Transcript PreLecture 01

Welcome to Physics 101!
Lecture 01: Introduction to Forces
Newton’s Laws
 Forces
 Free Body Diagrams

Newton’s 1st Law of Motion

If the sum of all external forces on an object is zero, then its
speed and direction will not change. (Know as the Law of
Inertia.)

Example: air hockey table
 When the puck travels with no net force on it, it moves in
a straight line with constant speed.
 The only time the speed or direction changes is when a
net force (from the walls or mallets) is exerted on the
puck!
Newton’s 2nd Law of Motion


If a nonzero net force is applied to an object its motion will
change. (The change is described by the equation F= ma.)
Example: dropping a book
 When you drop a Calculus book from your 10th story
dorm window, there is a net force acting on it (gravity)
thus it begins to accelerate toward the street below!
Newton’s 3rd Law of Motion

In an interaction between two objects, the forces that each
exerts on the other are equal in magnitude and opposite in
direction.

Example: rowing a boat
 When you row, you are pushing the water backwards…
 Thus the water pushes your paddle (and therefore you)
forward!
Forces

Quantifies “interactions” between objects.

4 Fundamental Forces
Gravity
Electromagnetic
Strong Nuclear
Weak Nuclear

Forces are Vectors
Have magnitude and direction
SI units are Newtons (N)
Gravity

Any two objects with mass are attracted.
m1m2
F G 2
r

On the surface of the earth, the force of attraction is
referred to the force of gravity or weight.
 rearth = 6.4x106 m, mearth = 6x1024 kg, G = 6.67x10-11 N-m2/kg2
 Near the earth, the radius of the earth can be used for r.

Example: a person’s weight
 Consider a person with a typical mass of 80 kg.
 Near the earth, his weight (or force of gravity pulling him down
toward the center of the earth) would be Fg = 784 N.
Contact Forces

An object in contact with a surface may have a normal force and
force of friction acting on it.
 The normal force is always perpendicular to the surface. There is no
equation for the normal force – its magnitude depends on the situation.
 The frictional force is always parallel to the surface (opposing motion).
The frictional force depends on the “coefficient of friction” between the
object and surface.
» Kinetic friction (object sliding on a surface): Ff = k*FN
» Static friction (no sliding): there is no definite equation for static friction, but
its maximum value is given by: Ff  s*FN

Example: a sled sliding across a snowy field
The ground is pushing up on the sled (to keep it from
falling into the earth), and the ground is pushing against
the sled’s motion (slowing it down a little).
Free Body Diagrams

The key to success in Physics 101!
Simple picture of just the object of interest.
Choose coordinate system (x,y).
Identify all forces acting on object and draw
arrows showing the directions.

FN
Example: a dresser being pushed
down the hall of your apartment as
you move in.
There are four forces: gravity, the
normal force, friction, and you
pushing on the dresser.
Ff
Fp
y
x
Fg
Summary

Newton’s Laws of Motion
Inertia
F=ma
Equal and Opposite Force Pairs

Forces:
Non-Contact: Gravity
Contact: Friction and Normal

Free Body Diagrams
Isolate One Object
Each Direction is Independent
Example: The Dresser


Let’s return to the dresser FBD and do some
calculations…(I will solve the horizontal direction
and you will solve the vertical direction)…
Given: the dresser has a
mass of 120 kg and you
are pushing on it with a
force of 345 N to make
it slide down the hall at
a constant velocity.
FN
Ff
Fp
y
x
Fg
Example: The Dresser
m = 120 kg and Fp = 345 N

Once the FBD is drawn and we have picked a
coordinate system we must apply Newton’s
Second Law:
FN
F = ma
Ff
We know the horizontal acceleration is zero (the velocity is constant).
F = 0
Fp
y
x
Fg
Example: The Dresser
m = 120 kg and Fp = 345 N

Now start to substitute in what is known:
FN
F = 0
Ff
We know there are two forces in the x-direction (the push and friction).
Fp – Ff = 0
Fp
y
x
Fg
Example: The Dresser
m = 120 kg and Fp = 345 N

Solve for the unknown:
Fp – Ff = 0
FN
The force of the push is given as 345 N.
Ff
Fp
(345 N) – Ff = 0
y
Ff = 345 N
x
Fg
Example: The Dresser

Follow-up: now we can find the coefficient
of kinetic friction between the dresser and
the floor…
FN
Ff = k*FN
Ff
(345 N) = k*(1176 N)
k = 0.29
Fp
y
x
Fg