Transcript The “block on a slope” problem: Consider a crate on a

The “block on a slope” problem:
Consider a crate resting on a loading
ramp, that makes an angle of 20o with
the ground. If the crate has a mass of
75kg, what is the minimum value of the
coefficient of friction necessary to keep
the block from sliding down the ramp?
Step 1: Set up the system.
Step 2: Identify/Draw all forces
Notice how N, the normal force, does not exactly oppose
gravity as it does on a horizontal surface. Despite this N is
still “normal” to the surface.
Step 3: Setting up your
coordinates.
Find and Add all forces in the system parallel and
perpendicular to the sloped surface respectively.
N and Ff are already perpendicular and parallel, respectively, to
the surface. Now we just need to find the components of the
crate’s weight that are in these directions.
I hope you like geometry...
Still working on step 3...
Note that the blue force, or the gravity that acts down (parallel)
to the ramp is a vector that represents a force acting on the
Center of mass of the box. As a vector it can be moved,
although I have not done so here.
More Step 3…add
parallel/perpendicular forces &
SOLVE
Perpendicular:
F = ma, a = 0 (at rest)
N–F=0
N - mgcos = 0
N = mgcos
Parallel:
F=ma, a = 0 (at rest)
F = Ff - F
F = N - (mg)sin
F = (mg)cos  - (mg)sin
(mg)cos  - (mg)sin = 0
(mg)cos  = (mg)sin
 = sin  / cos 
 = tan 
Step 4: From here we can determine the
FORCE of friction, or in other scenario’s
we could go after the acceleration, if the
crate were sliding…
Be sure to indicate the direction of any forces, or
accelerations, since they are vectors. Use descriptive words
like up the ramp or down the ramp, since these coincide with
our new (tilted) coordinate system.
How NOT fun was that??!! Actually I though it
was great, but as you know, I am a big dork!
Finding the effect (or the deflection)
of gravity due to a ramp.
• Way back in unit 1, we said that the
acceleration of a ball down an inclined
plane is not 9.81 m/s2, but rather it is a
smaller value based on the angle of
inclination of the ramp.
• If the crate in the pervious problem were
allowed to slide, how would you find the
acceleration (hint: use Newton’s 2nd Law).
• So the acceleration only depends on the
angle of inclination and the value g!
Changing the accleration
• As one RAISES the angle of the incline, the
acceleration of any object on the incline
becomes _____________, approaching
______________ m/s2 at 90 degrees.
• Conversely, as one LOWERS the angle of
the incline, the acceleration of any object on
the incline becomes _____________,
approaching ______________ m/s2 at zero
degrees.
Moving right along...
What if we do something nasty like add a tension
to the system? And another mass? And…
Easy! Just diagram the forces and add up the
parallel and perpendicular components
just as we did here!
Remember:
Friction always OPPOSES motion.
You can identify it as positive or negative as long as
you are consistent.
For example:
eeewwww…yuk!
Hey just be glad the pulley doesn’t have friction,
and that the tension in the cord is parallel to the incline!
Circular Motion
• Uniform Circular Motion is rotational
motion where the angular velocity of an
object is constant.
• Because we are moving in circles and
changing the direction of the velocity
vector, there must be an acceleration
present. We will discuss this further, later in
this unit.
• The key to understanding all types of
circular motion (uniform and accelerated) is
to understand that all of the circular
variables have the same relationships as
their linear counterparts.
Understanding angles:
Degrees and Radians
• Degrees and Radians are two types of
angular measurement.
• 1 rad = approx. 57.3o
• 1o = approx. 1.74 x 10-2 rad
xo
y radians
• To convert, set up a ratio where 360o  2 radians
• These are approximations because of the 
factor in the conversion.
Angular displacement
• Roughly, displacement refers to how far
something moves from its starting position.
• Since revolutions in a circle or on a wheel
would continue to bring a reference point on
the circle back to the same spot (x=0)
angular displacement refers to the quantity
of angles the reference point has swung
through.
• The symbol theta () is used for angular
displacement.
Sometimes this is
also called “s,” the
“arc length.”
Angular velocity: How fast something spins.
• Linear velocity is equal to change in
displacement over change in time.
Unfortunately, linear velocity, as it relates to
displacement is not terribly meaningful
when describing movement in circles.
• Why do you suppose this is?
• Angular velocity describes how fast
something spins and thus must be related to
angular displacement.
• Angular velocity is equal to the net change
in angular position divided by the time
taken making that change.
•  = /t
Angular acceleration
• Angular acceleration, like linear
acceleration, refers to how fast a velocity
changes.
• Angular acceleration is equal to the change
in angular velocity over time.
•  = /t
Linear and Angular Motion
Quantity Linear Angular Relationships
Symbol Symbol
Displacement
x
x = r

Velocity
v

v = r
Acceleration
a

at = r
What’s up with this “t” going on here? This refers to the
tangential component of the acceleration. That is to say
the acceleration that acts around the circle.
Relationships:
 = /t
 = /t
 = o + t
 = o +t
 = ot + ½t2
2 - o2 = 2
Look familiar?
Now it gets icky...
• The period of revolution is the amount of
time it takes the rotating object to make one
complete revolution.
2r
T
v
When you know the
linear components.
T
2

When you know the
angular components.
Centripetal Acceleration
• A rotating body is in many respects like a projectile (a
projectile with a high enough horizontal velocity is a
satellite, an object that orbits, or rotates around the
earth).
• Rotating bodies have a component of acceleration that
constantly pulls the object along the radius towards the
center of the circle or orbit. This is the centripetal
acceleration.
• Centripetal acceleration, since it is perpendicular to the
velocity vector at all times, does not change the
magnitude of said vector, only the direction. The
concept is similar to the change in direction that a
projectile undergoes during flight.
The centripetal acceleration,
ac, ensure that the object continues
to “fall”towards the center
of the circle as it
rotates.
at  r
Tangential
2
v
ac 
  2r
r
Angular
The tangential
acceleration, at,
and the tangential velocity
and displacement relate respectively to the object’s acceleration,
velocity and displacement around the circle.
Centripetal Force
Centripetal force is simply to force associated with and
causing the centripetal acceleration, ac.
2
mv
F  ma
r
Any object traveling in a circle or a circular path
experiences a centripetal force.
The car problem:
You are in the backseat of your parents sedan.
When the car turns sharply to the right (as in making a 90o turn at a
corner) your body is forced to the left. If a centripetal force is
actually pulling you and the car radially towards the inside of the
turn, why do you feel a force pushing you away from the center of
rotation? (hint: think about Newton’s 1st Law!)
Centripetal Force:
The Imaginary Force
• The centripetal force is, in physics, what we call
an imaginary force. This does NOT mean that
the force is non-existant, but rather this means
that the force we identify mathematically as
“centripetal is ALWAYS caused by some other
force.
– E.g., a contact (normal) force, a friction or a tension.
• This means that mv2/r is always equal to
something else that you must define based on the
circumstances of the system you are studying.
Torque
• Torque is simply a force applied to an object
at some radial distance from the center of
rotation. If the torque is large enough to
overcome the rotational inertia of the object
it will cause the object to rotate.
• The “radial distance” is sometimes called
the lever arm.
• Torque, like force, is governed by Newton’s
Second Law:   
Force
Radius
    Fr
Torque: Lower case “tau”
Angular acceleration
Moment of inertia:
describes the arrangement
of mass around a central point,
often something you look up...
Torques work in equilibrium just like forces,
and all of the same rules apply. E.g., When the net torque
on an object is zero it is either resting, or rotating with
constant rotational velocity.
    Fr
The torque used to create or sustain a rotation is equal to
the applied force multiplied by the radius at which that
force is applied if, and only if, the force is applied such
that is is perpendicular to the radius.
We will make this assumption in the course when solving
problems unless otherwise stated.
Some problems for your mental stimulation pleasure...
1) A wheel rotates at a rate of 8 rad/s for 16 s. What is the angular
displacement of the wheel? How many revolutions as it made?
2) If the wheel in problem 1 is actually a car tire with a diameter of
0.72 m (roughly 32 inches), how far has the car traveled?
3) a) Calculate the centripetal acceleration on a roller coaster car of mass m =
100 kg that goes through a circular loop of radius 25m at 55 m/s.
b) what is the centripetal force on the car and occupants?
c) what is the angular velocity of the roller coaster?
4. A 0.50 m crowbar is used to rip open a wooden crate. What force is
necessary to apply to the crowbar if the torque necessary to strip boards off the
crate is 47 Nm?
5) If the angular acceleration caused by the torque in problem 4 is 0.2 rad/s2,
what is the moment of inertia for the crowbar?