1. Force, Mass, & Newton’s 1st, 2nd, 3rd Laws of Motion (4-1 through 4-5)
Status: Unit 3
Force, Mass, & Newton’s 1st, 2nd, 3rd Laws of Motion (4-1 through 4-5)
Weight – the Force of Gravity: and the Normal Force, Problem solving (4-6, 4-8, 4-7)
Solving Problems w/ Newton’s Laws, Problems w/ Friction (4-7, 5-1)
Problems w/ Friction and Terminal Velocity revisited (5-1, 5-5)
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Physics 253

2. Free Body Diagrams A key element in understanding motion
A free-body diagram has:
A convenient coordinate system,
Representative vectors
For all forces acting on a body
Including those that are unknown.
If translational only, at the center of the body.
Descriptive labels for each force vector.
Doesn’t show forces the body exerts on other objects.
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3. The Plan
The best way to get a handle on the analysis of motion using Newton’s laws is by doing problems in class and at home.
With each problem we add complexity and new forces.
So far we’ve encountered
one fundamental force (gravity)
one non-fundamental force (the normal force).
Today we’ll introduce two more non-fundamental forces
Tension
Friction.
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4. Back to Mechanics: A Rope under Tension
A flexible rope, cord, or wire pulling on an object is said to be under tension and exerts a force FT.
Before we begin, we make a simple assumption that any such device is massless.
As a consequence the rope transmits force undiminished from one end to the other.
This is apparent from SF = ma. Since the mass is zero the net force on the cord is zero, so the force on the two ends must sum to zero.
Where does this other force come, from the object pulled!
-FT FT FT
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5. An Example: Two Boxes and a Cord.
Two boxes resting on a frictionless surface are connected by a massless cord.
The boxes have masses of 12.0 and 10.0 kg.
A horizontal force of FP=40.0 N is applied by pulling on the lighter box.
What is the acceleration of each box and the tension in the cord?
This problem adds a force and 2 dimensions!
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6. Box 1 freebody diagram has four forces:
From person pulling: FP
Tension from cord: FT
Weight: W1 = m1g
Normal Force: FN1
From the 2nd Law in the x direction
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7. Box 2 freebody diagram has three forces:
Tension from cord: FT
Weight: W2 = m2g
Normal Force: FN2
From the 2nd Law in the x direction
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8. The boxes must have identical acceleration a1=a2 otherwise the cord would part or bunch-up which is counter to our experience.
It makes sense for the tension to be less than the pulling force since the tension accelerates only the second box
Note how analyzing two free-body diagrams allows us to understand an “inner” force.
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9. Atwood’s Machine
A classic problem: Two suspended masses connected by a cable passing over a pulley.
Consider an elevator of mass m1 and a counter-weight of mass m2, where m1 > m2.
In terms of the masses (to keep it general) calculate the acceleration of the elevator and the tension in the cable.
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10. The Rev. George Atwood ( ) was a tutor at Trinity College, Cambridge when he published “A Treatise on the Rectilinear Motion and Rotation of Bodies, with a Description of Original Experiments Relative to the Subject” in 1784.
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11. Once again in order to avoid stretching or bunching of the cable the two masses must have equal & opposite acceleration: a1=-a2
A massless cable also ensures that the tension is equal on both sides of the cable or on both objects.
Since the elevator mass is greater than the CW the elevator will be moving down.
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12. Limiting cases are correct:
Masses equal: a=0 or it doesn’t move!
Second mass zero: a=g and the elevator is in free fall.
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13. We can use the equation for acceleration to derive the tension
Some limiting cases
Masses equal: FT = mg =W
Either mass zero: tension is zero and we’re in free fall again!
Note how the algebraic form helps us understand the physics.
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14. The mechanical advantage of a pulley
Apply the 2nd Law to the freebody diagram at zero acceleration.
In the y direction
So for N loops the force required is mg/N!
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15. The Mechanical Advantage of Two Dimensional Tension
Because tension is always a long a rope (otherwise it would buckle) pushing in the middle applies a much stronger force on the ends.
We can show this with SF=ma.
In the figure below if FP=300 N, what force is pulling on the vehicle (and tree)?
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16. Let’s look at the x direction:
In other words the magnitudes of the forces on the rope at the fulcrum are equal.
Looking at the y direction
The technique increases
the force on the car
six-fold. Note the smaller
q the larger the effect!
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17. A List of Forces Fundamental Forces Non Fundamental Forces
Gravitational Force – the oldest, weakest, and least understood
Electroweak Force – a unification of
Electromagnetism (electric and magnetic)
Weak force (radioactive force)
The Strong Force – the nuclear force and strongest of the three.
Non Fundamental Forces
Normal Force
Friction
Static
Kinetic
Tension
Most non-fundamental forces are real-world manifestations of the gravitational and electroweak force.
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18. Origin of Forces: Standard Model of Particle Physics
The essence: Bits of matter stick together by exchanging stuff.
The great achievement of particle physics is a model that describes all particles and particle interactions. The model includes:
6 quarks (the particles in the nucleus) and their antiparticles.
6 leptons (of which the electron is an example) and their antiparticles
4 force carrier particles
Precisely: “All known matter is composed of composites of quarks and leptons which interact by exchanging force carriers.”
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19. The Stuff: Standard Model Interactions Mediated by Boson Exchange
10-37 weaker
than EM, not
explained
Explained by Standard Model
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20. The Bits: Periodic Table of Fundamental Particles
All point-like down to
10-18 m
Families reflect
increasing mass and
a theoretical
organization
u, d, n, e are
“normal matter”
These all interact by
exchanging bosons
Mass 
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21. Two more problems, First the Accelerometer
We can apply the 2nd law to build a crude accelerometer. Consider the scenario below.
When the car is accelerating the pendulum makes an angle, q, with the vertical.
How does q depend on the acceleration?
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22. There are two forces, tension in the string, FT, and the weight, mg.
Since there is no accel-eration in the vertical direction:
And in thex direction there is an acceleration, thus
An acceleration of 1.2m/s2 = (2.6mi/hr)/s would correspond to 7.0o
Note this accelerometer
is independent of tension
and mass. q
FTcos(q) +y +x
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23. A box on an incline
This is another classic problem which we will make more realistic next lecture with friction. For now assume an ideal surface.
A box of mass m is on an inclined plane that makes an angle q with the horizontal.
What is the normal force? What is the acceleration?
Evaluate for a mass m=10kg and an angle of q=30o.
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24. We then have two forces:
Since the box never “lifts off” the incline we put the x axis for our free-body problem along the incline. This simplifies the problem since there will be no acceleration in the y direction.
We then have two forces:
The normal force to the plane
Gravity which as shown in the figure we resolve into x and y compoments.
Next we apply SF=ma
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25. First in the y direction:
In the limit:
No angle the normal force is the weight as expected.
At 90 degrees, FN=0, the incline is straight up and we are in free-fall, no normal force.
For the values given
Now the x direction:
The acceleration is always less than g.
In the limit
No angle the acceleration is zero.
At 90 degrees, a=g, the incline is straight up, and we are in free-fall
For the values given
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26. We’ve solved a number of interesting problems:
Boxes under tensions
Atwood’s machine
Cables as levels
Accelerometer
Box on an incline
Although simple these problems show the great power of Newton’s Laws
Next lecture we’ll expand the scope to encompass friction.
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