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Newton’s Laws
The Study of Dynamics
Isaac Newton Came up with 3 Laws of Motion to
explain the observations and analyses of Galileo and Johannes Kepler.
Invented Calculus. Published his Laws in 1687 in the book
Mathematical Principles of Natural Philosophy.
What is Force? A force is a push or pull on an
object. Forces cause an object to
accelerate… To speed up To slow down To change direction
Newton’s First Law The Law of Inertia. A body in motion stays in motion at
constant velocity and a body at rest stays at rest unless acted upon by an external force.
This law is commonly applied to the horizontal component of velocity, which is assumed not to change during the flight of a projectile.
The First Law is Counterintuitive
Aristotle firmly believed this.But Physics 1 students know better!
A force diagram illustrating no net force
A force diagram illustrating no net force
A force diagram illustrating no net force
A force diagram illustrating no net force
Another example illustrating no net force
Newton’s Second Law A body accelerates when acted
upon by a net external force. The acceleration is proportional to
the net force and is in the direction which the net force acts.
This law is commonly applied to the vertical component of velocity.
Newton’s Second Law ∑F = ma
where ∑F is the net force measured in Newtons (N)
m is mass (kg) a is acceleration (m/s2)
Newton (SI system) 1 N = 1 kg m /s2
1 N is the force required to accelerate a 1 kg mass at a rate of 1 m/s2
Pound (British system) 1 lb = 1 slug ft /s2
Units of force
The problem of weight
Are weight and mass the same thing?
No. Weight can be defined as the force due to gravitation attraction.
W = mg
Newton’s Third Law
For every action there exists an equal and opposite reaction.
If A exerts a force F on B, then B exerts a force of -F on A.
Step 1: Draw the problem
Larry pushes a 20 kg block on a frictionless floor at a 45o angle below the horizontal with a force of 150 N while Moe pulls the same block horizontally with a force of 120 N. What is acceleration?
Working a Newton’s 2nd Law Problem
20 kgFL FM
Step 2: Diagram
Force diagram
Working a Newton’s 2nd Law Problem
Free Body diagram
20 kgFL FM
FG
N
FL
FM
FG
N
Step 3: Set up equationsF = maFx = max
Fy = may
Working a Newton’s 2nd Law Problem
Always resolve two-dimensional problems into two one-dimensional problems.
Step 4: Substitute Make a list of givens from the word problem.
Substitute in what you know.
Working a Newton’s 2nd Law Problem
Step 5: Solve Plug-n-chug. Calculate your unknowns. Sometimes you’ll need to do
kimematic calculations following the Newton’s 2nd law calculations.
Working a Newton’s 2nd Law Problem
Gravity as an accelerating force
A very commonly used accelerating force is gravity. Here is gravity in action. The acceleration is g.
Gravity as an accelerating force
In the absence of air resistance, gravity acts upon all objects by causing the same acceleration…g.
The pulley lets us use gravity as our accelerating force… but a lot slower than free fall. Acceleration here is a lot lower than g.
Gravity as an accelerating force
2-Dimensional problem
Larry pushes a 20 kg block on a frictionless floor at a 45o angle below the horizontal with a force of 150 N while Moe pulls the same block horizontally with a force of 120 N.a) What is the acceleration?b) What is the normal force?
20 kgFL FM
FG
N
Flat surfaces – 1 D
N = mg for objects resting on horizontal surfaces.
mg
N
Applied forces affect normal force.
applied forcefriction
weightnormal
N = applied force
Elevator Ride – going up!
mg
N
Groundfloor
Normal feeling
V = 0A = 0
mg
N
Juststarting up
Heavy feeling
V > 0A > 0
mg
N
Betweenfloors
Normal feeling
V > 0A = 0
mg
N
Arriving attop floor
Light feeling
V > 0A < 0
Elevator Ride – going down!
mg
N
Topfloor
Normal feeling
V = 0A = 0
mg
N
Arriving atGround floor
Heavy feeling
V < 0A > 0
mg
N
Betweenfloors
Normal feeling
V < 0A = 0
mg
N
Beginningdescent
Light feeling
V < 0A < 0
Ramps – 2 D
mg
The normal force is perpendicular to angled ramps as well. It’s always equal to the component of weight perpendicular to the surface.
N
mgcos
mgsin
N = mgcos
Ramps – 2 D
mg
How long will it take a 1.0 kg block to slide down a frictionless 20 m long ramp that is at a 15o angle with the horizontal?
N
mgcos
mgsin
N = mgcos
Determination of the Coefficients of FrictionCoefficient of Static Friction1) Set a block of one material on an
incline plane made of the other material.
2) Slowly increase angle of plane until the block just begins to move. Record this angle.
3) Calculate s = tan.
Friction The force that opposes a sliding motion.
Enables us to walk, drive a car, etc.
Due to microscopic irregularities in even the smoothest of surfaces.
There are two types of friction
Static frictionexists before sliding occurs
Kinetic frictionexists after sliding occurs
In general fk <= fs
Friction and the Normal Force
The frictional force which exists between two surfaces is directly proportional to the normal force.
That’s why friction on a sloping surface is less than friction on a flat surface.
Static Friction fs sN
fs : static frictional force (N) s: coefficient of static friction
N: normal force (N) Static friction increases as the
force trying to push an object increases… up to a point!
A force diagram illustrating Static Friction
Applied Force
Frictional Force
Normal Force
Gravity
A force diagram illustrating Static Friction
Bigger Frictional Force
Normal Force
Gravity
Bigger Applied Force
A force diagram illustrating Static Friction
Frictional Force
Normal Force
GravityEven Bigger Applied Force
The forces on the book are now UNBALANCED!
Static friction cannot get any larger, and can no longer completely oppose the applied force.
Kinetic Friction fk = kN
fk : kinetic frictional force (N) k: coefficient of kinetic friction N: normal force (N)
Kinetic friction (sliding friction) is generally less than static friction (motionless friction) for most surfaces.
Determination of the Coefficients of FrictionCoefficient of Kinetic Friction
1) Set a block of one material on an incline plane made of the other material.
2) Slowly increase angle of plane until the block just begins to move at constant speed after giving it a slight tap. Record this angle.
3) Calculate k = tan.
Magic Pulleys
m1
m2
T
mg
N
mgT
-x
x
Mass 1 (10 kg) rests on a frictionless table connected by a string to Mass 2 (5 kg). Find (a) the acceleration of each block and, (b) the tension in the connecting string.
Pulley problem
m1
m2
Mass 1 (10 kg) rests on a table connected by a string to Mass 2 (5 kg) as shown. What must the minimum coefficient of static friction be to keep Mass 1 from slipping?
Pulley problem
m1
m2
Mass 1 (10 kg) rests on a table connected by a string to Mass 2 (5 kg). If s = 0.3 and k = 0.2, what is a) the acceleration and b) the tension in the string?
Pulley problem
m1
m2
Tension A pulling force. Generally exists in a rope,
string, or cable. Arises at the molecular level,
when a rope, string, or cable resists being pulled apart.
Step 1: Identify the body to analyze.
Working a Newton’s Law Problem
This may not be all that easy! It may be a knot, a nail, a hinge, a
person, an “object” or a “particle”. It is the focus of your subsequent
analysis.
Step 2: Select a reference frame.
Working a Newton’s Law Problem
This should be an inertial reference frame which may be moving but not accelerating.
Think of this as a coordinate system with a specific origin!
Step 3: Make a diagram of forces. Force diagram
Working a Newton’s Law Problem
Free Body diagram
F1F2
F1
F2
Step 4: Set up force equations.
F = maFx = max
Fy = may
Working a Newton’s Law Problem
Always resolve two-dimensional problems into two one-dimensional problems.
Step 5: Calculate!
Substitute in what you know into the second law equations.
Calculate unknown or unknowns.
Working a Newton’s Law Problem
Ramp (frictionless)
mg
The normal force is perpendicular to angled
ramps as well. It’s usually equal to the
component of weight perpendicular to the
surface.
N
mgcos
mgsin
N = mgcos
Ramp (frictionless)
mg
What will acceleration be in this situation?F= mamgsin = magsin = a
N
mgcos
mgsin
N = mgcos
Ramp (frictionless)
mg
How could you keep the block from accelerating?
N
mgcos
mgsin
N = mgcos
T
Tension (static 1D)
The horizontal and vertical components of the tension are equal to zero if the system is not accelerating.
15 kg
T
mg
F = 0T = mg
Tension (static 2D)
The horizontal and vertical components of the tension are equal to zero if the system is not accelerating.
15 kg
30o 45o
T1T2
T3
Fx = 0Fy = 0T3
mg
Tension (elevator)
When an elevator is still, the tension in the cable is equal to its weight.M
T
Mg
Tension (elevator)
What about when the elevator is just starting to head upward from the ground floor?M
T
Mg
Tension (elevator)
What about when the elevator is between floors?
M
T
Mg
Tension (elevator)
What about when the elevator is slowing at the top floor?
M
T
Mg
Tension (elevator)
What about if the elevator cable breaks?
M Mg
-x
x
Magic pulleys simply bend the coordinate system.
Pulley problems
m1
m2
m1g
NT T
m2g
F = mam2g= (m1+m2)a
Tension is determined by examining one block or the other
Pulley problems
m 1 m2
m1g
NT
T
m2g
F2m2am2g - T = m2a
F1m1am1gsin = m1a