Introduction to Physics Dynamics Notes
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Transcript of Introduction to Physics Dynamics Notes
7/29/2019 Introduction to Physics Dynamics Notes
http://slidepdf.com/reader/full/introduction-to-physics-dynamics-notes 1/9
2305-Dynamics Notes
Dynamics is the study of forces.
Newton’s First Law of Motion, the law of inertia, states that an object at rest will
remain at rest and an object in motion will remain in motion with a constant velocity(straight line motion with the same speed) unless acted on by a net external force.
Newton’s Second Law of Motion is most easily stated as an equation
F = ma
which is that the net force (the sum of all the forces) on an object equals the
object’s mass times the acceleration.
You should always start these problems by drawing a Free Body Diagram (FBD) bydrawing vectors for each force acting on the object.
In many problems, you will need to resolve the forces in components by doing F=ma
for the horizontal values and F=ma for the vertical values.
The easiest set-up is to choose the direction of acceleration to be the positive axis.
This means that if you are on a ramp, you will pick the direction up or down the
ramp to be your positive x axis and your y axis will be perpendicular to your ramp.
In circular motion, you will choose the direction towards the center of the circle to
be positive.
Newton’s Third Law of Motion- Forces always occur in equal and opposite pairs. If
object A exerts a force on object B, object B exerts and equal and opposite
(direction) force on object A.
It’s important to note here that the pair of forces act on different objects, so
they do not cancel each other.
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Weight versus Mass
Mass is a measure of the amount of matter in an object.
Weight, on the other hand, is the amount of force that pulls on you due to the
gravity on Earth, or some other massive object.
Since weight is a force, we look to F=ma and we find that Weight= mg, or
W=mg
Our awareness of weight comes from the forces that balance it. This force, called
apparent weight, is the weight that you feel. It depends on your circumstances
(where you are, how you’re accelerating, etc). If there is no force to balance your
weight, that state is called weightlessness.
Normal Force (N) is a supporting force which pushes an object perpendicular to
the surface that it rests on. In the above picture, the surface is horizontal, so the
normal force pushes upward.
If the above block is at rest,
F y = ma yN – mg = 0
N = mg
Note that the coordinate system has been set up in the standard way (up is +y,
right is +x) Since I used a negative sign to state that mg was downward, I do not
need to include the negative in g = 9.8 m/s2.
The normal force will not always be equal to mg, however.
mg
N
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Take a case where you (m = 80 kg) are standing on an elevator, moving upward and
decelerating with a magnitude of 8 m/s2. First we establish our coordinate system
so that down is positive
F y = ma y
Mg- N= m (8)N = mg - m(8) = (85)(9.8 – 8) = 144 N
Or take a case where there is a block being moved on the floor with a rope of
tension, T.
If the above block stays on the ground
Fx = max F y = ma y
T cos max N + Tsin-mg = 0
ax = N = mg - Tsin
mg
N
mg
a
N
T
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Inclined Planes
For inclined plane problems, you should establish your coordinate system so that
the x axis is aligned with the incline plane and the y axis is perpendicular to the
plane.
Fx = max F y = ma y
mg sin = max N - mg cos = 0
Many other problems (pendulum hanging backwards or forward in an accelerating
train, sign hanging at rest from multiple wires at different angles) can all be solved
simply in similar ways by choosing a coordinate system and breaking up your vectors
into x and y components.
Systems with multiple objects
In pulley systems and other systems with multiple bodies, you can either treat the
objects as one system and one FBD or treat them each individually with a FBD for
each object.
To find the acceleration of the blocks, I would treat the boxes as one system
(because they all have the same acceleration) so that T 1 is pulling all of the masses.
m
mg
N
M3 M2 M1 T1T2T3
x
y
mg
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F = maT 1 = (M1 + M2 + M3) a
To find the tension in T 2 and T 3, however, we could draw a FBD for each block.
F = ma F = ma
T 3 = M3a T 2 – T 3 = M2a
Or, we could treat M1 and M2 as one system and get
F = ma
T 2 = (M2 + M3) a
Pulleys:
Atwood Machine
You first need to draw your FBD for each mass. Let’s assumer that M2 is heavier
(if we assume wrong, we will end up with a negative acceleration at the end). Next
M3 M2 T2T3 T3
M1 + M2 + M3 T1
M1 M2
M2gM1g
TT
+a+a
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we need to establish our coordinate system. Since the blocks have the same
acceleration and tension, and M1 is going up while M2 is going down we need to
rotate our coordinate system so that both accelerations have the same sign.
F = ma F = ma
T - M1g = M1a M2g - T = M2a
OR: If you treat the blocks as one system:
F = ma
M2g – M1g = (M1 + M2)a
You can use this process for many different pulley type problems.
Friction
Static friction is the friction between objects at rest. The static friction force
can vary from 0 to some maximum force depending on the situation. For instance, ifthe maxiumum static friction force is 10 N and I push the object on a level surface
forward with 4 N and it doesn’t move, that means static friction only pushes back
with 4 N. If it pushed with all 10N, the object would have started moving
backwards!
Kinetic (sliding) friction is the friction between objects that are moving
0 < fs sN (static friction)
fk = kN (kinetic friction)
f= friction force = coefficient of friction N = normal force
Note that the area of contact doesn’t affect the friction force
Please make sure that you calculate the normal force using F=ma. Don’t just assume
that it is mg, because in many cases, it won’t be.
Circular Motion
Fc ac
v
r
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If an object is traveling in a circle, then the velocity isn’t constant. That means
that there is an acceleration inward, which means that there is a net force inward.
The Centripetal Force, Fc isn’t a force on its own; it is always the resultant force
of all of the forces in that direction.
Fc = mv2/r so ac = v2/r
For instance, take a rollercoaster loop. At the top of the loop, take the positive
direction to be towards the center of the loop (downwards).
F = ma
N + mg = m(v2/r)
at the bottom of the loop, take the positive direction to be towards the center of
the loop (upwards). F = ma
N –mg = mv2/r
Somewhere in the middle of the loop, you need to sum the forces in the x and y
direction
Fx = max F y = ma y
Nx = mv2/r N y -mg = 0
N2= Nx2 + N y
2
Banked Curves
mg
N
r
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In banked curve problems, you need to set your coordinate system such that x is
horizontal and y is vertical. (This is different than normal inclined planes, where x
is aligned along the plane because the centripetal force is horizontal.)
Fx = max F y = ma y
Nsin = mv2/r N cos -mg = 0
These equations apply for a banked curve without friction. In these cases, there is
some minimum velocity that the car has in order to stay in its circle without slipping
downward.
If friction is involved, its direction is dependent on the situation.
For instance, if the car is traveling faster than the minimum velocity (explained
above), then it has a tendency to move upwards on the slope, so the friction acts
down the slope.
If the car is traveling slower than the minimum velocity, then the car has a
tendency to slip down the slope, so friction must act up the slope.
Draw the free body diagram with friction, resolve it into x and y components and
solve as usual.
If you don’t know whether or not the velocity is faster or slower than the minimum
velocity, then just arbitrarily choose a direction for the friction and solve. If you
find that the friction is positive, then you choose your direction correctly. If the
frictional force is negative, then you choose the wrong direction.
Drag Force - A retarding force that tends to reduce the speed of an object. As
an object falls, air resistance is the drag force.
Fd = bv 2 for high speeds
Fd = bv for low speeds
b is drag constant, which would be determined by the object and the fluid through
which it is moving.
While falling, the object’d FBD produces the equation:
mg – Fd = ma
At terminal velocity, the object’s drag force equals the objects weight.
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Fd = mg
This, of course, means that there is no net force, thus no acceleration, so the
object isn’t speeding up or slowing down, but instead continuing with a constant
(terminal) velocity.