Transcript Mass on a plane with friction

Using Newton’s 2nd Law
We have learned that Newton’s 2nd law
is F = ma
We have learned some of the forces
that can be acting on an object: weight,
tension, normal, friction.
We have learned how to draw free body
diagrams to help us solve problems.
Now it is time to put it all together.
Newton’s 2nd Law in 3D
F = ma must be true in all directions.
Fx = max, Fy = may, Fz = maz.
If the object isn’t moving in the ydirection, for example, then ay = 0 and
Fy = 0.
If there is more than 1 force acting in
the y-direction, for example, then
∑Fy=may
Mass on a plane with friction
Forces acting on the
pink box are:
Weight – pointing
straight down
Normal – pointing up
perpendicularly from
the 2 surfaces.
friction – pointing in the
opposite direction of
motion (opposing
motion).
Mass on a plane with friction
The Weight has a component parallel
to the plane and pointing down the
plane, that we will call Wparallel:
Wparallel = mgsinθ
For an object on a non-horizontal
surface, we know that N = mgcosθ.
And we know that f=μN= μmgcosθ.
Mass on a plane with friction- not moving
If the mass is not moving, we know
that the force pulling it down the plane
(Wparallel) equals the force opposing the
motion (friction).
mgsinθ = μmgcosθ
But which μ is it? Static or kinetic?
Since the mass isn’t moving, it must be
static.
Mass on a plane with friction- not moving
Therefore:
mgsinθ = μsmgcosθ
Notice that mass and g will cancel out, and
this equation becomes:
sinθ = μscosθ
Rearranging this equation becomes:
μs = cosθ/sinθ = tanθ
This can be used to find μs for a non-moving
object on an inclined plane.
Mass on a plane with friction- not moving
Example
What is the friction acting on a 15 kg mass
resting on a plane inclined at 25° to the
horizontal?
Remember if it isn’t moving, then
Friction = Wparallel (and Wparallel = mgsinθ)
Therefore: Friction = mgsinθ
Friction = (15 kg)(9.8 m/s2) sin 25 °
Friction = 62 N
Mass on a plane with friction- not moving
Example 2
What is the coefficient of static friction acting
on the same mass?
Remember Friction = μsmgcosθ
62 N = μs (15 kg)(9.8 m/s2) cos 25 °
μs = .47
Remember coefficients of friction are unitless!
Mass on a plane with friction- moving
But what if the mass is accelerating?
If we sum up the forces acting
parallel to the line of motion (and
choosing our signs to indicate up or
down direction):
Σ F = -mgsinθ + μmgcosθ
But which μ is it? Static or kinetic?
Because the object is moving, it will be
kinetic friction.
Mass on a plane with friction- moving
Σ F = -mgsinθ + μkmgcosθ
By Newton’s 2nd Law, F = ma, so
-mgsinθ + μkmgcosθ = ma
Notice mass can cancel out (but not g):
-gsinθ + μkgcosθ = a
This equation can be used to find the
acceleration parallel to the plane of a mass
on an inclined plane.
Note that acceleration should turn out to be
negative, indicating motion down the plane.
Mass on a plane with friction- moving
Example
What is the acceleration parallel to the plane
of an object sliding down a plane that is
inclined at 48°? The coefficient of kinetic
friction between the object and the plane is
.26.
a = -gsinθ + μkgcosθ
a = - (9.8 m/s2)sin48° + (.26)(9.8
m/s2)cos48°
a = -5.6 m/s2
Mass on a plane with friction- moving
Example 2
How long will it take the same object to travel down
.50 meters of the plane? Assume the acceleration is
constant.
a = -5.6 m/s2
vi = 0 m/s
d = -.50 m (negative cuz “down”)
t=?
d = vit + ½ at2
-.50 m = (0) + ½ (-5.6 m/s2) t2
t = .42 seconds
2 attached masses on an inclined plane
Mass 1 is the mass
on the plane.
Mass 2 is the
hanging mass.
Now we have an
added force of
Tension, pointing
away from each
mass along the
rope.
2 attached masses on an inclined plane
The forces acting on mass 1 (the mass on the plane)
are the same as for a mass on a plane with the
addition of Tension:
-m1gsinθ + μkm1gcosθ + T = m1a
Where a is acceleration parallel to the plane.
The forces acting on mass 2 (the hanging mass) are
only in the y direction. If we ignore air friction, they
are:
Σ Fy = -m2g + T = m2a
If we assume a massless, non-stretching rope, we
can assume that this a is the same as the
acceleration of mass 1.
2 attached masses on an inclined plane
Solving one equation for T and plugging into the
other equation yields the following “Big Ugly”
Equation:
For the magnitude of the acceleration down a
plane:
a = (m1gsinθ - μkm1gcosθ - m2g) / (m1 + m2)
For the magnitude of the acceleration up a plane:
a = (-m1gsinθ - μkm1gcosθ + m2g) / (m1 + m2)
2 attached masses on an inclined plane
But how do you determine if acceleration will be up
or down the plane?
Compare the values of m1gsinθ and m2g to determine
which is bigger.
If m1gsinθ is bigger it will pull the mass down the
plane.
If m2g is bigger it will pull the mass up the plane.
In general, a big m1 and a big θ lead to acceleration
down the plane. A small m1 and a small θ lead to
acceleration up the plane.
2 attached masses on an inclined plane
Example
If m1 is 15 kg, m2 is 35 kg, μk is .13, and θ is 12°,
what will be the acceleration of the mass?
First determine if mass 1 will accelerate up or down
the plane by comparing m1gsinθ and m2g.
m1gsinθ = (15kg)(9.8 m/s2)sin 12° = 31 N
m2g = (35 kg )(9.8 m/s2) = 343 N
So it will accelerate up the plane.
a = (-m1gsinθ - μkm1gcosθ + m2g) / (m1 + m2)
a = 5.9 m/s2
2 attached masses on an inclined plane
Example 2
For the previous example, what will be the tension in
the rope?
For problems such as these, the easiest way to solve
is to sum up the forces acting on the hanging mass.
This way we can ignore friction and angles.
The forces acting on the hanging mass are tension
pointing up, and weight pointing down. Note that
acceleration will be negative since mass 2 is
falling!
Σ F = T – m2g = m2a
T – (35 kg) (9.8 m/s2) = (35 kg) (-5.9 m/s2)
T = 140 N
2 attached masses in an Atwood machine
An Atwood machine
is simply 2 masses
attached to a string
that passes thru a
pulley.
Notice the FBD’s for
each mass have
been drawn for you.
Atwood Machine
If mass 1 is smaller than mass 2, mass 1 will accelerate
upwards and mass 2 will accelerate downwards. Again
assuming a massless non-stretching rope, we can say that these
accelerations will be equal.
Sum up the forces acting on mass 1:
Σ F1 = T – m1g = m1a
Note that this acceleration is positive because mass 1 is moving
up.
Sum up the forces acting on mass 2:
Σ F2 = T – m2g = -m2a
Note that this acceleration is negative because mass 2 is moving
down.
Atwood Machine
Solving each equation for Tension and
setting them equal to each other:
m1g + m1a = m2g - m2a
This equation can be manipulated to
solve for m1, m2, or a. Remember that
in our derivation of this equation, mass
2 was the heavier mass.
Atwood Machine - Example
If mass 1 is 12 kg and mass 2 is 15 kg, what will be
the acceleration of each mass?
m1g + m1a = m2g - m2a
Rearranging this equation so that a is on one side:
m2a + m1a = m2g – m1g
Factoring out an a:
a(m2 + m1) = m2g – m1g
Solving for a:
a = (m2g – m1g) / (m2 + m1)
Atwood Machine – Example 2
a = (m2g – m1g) / (m2 + m1)
Plug in your values:
a =[(15 kg)(9.8.m/s2) – (12 kg)(9.8
m/s2)] / (15 + 12 kg)
a = 1.1 m/s2
Note that for Atwood machines, if you
get a = 9.8 m/s2 you have done
something wrong!
Atwood Machine - Example
What is the tension in the rope?
Use one of the Σ equations derived
earlier:
Σ F1 = T – m1g = m1a
T = 12 kg(9.8 m/s2) + (12 kg)(1.1 m/s2)
T = 130 N
Atwood Machine – Example 3
How fast will mass 2 be falling at the end of 3.0
seconds if released from rest?
a = - 1.1 m/s2 (negative cuz mass 2 is falling)
vi = 0 m/s
t = 3.0 s
vf = ?
vf = vi + at
vf = 0 m/s + (- 1.1 m/s2 )(3.0 s)
vf = - 3.3 m/s
Elevator problems
Elevators are a classic problem in physics.
Weight will still be a force, since gravity will still be
attracting the body.
But if the elevator is accelerating (not moving at a
constant velocity) then the acceleration will also be a
force.
The question is: will it be a positive or a negative
force?
Elevator accelerating upwards
A man, mass 50 kg, is inside an
elevator which begins to accelerate
upward from rest at 2 m/s2.
There are 2 forces acting on the
man. His weight pulls down. But the
scale also pushes up on the man.
Since the net acceleration is 2 m/s2
upward, Newton's Second Law says
Fnet = ma
Fnet = (50 kg)(2 m/s2)
Fnet = 100 N (upward because accel
is upward)
Elevator accelerating upwards
Since the net force is 100 N, the upward force (of the
scale) and downward force (of the weight) must
equal a 100 N upward force.
Fnet = F scale – W
100 N = F scale - (50 kg)(9.8 m/s2)
F scale = 600 N
Therefore, the scale will read 600 N
Note that this is a heavier reading than if the elevator
were at rest. This is often called the “apparent
weight”.
Elevator accelerating downwards
A man, mass 50 kg, is inside an
elevator which begins to accelerate
downward from rest at 2 m/s2.
The same 2 forces are acting on the
man (weight pulls down and scale
pushes up)
Since the net acceleration is 2 m/s2
downward, Newton's Second Law
says
ΣF = ma
ΣF = (50 kg)(-2 m/s2)
ΣF = -100 N (downward)
Elevator accelerating downwards
Since the sum force is -100 N, the upward
force (of the scale) and downward force (of
the weight) must equal a 100 N downward
force.
ΣF = F scale – W
-100 N = F scale - (50 kg)(9.8 m/s2)
F scale = 400 N
Therefore, the scale will read 400 N
Note that this is a lighter reading than if the
elevator were at rest.
Elevator
The force that the scale exerts on the man is the
normal force. To sum up:
N = mg if the elevator is at rest or moving at
constant velocity
N = mg + ma if the elevator has an upward
acceleration
N = mg - ma if the elevator has a downward
acceleration
Attached masses – unequal tension
Consider 3 masses, each
attached to a massless, nonstretching string, then pulled
upward so that they
accelerate.
The accelerations will be
equal (a1=a2=a3).
But the tensions will NOT be
equal.
Attached masses – unequal tension 2
Sum up the forces on each
mass:
ΣF3: T3 – m3g = m3a
Mass 2 will have 2 tensions
acting on it, T3 pulling down
and T2 pulling up, so:
ΣF2: -T3 + T2 – m2g = m2a
Similarly for mass 1:
ΣF1: -T2 + T1 – m1g = m1a
Attached masses – unequal tension 3
Solve each summation for T:
ΣF3: T3 – m3g = m3a
T3 = m3g + m3a
T3 = m3(g + a)
ΣF2: -T3 + T2 – m2g = m2a
T2 = T3 + m2g + m2a
T2 = T3 + m2(g + a)
But T3 = m3(g + a)
So T2 = m3(g + a) + m2(g + a)
Attached masses – unequal tension 4
Solve each summation for T (cont’d)
ΣF1: -T2 + T1 – m1g = m1a
T1 = T2 + m1g + m1a
T1 = T2 + m1(g + a)
But T2 = m3(g + a) + m2(g + a)
So T1 = m3(g + a) + m2(g + a) + m1(g +
a)
See the pattern?
Attached masses – unequal tension 5
Each mass adds a tension m(g+a) to
the string pulling the box above it.
References for images
http://content.answers.com/main/content/wp/en/thumb/d/d8/1
80px-Free_body_diagram_mod.png
http://panda.unm.edu/Courses/Price/Phys160/F11-2.jpeg
http://www.batesville.k12.in.us/Physics/PhyNet/Mechanics/Newt
on2/ElevatorProblem.html
http://dept.physics.upenn.edu/courses/gladney/mathphys/java/
sect3/phys_lecture_3.html