from the
center, straight down
Force Normal(FN) – A surface force always
drawn perpendicular to a surface.
Tension(T or FT) – force in ropes and always
drawn AWAY from object.
Friction(Ff)- Always drawn opposing the
motion.
INERTIA – a quantity of matter, also called MASS.
Italian for “LAZY”. Unit for MASS = kilogram.
Weight or Force due to Gravity is how your MASS
is effected by gravity.
W  mg
NOTE: MASS and WEIGHT are NOT the same thing. MASS never changes
When an object moves to a different planet.
What is the weight of an 85.3-kg person on earth? On Mars=3.2 m/s/s)?
W  mg  W  (85.3)(9.8)  835.94 N
WMARS  (85.3)(3.2)  272.96 N
An object in motion remains in motion in a
straight line and at a constant speed OR an
object at rest remains at rest, UNLESS acted
upon by an EXTERNAL (unbalanced) Force.
There are TWO conditions here and one constraint.
Condition #1 – The object CAN move but must be at a CONSTANT SPEED
Condition #2 – The object is at REST
Constraint – As long as the forces are BALANCED!!!!! And if all the forces
are balanced the SUM of all the forces is ZERO.
The bottom line: There is NO ACCELERATION in this case AND the object
must be at EQILIBRIUM ( All the forces cancel out).
acc  0   F  0
If an object is NOT at rest or moving at a
constant speed, that means the FORCES are
UNBALANCED. One force(s) in a certain
direction over power the others.
THE OBJECT WILL THEN ACCELERATE.
The acceleration of an object is directly
proportional to the NET FORCE and
inversely proportional to the mass.
a  FNET
a
1
a
m
FNET
 FNET  ma
m
FNET   F
Tips:
•Draw an FBD
•Resolve vectors into components
•Write equations of motion by adding and
subtracting vectors to find the NET FORCE.
Always write larger force – smaller force.
•Solve for any unknowns
A 10-kg box is being pulled across the table to
the right by a rope with an applied force of
50N. Calculate the acceleration of the box if a
12 N frictional force acts upon it.
FN
Ff
mg
Fa
In which direction,
is this object
accelerating?
The X direction!
So N.S.L. is worked
out using the forces
in the “x” direction
only
FNet  ma
Fa  F f  ma
50  12  10a
a  3.8 m / s
2
Newton’s Third Law
“For every action there is an EQUAL and
OPPOSITE reaction.


This law focuses on action/reaction pairs (forces)
They NEVER cancel out
All you do is SWITCH the wording!
•PERSON on WALL
•WALL on PERSON
N.T.L
This figure shows the force during a
collision between a truck and a train. You
can clearly see the forces are EQUAL
and OPPOSITE. To help you understand
the law better, look at this situation from
the point of view of Newton’s Second
Law.
FTruck  FTrain
mTruck ATruck  M TrainaTrain
There is a balance between the mass and acceleration. One object usually
has a LARGE MASS and a SMALL ACCELERATION, while the other has a
SMALL MASS (comparatively) and a LARGE ACCELERATION.
N.T.L Examples
Action: HAMMER HITS NAIL
Reaction: NAIL HITS HAMMER
Action: Earth pulls on YOU
Reaction: YOU pull on the earth
A pictorial representation of forces complete
with labels.
FN
T
Ff
T
W1,Fg1
or m1g
m2g
•Weight(mg) – Always
drawn from the center,
straight down
•Force Normal(FN) – A
surface force always drawn
perpendicular to a surface.
•Tension(T or FT) – force in
ropes and always drawn
AWAY from object.
•Friction(Ff)- Always drawn
opposing the motion.
Ff
FN
mg
Since the Fnet = 0, a system moving at a
constant speed or at rest MUST be at
EQUILIBRIUM.
TIPS for solving problems
• Draw a FBD
• Resolve anything into COMPONENTS
• Write equations of equilibrium
• Solve for unknowns
A 10-kg box is being pulled across the table to the
right at a constant speed with a force of 50N.
a)
b)
Calculate the Force of Friction
Calculate the Force Normal
FN
Ff
mg
Fa
Fa  F f  50 N
mg  Fn  (10)(9.8)  98N
Suppose the same box is now pulled at an angle of 30
degrees above the horizontal.
a)
Calculate the Force of Friction
b)
Calculate the Force Normal
Fax  Fa cos   50 cos 30  43.3N
F f  Fax  43.3N
FN
Ff
Fa
Fay
30
Fax
mg
FN  mg!
FN  Fay  mg
FN  mg  Fay  (10)(9.8)  50 sin 30
FN  73N
A mass, m1 = 3.00kg, is resting on a frictionless horizontal table is connected
to a cable that passes over a pulley and then is fastened to a hanging mass,
m2 = 11.0 kg as shown below. Find the acceleration of each mass and the
tension in the cable.
FNet  ma
FN
m2 g  T  m2 a
T  m1a
T
T
m1g
m2 g  m1a  m2 a
m2 g  m2 a  m1a
m2 g  a (m2  m1 )
m2g
a
m2 g
(11)(9.8)

 7.7 m / s 2
m1  m2
14
FNet  ma
m2 g  T  m2 a
T  m1a
FNET
FNet  ma 
m
a
Rise
Slope 
Run
T  (3)(7.7)  23.1 N
TWO types of Friction


Static – Friction that keeps an object at rest
and prevents it from moving
Kinetic – Friction that acts during motion
Force of Friction

The Force of Friction is F f  FN
  constant of proportion ality
directly related to the
Force Normal.
  coefficien t of friction
 Mostly due to the fact
Fsf   s FN
The coefficient of
that BOTH are surface
forces
Fkf   k FN
friction is a unitless
constant that is
specific to the
material type and
usually less than
one.
Note: Friction ONLY depends on the MATERIALS sliding against
each other, NOT on surface area.
Example
A 1500 N crate is being pushed
a) What is the coefficient of kinetic
across a level floor at a
friction between the crate and the
constant speed by a force F of
floor?
600 N at an angle of 20° below
the horizontal as shown in the
figure.
F f   k FN
Fa
FN
Fay
F f  Fax  Fa cos   600(cos 20)  563.82 N
FN  Fay  mg  Fa sin   1500
20
FN  600(sin 20)  1500  1705.21N
Fax
563.82   k 1705.21
 k  0.331
Ff
mg
Example
FN
If the 600 N force is instead pulling the
block at an angle of 20° above the
horizontal as shown in the figure,
what will be the acceleration of the
crate. Assume that the coefficient of
friction is the same as found in (a)
FNet  ma
20
Fax
Ff
Fax  F f  ma
Fa cos   FN  ma
Fa cos    (mg  Fa sin  )  ma
600 cos 20  0.331(1500  600 sin 20)  153.1a
563.8  428.57  153.1a
a  0.883 m / s 2
Fa
mg
Fay
Inclines

Ff
FN
mg cos 


mg 
mg sin 


Tips
•Rotate Axis
•Break weight into components
•Write equations of motion or
equilibrium
•Solve
Example
Masses m1 = 4.00 kg and m2 = 9.00 kg are connected by a light string that passes over
a frictionless pulley. As shown in the diagram, m1 is held at rest on the floor and m2 rests
on a fixed incline of angle 40 degrees. The masses are released from rest, and m2 slides
1.00 m down the incline in 4 seconds. Determine (a) The acceleration of each mass (b)
The coefficient of kinetic friction and (c) the tension in the string.
T
FN
Ff
m2gcos40
m2g
m1
40
m2gsin40
m1g
T  m1 g  m1a  T  m1a  m1 g
m2 g sin   ( F f  T )  m2 a
40
T
FNET  ma
Example
FNET  ma
T  m1 g  m1a  T  m1a  m1 g
m2 g sin   ( F f  T )  m2 a
x  voxt  1 at 2
2
1  0  1 a ( 4) 2
2
a  0.125 m / s 2
T  4(.125)  4(9.8)  39.7 N
m2 g sin   F f  T  m2 a
m2 g sin   F f  (m1a  m1 g )  m2 a
m2 g sin    k FN  m1a  m1 g  m2 a
m2 g sin    k m2 g cos   m1a  m1 g  m2 a
m2 g sin   m1a  m1 g  m2 a   k m2 g cos 
k 
m2 g sin   m1a  m1 g  m2 a
m2 g cos 
k 
56.7  0.5  39.2  1.125
 0.235
67.57