Dynamics
Things to Remember from Last Year
Newton's Three Laws of Motion
Inertial Reference Frames
Mass vs. Weight
Forces we studied:
weight / gravity
normal force
tension
friction (kinetic and static)
Drawing Free Body Diagrams
Problem Solving
Newton's Laws of Motion
1. An object maintains its velocity (both speed and
direction) unless acted upon by a nonzero net force.
2. Newton’s second law is the relation between
acceleration and force.
ΣF = ma
3. Whenever one object exerts a force on a second
object, the second object exerts an equal force in
the opposite direction on the first object.
1
Which of Newton's laws best explains why motorists should buckleup?
A
First Law
B
Second Law
C
Third Law
D
Law of Gravitation
2
You are standing in a moving bus, facing forward, and you
suddenly fall forward. You can infer from this that the bus's
A
velocity decreased.
B
velocity increased.
C
speed remained the same, but it is turning to the right.
D
speed remained the same, but it is turning to the left.
3
You are standing in a moving bus, facing forward, and you
suddenly fall forward as the bus comes to an immediate stop. What
force caused you to fall forward?
A
gravity
B
normal force due to your contact with the floor of the bus
C
force due to friction between you and the floor of the bus
D
There is not a force leading to your fall.
4
When the rocket engines on the spacecraft are suddenly turned off,
while traveling in empty space, the starship will
A
stop immediately.
B
slowly slow down, then stop.
C
go faster.
D
move with constant speed
5
A net force F accelerates a mass m with an acceleration a. If the
same net force is applied to mass 2m, then the acceleration will be
A
4a
B
2a
C
a/2
D
a/4
6
A net force F acts on a mass m and produces an acceleration a.
What acceleration results if a net force 2F acts on mass 4m?
A
a/2
B
8a
C
4a
D
2a
7
An object of mass m sits on a flat table. The Earth pulls on this
object with force mg, which we will call the action force. What is
the reaction force?
A
The table pushing up on the object with force mg
B
The object pushing down on the table with force mg
C
The table pushing down on the floor with force mg
D
The object pulling upward on the Earth with force mg
8
A 20-ton truck collides with a 1500-lb car and causes a lot of
damage to the car. Since a lot of damage is done on the car
A
the force on the truck is greater then the force on the car
B
the force on the truck is equal to the force on the car
C
the force on the truck is smaller than the force on the car
D
the truck did not slow down during the collision
Inertial Reference Frames
Newton's laws are only valid in inertial reference frames:
An inertial reference frame is one in which Newton’s first law is valid.
This excludes rotating and accelerating frames.
Mass and Weight
MASS is the measure of the inertia of an object, the resistance of an
object to accelerate.
WEIGHT is the force exerted on that object by gravity. Close to the
surface of the Earth, where the gravitational force is nearly constant,
the weight is:
FG = mg
Mass is measured in kilograms, weight Newtons.
Normal Force and Weight
FN
The Normal Force, FN, is
ALWAYS perpendicular to
the surface.
Weight, mg, is ALWAYS
directed downward.
mg
9
The acceleration due to gravity is lower on the Moon than on Earth.
Which of the following is true about the mass and weight of an
astronaut on the Moon's surface, compared to Earth?
A
Mass is less, weight is same
B
Mass is same, weight is less
C
Both mass and weight are less
D
Both mass and weight are the same
10
A 14 N brick is sitting on a table. What is the normal force supplied
by the table?
A
14 N upwards
B
28 N upwards
C
14 N downwards
D
28 N downwards
Kinetic Friction
Friction forces are
ALWAYS parallel to the
surface exerting them.
Kinetic friction is always
directed opposite to
direction the object is
sliding and has
magnitude:
fK = μkFN
v
fK
11
A 4.0kg brick is sliding on a surface. The coefficient of kinetic
friction between the surfaces is 0.25. What it the size of the force of
friction?
A
8.0
B
8.8
C
9.0
D
9.8
12
A brick is sliding to the right on a horizontal surface.
What are the directions of the two surface forces: the friction force
and the normal force?
A
right, down
B
right, up
C
left, down
D
left, up
Static Friction
Static friction is always
equal and opposite the
Net Applied Force
acting on the object (not
including friction).
Its magnitude is:
fS ≤ μSFN
fS
FAPP
13
A 4.0 kg brick is sitting on a table. The coefficient of static friction
between the surfaces is 0.45. What is the largest force that can be
applied horizontally to the brick before it begins to slide?
A
16.33
B
17.64
C
17.98
D
18.12
14
A 4.0kg brick is sitting on a table. The coefficient of static friction
between the surfaces is 0.45. If a 10 N horizontal force is applied to
the brick, what will be the force of friction and will the brick move?
A
16.12, no
B
17.64, no
C
16.12, yes
D
17.64, yes
Tension Force
When a cord or rope pulls on an object, it is
said to be under tension, and the force it
exerts is called a tension force, FT.
a
FT
mg
15
A crane is lifting a 60 kg load at a constant velocity. Determine the
tension force in the cable.
A
568 N
B
578 N
C
504 N
D
600 N
16
A system of two blocks is accelerated by an applied force of
magnitude F on the frictionless horizontal surface. The tension in the
string between the blocks is:
A
6F
B
4F
C
3/5 F
D
1/6 F
E
1/4 F
6 kg
4 kg
F
wo Dimensions
T
Adding Orthogonal Forces
Since forces are vectors, they add as vectors.
The simplest case is if the forces are perpendicular (orthogonal) with
another.
Let's add a 50 N force at 0o with a 40 N force at 90o.
Adding Orthogonal Forces
1. Draw the first force vector, 50 N
at 0o, beginning at the origin.
90o
180o
0o
270o
Adding Orthogonal Forces
1. Draw the first force vector, 50 N
at 0o, beginning at the origin.
90o
2. Draw the second force vector,
40 N at 90o, with its tail at the tip of
the first vector.
180o
0o
270o
Adding Orthogonal Forces
1. Draw the first force vector, 50 N
at 0o, beginning at the origin.
90o
2. Draw the second force vector,
40 N at 90o, with its tail at the tip of
the first vector.
3. Draw Fnet from the tail of the
first force to the tip of the last
force.
180o
0o
270o
Adding Orthogonal Forces
We get the magnitude of the resultant
from the Pythagorean Theorem.
90o
c2 = a2 + b2
Fnet2 = (50N)2 + (40N)2
= 2500N2 + 1600N2
= 4100N2
180o
0o
Fnet = (4100N2)1/2
= 64 N
270o
Adding Orthogonal Forces
We get the direction of the net Force from the
inverse tangent.
90o
tan(θ) = opp / adj
tan(θ) = (40N) / (50N)
tan(θ) = 4/5
tan(θ) = 0.8
180o
0o
θ = tan-1(0.80) = 39o
Fnet = 64 N at 39o
270o
Decomposing Forces into Orthogonal Components
90o
Let's decompose a 120 N force at
34o into its x and y components.
F
θ
180o
0o
270o
Decomposing Forces into Orthogonal Components
90o
We have Fnet, which is the
hypotenuse, so let's first find the
adjacent side by using
cosθ = adj / hyp
cosθ = Fx / Fnet
Fx = Fnet cosθ
= 120N cos34o
= 100 N
F
θ
180o
0o
Fx
270o
Decomposing Forces into Orthogonal Components
90o
Now let's find the opposite side by
using
sinθ = adj / hyp
sinθ = Fy / Fnet
Fy = Fnet sinθ
= 120N sin34o
= 67 N
Fy
F
θ
180o
0o
Fx
270o
Adding non-Orthogonal Forces
Now that we know how to add orthogonal forces.
And we know now to break forces into orthogonal components, so we can
make any force into two orthogonal forces.
We combine those two steps to add any number of forces together at any
different angles.
Adding non-Orthogonal Forces
Let's add together these three forces:
F1 = 8.0 N at 50o
F2 = 6.5 N at 75o
F3 = 8.4 N at 30o
First, we will do it graphically, to show the principle for how we will
do it analytically.
Then, we will do it analytically, which is easy once you see why it
works that way.
Adding non-Orthogonal Forces
90o
First, here are the vectors
all drawn from the origin.
F1 = 4.0 N at 50o
F2 = 6.5 N at 75o
F3 = 8.4 N at 30o
180o
0o
270o
Adding non-Orthogonal Forces
90o
Now we arrange them tail
to tip.
F3
F2
F1
F1 = 4.0 N at 50o
F2 = 6.5 N at 75o
F3 = 8.4 N at 30o
180o
0o
270o
Adding non-Orthogonal Forces
90o
Then we draw in the
Resultant, the sum of the
vectors.
F3
F2
F
F
F1
F2
F3
F1
F1 = 4.0 N at 50o
F2 = 6.5 N at 75o
F3 = 8.4 N at 30o
180o
0o
270o
Adding non-Orthogonal Forces
90o
F3
Now, we graphically break
each vector into its
orthogonal components.
F3y
F3x
F2
F
F2y
F2x
F1
F1y
F1 = 4.0 N at 50o
F2 = 6.5 N at 75o
F3 = 8.4 N at 30o
180o
0o
F1x
270o
Adding non-Orthogonal Forces
90o
Then, we remove the
original vectors and note
that the sum of the
components is the same
as the sum of the vectors.
F3y
F3x
F
F2y
F2x
F1y
F1 = 4.0 N at 50o
F2 = 6.5 N at 75o
F3 = 8.4 N at 30o
180o
F1x
270o
0o
Adding non-Orthogonal Forces
90o
This is the key result:
F3y
The sum of the xcomponents of the vectors
equals the x- component of
the Resultant.
F
As it is for the ycomponents.
F2y
Fy
F1y
F1x
F1 = 4.0 N at 50o
F2 = 6.5 N at 75o
F3 = 8.4 N at 30o
F2x
180o
F3x
0o
Fx
270o
Adding non-Orthogonal Forces
This explains how we will add vectors analytically.
1. Write the magnitude and direction (from 0 to 360 degrees) of each vector.
2. Compute the values of the x and y components.
3. Add the x components to find the x component of the Resultant.
4. Repeat for the y-components.
5. Use Pythagorean Theorem to find the magnitude of the Resultant.
6. Use Inverse Tangent to find the Resultant's direction.
Adding non-Orthogonal Forces
First make a table and enter what you know.
F1 = 8.0 N at 50o
F2 = 6.5 N at 75o
F3 = 8.4 N at 30o
Force
F1
F2
F3
Fnet
Magnitud Directio
n
e
x-comp. y-comp.
(Fcosθ) (Fsinθ)
Adding non-Orthogonal Forces
Then calculate the x and y components.
F1 = 8.0 N at 50o
F2 = 6.5 N at 75o
F3 = 8.4 N at 30o
Force
Magnitud Directio
n
e
F1
8.0 N
50o
F2
6.5 N
75o
F3
8.4 N
30o
Fnet
x-comp. y-comp.
(Fcosθ) (Fsinθ)
Adding non-Orthogonal Forces
Then add up the x and y components.
F1 = 8.0 N at 50o
F2 = 6.5 N at 75o
F3 = 8.4 N at 30o
Force
Magnitud Directio
n
e
F1
8.0 N
50o
5.14 N
6.13 N
F2
6.5 N
75o
1.68 N
6.28 N
F3
8.4 N
30o
7.27 N
4.2 N
Fnet
x-comp. y-comp.
(Fcosθ) (Fsinθ)
Adding non-Orthogonal Forces
Then use Pythagorean Theorem and tan-1 to determine Fnet.
F1 = 8.0 N at 50o
F2 = 6.5 N at 75o
F3 = 8.4 N at 30o
Force
Magnitud Directio
n
e
F1
8.0 N
50o
5.14 N
6.13 N
F2
6.5 N
75o
1.68 N
6.28 N
F3
8.4 N
30o
7.27 N
4.20 N
14.09 N
16.61 N
Fnet
x-comp. y-comp.
(Fcosθ) (Fsinθ)
Adding non-Orthogonal Forces
Let's add together these three forces:
F1 = 8.0 N at 50o
F2 = 6.5 N at 75o
F3 = 8.4 N at 30o
Force
Magnitud Directio
n
e
x-comp. y-comp.
(Fcosθ) (Fsinθ)
F1
8.0 N
50o
5.14 N
6.13 N
F2
6.5 N
75o
1.68 N
6.28 N
F3
8.4 N
30o
7.27 N
4.20 N
Fnet
22.9 N
50o
14.09 N
16.61 N
Adding non-Orthogonal Forces
Now use the same procedure to add these forces.
DO NOT IGNORE NEGATIVE SIGNS
F1 = 21 N at 60o
F2 = 65 N at 150o
F3 = 8.4 N at 330o
Adding non-Orthogonal Forces
F1 = 21 N at 60o
F2 = 65 N at 150o
F3 = 8.4 N at 330o
Forc
e
F1
Magnitude Direction
21 N
60o
F2
65 N
150o
F3
45 N
330o
Fnet
x-comp.
(Fcosθ)
y-comp. (Fsinθ)
Answers
F1 = 21 N at 60o
F2 = 65 N at 150o
F3 = 8.4 N at 330o
Forc
e
F1
Magnitude Direction
x-comp.
(Fcosθ)
y-comp. (Fsinθ)
21 N
60o
10.50 N
8.66 N
F2
65 N
150o
-56.29 N
32.50 N
F3
45 N
330o
38.97 N
-22.50 N
Fnet
131 N
290o
-6.82 N
18.66 N
17
Three forces act on an object. Which of the following is true in order
to keep the object in translational equilibrium?
I. The vector sum of the three forces must equal zero.
II. The magnitude of the three forces must be equal.
III. All three forces must be parallel.
A
I only
B
II only
C
I and III only
D
II and III only
E
I, II, and III
Force and friction acting on an object
Previously, we solved
problems with multiple
forces, but they were either
parallel or perpendicular.
For instance, draw the free
body diagram of the case
where a box is being pulled
along a surface, with friction,
at constant speed.
Force and friction acting on an object
Now find the acceleration
given that the applied
force is 20 N, the box
has a mass of 3.0kg, and
the coefficient of kinetic
friction is 0.20.
FN
FAPP
f
mg
Force and friction acting on an object
FN
x - axis
FAPP
f
ΣF = ma
ΣF = ma
FAPP - fk = ma
FN - mg = 0
FAPP - μkFN = ma
mg
FAPP = 20N
m = 3.0kg
μk = 0.20
y - axis
FN = mg
FAPP - μkmg = ma
FN = (3.0kg)(10m/s2)
a = (FAPP - μkmg)/m
FN = 30N
a = (20N - (0.20)(30N))/3.0kg
a = (20N - 6.0N)/3.0kg
a = (14N)/3.0kg
a = 4.7 m/s2
Forces at angles acting on an object
Now we will solve problems
where the forces act at an
angle so that it is not parallel
or perpendicular with one
another.
First we do a free body
diagram, just as we did
previously.
Forces at angles acting on an object
The next, critical, step is to
choose axes.
Previously, we always used
vertical and horizontal axes,
since one axis lined up with
the forces...and the
acceleration.
FN
f
mg
Now, we must choose axes so that all the acceleration is along one axis, and there
is no acceleration along the other.
You always have to ask, "In which direction could this object accelerate?" Then
make one axis along that direction, and the other perpendicular to that.
What's the answer in this case?
Forces at angles acting on an object
This time vertical and
horizontal axes still
work...since we assume the
box will slide along the
surface.
However, if this assumption is
wrong, we will get answers
that do not make sense, and
we will have to reconsider our
choice.
FN
f
mg
Forces at angles acting on an object
Now we have to break any
forces that do not line up with
our axes into components that
do.
FN
f
In this case, FAPP, must be
broken into
Fx and Fy
mg
Forces at angles acting on an object
Once we do that, we can
now proceed as we did
previously, using each
component appropriately.
FN
Fy
f
mg
Force and friction acting on an object
Let's use our work to find
the acceleration if the
applied force is 20 N at
37o above horizontal, the
box has a mass of 3.0kg,
and the coefficient of
kinetic friction is 0.20.
FN
Fy
f
mg
Force and friction acting on an object
FAPP = 20N at 37o
m = 3.0kg
μk = 0.20
FN
ΣF = ma
ΣF = ma
Fx - fk = ma
FN + Fy - mg = 0
Fx - μkFN = ma
FN = mg - Fy
Fcosθ - μkmg = ma
FN = mg - Fsinθ
a = (Fcosθ - μkFN)/m
FN = (3.0kg)(10m/s2)
- (20N)(sin37o)
FAPPy
f
mg
y - axis
x - axis
a = (20N cos37o
- (0.20)(18N))/3.0kg
FN = 30N - 12N
a = (16N - 3.6N)/3.0kg
FN = 18N
a = (12.4N)/3.0kg
Note that FN is lower due to the
force helping to support the
object
a = 4.1 m/s2
Normal Force and Friction
Friction was reduced because
the Normal Force was reduced;
the box's weight, mg, was
supported by the ycomponent of the force plus
the Normal Force...so the
Normal Force was
lowered...lowering friction.
FN
FAPPy
mg
Just looking at the y-axis
ΣF = ma
Fy
FN + Fy - mg = 0
FN = mg - Fy
FN = mg - Fsinθ
FN
mg
Normal Force and Friction
FN
What would happen with both
the Normal Force and Friction
in the case that the object is
being pushed along the floor
by a downward angled force?
f
mg
Normal Force and Friction
FN
In this case the pushing force
is also pushing the box into
the surface, increasing the
Normal Force as well as
friction.
f
Fy
Just looking at the y-axis
mg
FN
ΣF = ma
FN - Fy - mg = 0
FN = mg + Fy
mg
FN = mg + Fsinθ
Fy
The normal force on the box is:
18
A
mg
B
mg sin(θ)
C
mg cos(θ)
D
mg + F sin(θ)
E
mg – F sin(θ)
θ
Fapp
19
The frictional force on the box is:
A
μ(mg + Fsin(θ))
B
μ(mg - Fsin(θ))
C
μ(mg + Fcos(θ))
D
μ(mg - Fcos(θ))
E
μmg
θ
Fapp
20
A block of mass m is pulled along a horizontal surface at constant
speed v by a force Fapp , which acts at an angle of θ with the horizontal.
The coefficient of kinetic friction between the block and the surface is
μ.
The normal force exerted on the block by the surface is:
Fapp
A
mg - Fapp cosθ
B
mg - Fapp sinθ
C
mg
D
mg + Fapp sinθ
E
mg + Fapp cosθ
θ
v
m
21
A block of mass m is pulled along a horizontal surface at constant
speed v by a force Fapp , which acts at an angle of θ with the
horizontal. The coefficient of kinetic friction between the block and
the surface is μ.
The friction force on the block is:
Fapp
A
μ(mg - Fapp cosθ)
B
μ(mg - Fapp sinθ)
θ
v
C
μmg
D
μ(mg + Fapp sinθ)
E
μ(mg + Fapp cosθ)
m
Normal Force and Weight
FN
Previously we dealt
mostly with horizontal
(or, rarely, vertical
surfaces). In that case
FN, and mg were always
along the same axis.
Now we will look at the
more general case.
mg
Inclined Plane
On the picture, draw the
free body diagram for the
block.
Show the weight and the
normal force.
Inclined Plane
FN is ALWAYS perpendicular
to the surface.
FN
mg is ALWAYS directed
downward.
But now, they are neither
parallel or perpendicular to
one another.
mg
Choosing Axes
Previously, we used vertical
and horizontal axes. That
worked because problems
always resulted in an
acceleration that was along
one of those axes.
FN
We must choose axes along
which all the acceleration is
along one axis...and none
along the other.
mg
Choose of Axes
In this case, the block can only
accelerate along the surface of
the plane.
Even if there is no acceleration
in a problem, this is the ONLY
POSSIBLE direction of
acceleration.
So we rotate our x-y axes to
line up with the surface of the
plane.
FN
mg
Inclined Plane Problems
By the way, here's one way to see that θ is the both the angle of incline and the angle
between mg and the new y-axis.
Let's name the angle of the inclined plane θ
and the angle between mg and the x-axis θ'
and show that
θ = θ'
y-axis
FN
Since the angles in a triangle add to 180o,
and the bottom left angle is 90o, that means:
α + θ = 90o
θ' α
mg
θ
Inclined Plane Problems
Now look at the angles in the upper left corner of the triangle.
The surface of the inclined plane forms an
angle of 180o, so
y-axis
90o + θ' + α = 180o
FN
θ' + α = 90o
But we already showed that
α + θ = 90o
θ' α
θ' + α = 90o = α + θ
mg
θ' = θ
θ
Inclined Plane Problems
We need to resolve all forces which don't align with an axis into
components that will.
In this case, we resolve mg into its x and y components.
FN
FN
θ
θ
mg
mg
θ
Inclined Plane Problems
Now we just use Newton's Second Law, which is true for each axis.
x - axis
FN
ΣFx = max
ΣFy = may
mgsinθ = max
FN - mgcosθ = 0
a = gsinθ
FN = mgcosθ
θ
mg
y - axis
down the plane
Answer
ΣFx = max
θ
mg sin θ = ma
g sin θ = a
A 5 kg block slides down a frictionless incline at an angle
of 30 degrees.
a) Draw a free body diagram.
b) Find its acceleration.
(Use g = 10 m/s2)
a = g sin θ
a = 10 m/s2 sin (30o)
a = 10 m/s2 (0.5)
a = 5 m/s2
Inclined Plane Problems with Friction
We can add kinetic friction to our inclined plane example. The kinetic friction points
opposite the direction of motion.
We now have a second vector along the x axis. fk points in the negative direction
(recall fk = μk FN)
fk
FN
FN
fk
mg
θ
θ
mg
Inclined Plane Problems with Friction
x - axis
ΣFx = max
y - axis
ΣFy = may
FN = mg cosθ
mgsinθ - fk = ma
mgsinθ - μkFN = ma
mgsinθ - μkmgcosθ = ma
FN
fk
gsinθ - μkgcosθ = a
a = gsinθ - μkgcosθ
a = g(sinθ - μk cosθ)
θ
mg
Inclined Plane Problems with Friction
The general solution for objects sliding down an incline is:
a = g(sinθ - μkcosθ)
Note, that if there is no friction:
μk = 0
and we get our previous result for a frictionless plane:
a = gsinθ
Inclined Plane Problems with Friction
a = g(sinθ - μkcosθ)
If the object is sliding with constant velocity: a = 0
FN
fk
a = g(sinθ - μkcosθ)
0 = g(sinθ - μkcosθ)
a=0
0 = sinθ - μkcosθ
θ
mg
μk cosθ = sinθ
μk = sinθ / cosθ
μk = tanθ
Inclined Plane with Static Friction
We just showed that for an object sliding with
constant velocity down an inclined plane that:
μk = tanθ
FN
fs
a=0
Similarly, substituting μs for μk (at the maximum
static force; the largest angle of incline before
the object begins to slide) than:
μs = tanθmax
θ
mg
But this requires that the MAXIMUM ANGLE of
incline, θmax, be used to determine μs.
Answer
θ
ΣF = ma
mg sin θ - fK = 0
A 5 kg block slides down a incline at an angle of 30o
with a constant speed.
mg sin θ = fK
mg sin θ = μ FN
mg sin θ = μ mg cos θ
a) Draw a free body diagram.
b) Find the coefficient of friction
between the block and the incline.
(Use g = 10 m/s2)
μ = mg sin θ / mg cos θ
μ = tan θ
μ = tan 30 = 0.58
Answer
y-direction
ΣF = ma
FN - mg cos θ = 0
FN = mg cos θ
θ
A 5 kg block is pulled up an incline at an angle of 30o
at a constant velocity The coefficient of friction
between the block and the incline is 0.3.
x-direction
ΣF = ma
a) Draw a free body diagram.
b) Find the applied force.
(Use g = 10 m/s2)
Fapp - mg sin θ -μk FN = 0
Fapp - mg sin θ - fk = ma
Fapp = mg sin θ + μk mg cos θ
Fapp = 38 N
y-axis
ΣF = ma
FN - mg cos θ = 0
FN = mg cos θ
θ
A 5 kg block is pulled UP an incline at an
angle of 30 degrees with a force of 40 N.
The coefficient of friction between the
block and the incline is 0.3.
x-axis
ΣF = ma
a) Draw a free body diagram.
b) Find the block's acceleration.
(Use g = 10 m/s2)
Fapp - mg sin θ - μk FN = ma
Fapp - mg sin θ - fk = ma
Fapp - mg sin θ - μk mg cos θ = ma
a = (Fapp/m) - g sin θ - μk g cos θ
a = 0.4 m/s2
Inclined Plane Problems
If a mass, m, slides down a frictonless inclined plane, we have this general setup:
FN
It is helpful to rotate our
reference frame so that the +x
axis is parallel to the inclined
plane and the +y axis points
in the direction of FN.
mg
θ
y-direction
ΣF = ma
FN - mg cos θ = 0
FN = mg cos θ
θ
A 5 kg block remains stationary on an
incline. The coefficients of static and
kinetic friction are 0.4 and 0.3,
respectively.
x-direction
ΣF = ma
fs = mg sin θ
μs mg cos θ = mg sin θ
μs cos θ = sin θ
a) Draw a free body diagram.
b) Determine the angle that the block
will start to move.
(Use g = 10 m/s2)
μs = sin θ / cos θ
μs = tan θ
θ = tan-1 μs
θ = 21.8o
22
A block of mass m slides down a rough incline as shown. Which
free body diagram correctly shows the forces acting on the
block?
B
A
N
C
N
D
E
N
N
N
f
f
f
f
f
W
W
W
W
W
23
A block with a mass of 15 kg slides down a 43° incline as shown
above with an acceleration of 3 m/s2.
What is the normal force N applied
by the inclined plane on
the block?
A
95 N
B
100 N
C
105 N
D
110 N
E
115 N
24
A block with a mass of 15 kg slides down a 43° incline as shown
above with an acceleration of 3 m/s2.
The magnitude of the
frictional force along the
plane is nearly:
A
56 N
B
57 N
C
58 N
D
59 N
E
60 N
Static
Equilibrium
There is a whole field of problems called "Statics" that has to do
with cases where no acceleration occurs, objects remain at rest.
Anytime we construct something (bridges, buildings, houses,
etc.) we want them to remain stationary, not accelerate. So this
is a very important field.
The two types of static equilibrium are with respect to linear and
rotational acceleration, a balancing of force and of torque (forces
that cause objects to rotate). We'll look at them in that order.
Tension Force
Previously, we did problems where a rope supporting an
object exerted a vertical force straight upward, along the
same axis as the force mg was pulling it down. That led to
the simple case that if a = 0, then FT = mg
FT
mg
25
A uniform rope of weight 20 N hangs from a hook as shown above. A
box of mass 60 kg is suspended from the rope. What is the tension in
the rope?
A
20 N throughout the rope
B
120 N throughout the rope
C
200 N throughout the rope
D
600 N throughout the rope
E
It varies from 600 N at the bottom of the
rope to 620 N at the top.
Tension Force
But it is possible for two (or more) ropes to support an object
(a = 0) by exerting forces at angles. In that case:
T2
T1
The vertical components of the force exerted by each rope
must add up to mg.
And the horizontal components must add to zero.
mg
Tension Force
T2
T1
mg
So we need to break the forces into components that align
with our axes.
Tension Force
One result can be seen just from this diagram.
T2y
In order for the forces along the x-axis to cancel out, the
more vertical Tension must be much larger than the Tension
which is at a greater angle to the vertical.
T1y
T1x
T2x
mg
In problems with two supporting ropes or wires, the more
vertical has a greater tension.
Tension Force
20o
20o
50o
T1y
T1x
50o
T2x
mg
T2y
Let's calculate the tension in two ropes if T1 is at
an angle of 50o from the vertical and T2 is at an
angle of 20o from the vertical and they are
supporting a 8.0 kg mass.
Note, that in this case the horizontal component is
given by Tsinθ, where before it was given by Tcosθ.
To know which function to use you MUST draw the
diagram and find the sides opposite and adjacent
to the given angle.
Tension Force
20o
20o
50o
T1y
T2y
x - axis
ΣFx = max = 0
50o
T1x - T2x = 0
T1x
T2x
mg
T1sinθ1 = T2sinθ2
T1 = T2sinθ2/sinθ1
T1 = T2 (sin20o/sin50o)
T1 = T2 (0.34/0.77)
T1 = 0.44 T2
y - axis
ΣFy = may = 0
T1y + T2y - mg = 0
T1cosθ1 + T2cosθ2 = mg
(0.44T2)cos(50o) + T2cos(20o) = mg
(0.44T2)(0.64) + T2(0.94)
=(8kg)(9.8m/s2)
1.22 T2 = 78N
T2 = 78N / 1.22
T1 = 0.44 (64N)
T1 = 28N
T2 = 64N
Tension Force
θ
θ
Ty
Ty
θ
Tx
θ
Tx
mg
If the ropes form equal angles to the vertical, the
tension in each must also be equal, otherwise the
x- components of Tension would not add to zero.
Tension Force
x - axis
θ
y - axis
θ
Ty
Ty
θ
Tx
θ
ΣFx = max = 0
ΣFy = may = 0
T1x - T2x = 0
T1y + T2y - mg = 0
Tsinθ = Tsinθ
Tcosθ + Tcosθ = mg
2Tcosθ = mg
Tx
mg
which just confirms that if
the angles are equal, the
tensions are equal
T = mg / (2cosθ)
Note that the tension rises as
cosθ becomes smaller...which
occurs as θ approaches 90o.
It goes to infinity at 90o, which
shows that the ropes can never
be perfectly horizontal.
A lamp of mass m is suspended from two ropes of unequal length as
shown above. Which of the following is true about the tensions T1 and
T2 in the cables?
26
A
T1 > T2
B
T1 = T2
C
T1 < T2
D
T1 - T2 = mg
E
T1 + T2 = mg
T2
T1
27
A large mass m is suspended from two massless strings of an equal
length as shown below. The tension force in each string is:
A
½ mg cos(θ)
B
2 mg cos(θ)
C
mg cos(θ)
D
mg/cos(θ)
E
mg/2cos(θ)
θ θ
m
Torque and Rotational Equilibrium
Forces causes objects to linearly accelerate.
Torque causes objects to rotationally accelerate.
Rotational dynamics is a major topic of AP Physics C: Mechanics. It isn't
particularly difficult, but for AP B, we only need to understand the static case,
where all the torques cancel and add to zero.
First we need to know what torque is.
Torque and Rotational Equilibrium
Until now we've treated objects as points, we haven't been concerned with their
shape or extension in space.
We have assumed that any applied force acts through the center of the object and
it is free to accelerate. That does not result in rotation, just linear acceleration.
But if the force acts on an object so that it causes the object to rotate around its
center of mass...or around a pivot point, that force has exerted a torque on the
object.
Torque and Rotational Equilibrium
A good example is opening a door, making a
door rotate. The door does not accelerate in a
straight line, it rotates around its hinges.
Think of the best direction and location to push
on a heavy door to get it to rotate and you'll have
a good sense of how torque works.
Which force (blue arrow) placed at which
location would create the most rotational
acceleration of the green door about the black
hinge?
Torque and Rotational Equilibrium
The maximum torque is obtained from:
The largest force
At the greatest distance from the pivot
At an angle to the line to the pivot that
is closest to perpendicular
Mathematically, this becomes:
τ = Frsinθ
τ (tau) is the symbol for torque;
F is the applied force
r is the distance from the pivot
θ is the angle of the force to a line to the pivot
Torque and Rotational Equilibrium
τ = Frsinθ
F
When r decreases, so does the
torque for a given force. When r = 0,
τ = 0.
θ
r
When θ differs from 90o, the torque
decreases. When θ = 0o or 180o, τ =
0.
Two Physical Interpretations of Torque
τ = F(rsinθ)
τ = r(Fsinθ)
The torque due to F acting at angle θ
can be replaced by F acting at a
smaller distance rsinθ.
The torque due to F acting at angle θ
can be replaced by a smaller
perpendicular force Fsinθ acting at r.
F
These are all equivalent.
Fsinθ
F
θ
θ
r
F
r
rsinθ
r
θ
Fsinθ
Rotational Equilibrium
When the sum of the torques on an object is zero, the object is in rotational
equilibrium.
Define counter clockwise (CCW) as the positive direction for rotation and
clockwise (CW) as the negative.
For instance, what perpendicular force, F, must be applied at a distance of 7.0 m
for the pivot to exactly offset a 20N force acting at a distance of 4.0m from the
pivot of a door at an angle of 30o?
Rotational Equilibrium
20N
Στ = 0
30o
F1r1sinθ1 + F2r2sinθ2 =0
4.0m
F1
F1r1sinθ1 = - F2r2sinθ2
F1 = - F2r2sinθ2 / (r1sinθ1)
F1 = - (-20N)(4.0m)sin(30o) / ((7m)sin90o)
F1 = (80N-m)(0.50) / (7m)
F1 = 5.7 N
Rotational Equilibrium
1.0m
3.0m
4kg
4.0m
2kg
What mass must be added at distance 4.0m to put the above apparatus into
equilibrium?
Rotational Equilibrium
1.0m
3.0m
4kg
2kg
Στ = 0
F1r1sinθ1 + F2r2sinθ2 - F3r3sinθ2 = 0
m1gr1 + m2gr2 - m3gr3 = 0
m1r1 + m2r2 - m3r3 = 0
m3 = (m1r1 + m2r2) / r3
m3 = ((4kg)(3m) +(2kg)(1m)) / 4m
F1 = (14kg-m)) / (4m)
F1 = 3.5 kg
4.0m
15 kg
12 kg
A 12 kg load hangs from one end of a rope that passes over a small frictionless pulley. A 15 kg
counterweight is suspended from the other end of the rope. The system is released from rest.
a. Draw a free-body diagram for each object showing all applied forces in
diagram show the direction of the acceleration of that object.
b. Find the acceleration each mass.
c. What is the tension force in the rope?
d. What distance does the 12 kg load move in the first 3 s?
e. What is the velocity of 15 kg mass at the end of 5 s?
relative scale. Next to each
FT
FT
a.
a
a
m1
m2
15 kg
m1 = 12 kg
m2 = 15 kg
12 kg
m1g
m2g
b. ΣF = ma
ΣF = FT - m1g = m1a
FT = m1g + m1a
ΣF = FT - m2g = -m2a
FT = m2g - m2a
FT = FT
m1g + m1a = m2g - m2a
m1a + m2a = m2g - m1g
a(m1 + m2) = g(m2 - m1)
a = (g(m2 - m1)) / (m1 + m2)
= (10 m/s2 (15kg - 12kg)) / (12kg + 15kg)
= 30 m/s2 * kg / 27kg
a = 1.11 m/s2
c. FT = m1g + m1a or FT = m2g - m2a
FT = m1g + m1a
= 12kg(10 m/s2) + 12kg(1.11 m/s2)
= 120 N + 13.33 N
= 133.33 N
FT = m2g - m2a
= 15kg(10 m/s2) - 15kg(1.11 m/s2)
= 150 N - 16.67 N
= 133.33 N
d. x = xo + vot + 1/2 at2
xo = 0
vo = 0
x = 1/2 at2
= 1/2 (1.11 m/s2)(3s)2
=5m
e. v = vo + at
vo = 0
v = at
= (1.11 m/s2)(5s)
= 5.55 m/s
15 kg
12 kg
m1 = 12 kg
m2 = 15 kg
500 g
300 g
A 500 g block lies on a horizontal tabletop. The coefficient of kinetic friction between the block and the
surface is 0.25. The block is connected by a massless string to the second block with a mass of 300 g.
The string passes over a light frictionless pulley as shown above. The system is released from rest.
a. Draw clearly labeled free-body diagrams for each of the 500 g and the 300g masses. Include all forces
and draw them to relative scale. Draw the expected direction of acceleration next to each free-body
diagram.
b. Use Newton’s Second Law to write an equation for the 500 g mass.
c. Use Newton’s Second Law to write an equation for the 300 g mass.
d. Find the acceleration of the system by simultaneously solving the system
e. What is the tension force in the string?
of two equations.
m1 = 500 g = 0.5 kg
m2 = 300 g = 0.3 kg
μ = 0.25
d. FT - f = m1a
FT - m2g = - m2a
FT = f + m1a
FT = m2g - m2a
FN
a
FT
FT
f
a
f + m1a = m2g - m2a
m1a + m2a = m2g - f
a(m1 + m2) = m2g - f
a.
m1g
b. ΣF = ma
x-direction
ΣF = FT - f = m1a
f = μFn
f = μm1g
y-direction
ΣF = ma = 0
ΣF = FN - m1g = 0
FN = m1g
c. ΣF = ma
ΣF = FT - m2g = - m2a
m2g
a = (m2g - f) / (m1 + m2)
= (m2g - μm1g) / (m1 + m2)
= ((0.3kg)(10 m/s2) - (0.25)(0.5kg)(10m/s2)) / (0.5kg
+0.3kg)
= (3 - 1.25) kg*m/s2 / (0.8 kg)
= 2.1875 m/s2
e. FT = f + m1a or FT = m2g - m2a
FT = μm1g + m1a
FT = (0.25)(0.5kg)(10m/s2) + 0.5kg(2.1875 m/s2) = 2.34 N
FT = m2g - m2a
FT = 0.3kg(10m/s2) - 0.3kg(2.1875 m/s2)
= 2.34 N
F
A 10-kilogram block is pushed along a rough horizontal surface by a constant horizontal force F as
shown above. At time t = 0, the velocity v of the block is 6.0 meters per second in the same direction as
the force. The coefficient of sliding friction is 0.2. Assume g = 10 meters per second squared.
a. Calculate the force F necessary to keep the velocity constant.
m = 10 kg
v = 6 m/s
μ = 0.2
g = 10 m/s2
a. Calculate the force F necessary to keep the velocity constant.
x y
FN
fk
Fapp
mg
ΣF = ma ΣF = FN - mg = 0
Fapp - fk = 0
FN = mg
Fapp = fk
Fapp = μFN
Fapp = μmg
Fapp = (0.2)(10kg)(10m/s2)
Fapp = 20 N
A helicopter holding a 70-kilogram package suspended from a rope 5.0 meters long accelerates upward at a
rate of 5 m/s2. Neglect air resistance on the package.
a.
On the diagram, draw and label all of the forces acting on the package.
b.
Determine the tension in the rope.
c. When the upward velocity of the helicopter is 30 meters per second, the rope is cut and the helicopter
continues to accelerate upward at 5 m/s2.
Determine the distance between the helicopter and the package
2.0 seconds after the rope is cut.
m = 70 kg
a = 5 m/s2
x=5m
a.
On the diagram, draw and label all of the forces acting on the package.
FT
a
b.
Determine the tension in the rope.
ΣF = ma
FT - mg = ma
FT = mg + ma
FT = m (g+a)
FT = (70kg) (10 m/s2 + 5 m/s2)
FT = (70kg) (15 m/s2)
FT = 1050 N
mg
ahelicopter
m = 70 kg
a = 5 m/s2
x=5m
v = 30 m/s
c. When the upward velocity of the helicopter is 30 meters per
second, the rope is cut and the helicopter continues to accelerate
upward at 5.2 m/s2. Determine the distance between the helicopter and
the package 2 seconds after the rope is cut.
apackage
Method 1
Use the reference frame of the package after it is released.
The relative acceleration of the helicopter is given by:
arelative = ahelicopter - apackage
arelative = 5 m/s2 - (-10 m/s2)
arelative = 15 m/s2
Their relative initial velocity is zero, since they are connected together
The helicopter is 5.0 m above the package when released, the length of the rope.
y = y0 + v0 + 1/2at2
y = 5m + 0 + 1/2(15 m/s2)(2s)2
y= 35m
ahelicopter
m = 70 kg
a = 5 m/s2
x=5m
v = 30 m/s
c. When the upward velocity of the helicopter is 30 meters per
second, the rope is cut and the helicopter continues to accelerate
upward at 5.2 m/s2. Determine the distance between the helicopter
and the package 2 seconds after the rope is cut.
apackage
Method 2
Use the reference frame of the earth, but use the initial height of the package as zero.
Package Helicopter
y = y0 + v0t+ 1/2at2 y = y0 + v0t+ 1/2at2
y = 0 + (30 m/s)(2s) + 1/2(-10 m/s2)(2s)2 y = 5m + (30 m/s)(2s) + 1/2(+5 m/s2)(2s)2
y= 60 m - 20 m y= 5m + 60 m + 10 m
y= 40 m y= 75 m
Their separation is just the difference in their positions
Δy = yhelicopter - ypackage
Δy = 75m - 40m
Δy = 35m
1 kg
2 kg
4 kg
Three blocks of masses 1.0, 2.0, and 4.0 kilograms are connected by massless strings, one of which passes over a
frictionless pulley of negligible mass, as shown above. Calculate each of the following.
a.
The acceleration of the 4-kilogram block
b.
The tension in the string supporting the 4-kilogram block
c.
The tension in the string connected to the l-kilogram block
m1 = 1 kg
m2 = 2 kg
m3 = 4 kg
g = 9.8 m/s2
FT2
FN
FN
a
a
1 kg
FT1
FT1
2 kg
FT2
4 kg
a
m1g
a. The acceleration of the 4-kilogram
block
ΣF = ma
m1: m2: m3:
FT1 = m1a FT2 - FT1 = m2a FT2 - m3g = -m3a
FT2 - m1a = m2a FT2 - m3g = -m3a
FT2 = m1a + m2a FT2 = m3g - m3a
m1a + m2a = m3g - m3a
m1a + m2a + m3a = m3g
a(m1 + m2 + m3) = m3g
a = m3g / (m1 + m2 + m3)
a = (4kg)(9.8 m/s2) / (1kg + 2kg + 4kg)
a = 5.6 m/s2
m2g
m3g
m1 = 1 kg
m2 = 2 kg
m3 = 4 kg
g = 9.8 m/s2
b.
The tension in the string supporting the 4-kilogram block
FT2 = m1a + m2a
FT2 = 1kg(5.6 m/s2) + 2kg(5.6 m/s2)
FT2 = 16.8 N
or
FT2 = m3g - m3a
FT2 = 4kg(9.8 m/s2) - 4kg(5.6 m/s2)
FT2 = 16.8 N
c.
The tension in the string connected to the l-kilogram block
FT1 = m1a
FT1 = 1kg(5.6 m/s2)
FT1 = 5.6 N
10 kg
Case I
F
10 kg
Case II
5 kg
A 10-kilogram block rests initially on a table as shown in cases I and II above. The coefficient of sliding
friction between the block and the table is 0.2. The block is connected to a cord of negligible mass, which
hangs over a massless frictionless pulley. In case I a force of 50 newtons is applied to the cord. In case II an
object of mass 5 kilograms is hung on the bottom of the cord. Use g = 10 meters per second squared.
a. Calculate the acceleration of the 10-kilogram block in case I.
b. On the diagrams below, draw and label all the forces acting on each block in
c. Calculate the acceleration of the 10-kilogram block in case II.
d. Calculate the tension force in case II.
case II
m1 = 10 kg
m2 = 5 kg
μ = 0.2
FApp = 50 N
g = 10 m/s2
a.
Calculate the acceleration of the 10-kilogram block in case I.
ΣF = ma
FApp - fK = ma
FApp - μmg = ma
a = (FApp - μmg)/m
a = (50N - 0.2(10kg)(10m/s2))/10kg
a = 3 m/s2
b.
y
ΣF = ma
ΣF = FN - mg = 0
FN = mg
fk = μFN
fk = μmg
On the diagrams below, draw and label all the forces acting on each block in case II
FN
FT
f
FT
10 kg
5 kg
m2g
m1g
m1 = 10 kg
m2 = 5 kg
μ = 0.2
FApp = 50 N
g = 10 m/s2
c. Calculate the acceleration of the 10-kilogram block in case II.
ΣF = ma
FT - f = m1a FT - m2g = -m2a
FT - μm1g = m1a FT = m2g - m2a
FT = μm1g + m1a
μm1g + m1a = m2g - m2a
m1a + m2a = m2g - μm1g
a(m1 + m2) = g(m2 - μm1)
a = g(m2 - μm1)/(m1 + m2) = 10 m/s2(5kg - 0.2*10kg)/(5kg +10kg)
a = 2 m/s2
d. Calculate the tension force in case II.
FT - m2g = -m2a
FT = m2g - m2a
FT = m2(g - a) = 5kg (10 m/s2 - 2 m/s2) = 40 N
M1
M2
In the system shown above, the block of mass M1 is on a rough horizontal table. The string that attaches it to the block
of mass M2 passes over a frictionless pulley of negligible mass. The coefficient of kinetic friction mk between M1 and
the table is less than the coefficient of static friction ms
a. On the diagram below, draw and identify all the forces acting on the block of mass
b. In terms of M1 and M2 determine the minimum value of ms that will prevent the
M1.
blocks from moving.
The blocks are set in motion by giving M2 a momentary downward push. In terms of M1, M2, mk, and g, determine each
of the following:
c. The magnitude of the acceleration of M1
d. The tension in the string.
M1, M2, mk, ms, g
a.
On the diagram below, draw and identify all the forces acting on the block of mass M1.
FN
f
b.
FT
In terms of M1 and M2 determine the minimum value of ms that will prevent the blocks from moving.
The pulley system is not moving, so acceleration is zero.
M1g
ΣF = ma
x-direction for M1: FT - f = 0
FT = f
y-direction for M1: FN - M1g = 0
FN = M1g
y-direction for M2: FT - M2g = 0
FT = M2g
f = μsFN or msFN in this case
f = msM1g = M2g
ms = M2 / M1
The blocks are set in motion by giving M2 a momentary downward push. In terms of M1, M2, mk, and g, determine each
of the following:
c.
The magnitude of the acceleration of M1
ΣF = ma
x-direction for M1: FT - f = M1a
FT = f + M1a
y-direction for M1: FN - M1g = 0
FN = M1g
f = μkFN or mkFN in this case
f = mkM1g
y-direction for M2: FT - M2g = -M2a
FT = M2g - M2a
FT = f + M1a, FT = M2g - M2a
mkM1g + M1a = M2g - M2a
M1a + M2a = M2g - mkM1g
a(M1 + M2) = M2g - mkM1g
a = M2g - mkM1g / (M1 + M2)
d. The tension in the string.
FT = f + M1a or FT = M2g - M2a
FT = mkM1g + M1(M2g - mkM1g / (M1 + M2) )
FT = M2g - M2(M2g - mkM1g / (M1 + M2) )
or
m1
m2
60o
Two 10-kilogram boxes are connected by a massless string that passes over a massless frictionless pulley as
shown above. The boxes remain at rest, with the one on the right hanging vertically and the one on the left 2.0
meters from the bottom of an inclined plane that makes an angle of 60° with the horizontal. The coefficients of
kinetic friction and static friction between the Ieft-hand box and the plane are 0.15 and 0.30, respectively. You may
use g = 10 m/s2, sin 60° = 0.87, and cos 60° = 0.50.
a. What is the tension FT in the string.
b. On the diagram above, draw and label all the forces acting on the box that is
c. Determine the magnitude of the frictional force acting on the box on the
on the plane.
plane.
d. The string is then cut and the left-hand box slides down the inclined plane.
acceleration?
What is the magnitude of its
m1 = 10 kg
m2 = 10 kg
x=2m
θ = 60o , sin 60° = 0.87, cos 60° = 0.50
μk = 0.15, μs = 0.30
g = 10 m/s2
a. What is the tension FT in the string.
The acceleration is 0 because the system is not moving.
ΣF = ma
x-direction for m1: FT - f - m1gsinθ = 0
FT = f + m1gsinθ
y-direction for m1: FN - m1gcosθ = 0
FN = m1gcosθ
f = μsFN
f = μsm1gcosθ
FT = f + m1gsinθ
FT = μsm1gcosθ + m1gsinθ
FT = (0.30)(10kg)(10 m/s2)(0.50) + (10kg)(10 m/s2)(0.87)
FT = 102 N
m1 = 10 kg
m2 = 10 kg
x=2m
θ = 60o , sin 60° = 0.87, cos 60° = 0.50
μk = 0.15, μs = 0.30
g = 10 m/s2
b. On the diagram above, draw and label all the forces acting on the box that is on the plane.
FN
FT
m1
c. Determine the magnitude of the frictional force acting on the box on the plane.
y-direction for m1: FN - m1gcosθ = 0
FN = m1gcosθ
f = μsFN
f = μsm1gcosθ
f = 0.30(10kg)(10 m/s2)(0.50)
f = 15 N
f
m1g
60o
m1 = 10 kg
m2 = 10 kg
x=2m
θ = 60o , sin 60° = 0.87, cos 60° = 0.50
μk = 0.15, μs = 0.30
g = 10 m/s2
FN
f
m1
m2
a
m1g What is the magnitude of its acceleration?
d. The string is then cut and the left-hand box slides down the 60o
inclined plane.
ΣF = ma
x-direction of m1: f - m1gsinθ = -m1a
y-direction for m1: FN - m1gcosθ = 0
FN = m1gcosθ
f = μkFN = μkm1gcosθ
m1a = m1gsinθ - f
a = (m1gsinθ - μkm1gcosθ) / m1
a = ((10kg)(10 m/s2)(0.87) - (0.15)(10kg)(10 m/s2)(0.50)) / 10kg
a = 7.95 m/s2
m1
m2
M
v
θ
Masses ml and m2, are connected by a light string. They are further connected to a block of mass M by
another light string that passes over a pulley. Blocks l and 2 move with a constant velocity v down the inclined
plane, which makes an angle θ with the horizontal. The kinetic frictional force on block 1 is f and that on block
2 is 2f.
a. On the figure above, draw and label all the forces on block m1, m2, and M.
Express your answers to each of the following in terms of ml, m2, g, θ, and f.
b. Determine the coefficient of kinetic friction between the inclined plane and block 1.
c. Determine the value of the suspended mass M that allows blocks 1 and 2 to move with constant velocity
down the plane.
d. The string between blocks 1 and 2 is now cut. Determine the acceleration of block 1 while it is on the
inclined plane.
m2
m1
M
v
θ
a. On the figure above, draw and label all the forces on block m1, m2, and M.
FN
FN
f
f
FT1
a=0
a=0
m2
v
m1
FT1
v
θ
θ
m1g
FT2
v
M
a=0
Mg
m2g
FT2
Express your answers to each of the following in terms of ml, m2, g, θ, and f.
b. Determine the coefficient of kinetic friction between the inclined plane and block 1.
f = μFN
ΣF = ma
FN - m1gcosθ = 0
FN = m1gcosθ
f = μm1gcosθ
μ = f /(m1gcosθ)
c. Determine the value of the suspended mass M that allows blocks 1 and 2 to move with constant velocity down the plane.
m1: ΣF = ma M: FT2 - Mg = 0
FT1 - m1gsinθ - f = 0 FT2 = Mg
FT1 = m1gsinθ
m2: ΣF = ma
FT2 - FT1 - m2gsinθ - 2f = 0
Mg - m1gsinθ - f - m2gsinθ - 2f = 0
Mg = m1gsinθ + m2gsinθ + 3f
M = (m1gsinθ + m2gsinθ + 3f) / g
d. The string between blocks 1 and 2 is now cut. Determine the acceleration of block 1 while it is on the inclined plane.
FN
f
m1
a
x-direction: ΣF = ma
f - m1gsinθ = -m1a
m1gsinθ - f = m1a
a = (m1gsinθ - f) / m1
θ
m1g
Calculus Based
Fluid Resistance
When first learning dynamics discussed cases in which an object moves with
any friction. Then we began to take the frictional force into account between an
object and its surface. Now we are going to begin taking into account air
resistance.
v
Fresistance
Fluid Resistance
a
No resistance
t
v = vo + gt
vo = 0
g
v
a is constant
v increasing
t
mg = ma
y
a=g
t
Fluid Resistance
If we do not take into account air resistance then the object will fall with a
constant acceleration.
If air resistance however is taken into account then the objects
acceleration will eventually be cancelled out by the drag force.
Since the acceleration will essentially become zero, at some point in time
your velocity will be constant.
Fluid Resistance
With resistance
a=g
In this example the equation for the resistive
force is given as
F = kv, but in other problems the resisitive
force equation can be different.
vo=0
FR
FR
When the falling object stops accelerating it
reaches its terminal velocity.
28
If the equation for the force due to air resistance
is kv2. What is the terminal velocity reached by the object?
A
B
C
D
E
Velocity with respect to Time
Graphical Representation of the
Fluid Resistance Equation
x
v
a
t
t
t
Which of these is the graph that best represents the general velocity
versus time for air resistance?
29
A
v
B
v
t
t
C
D
v
t
v
t
We have learned before that distance is the integral of velocity so by
integrating the velocity equation we were able to come up with the distance
equation.
30
Which of these is the graph that best represents the general
position versus time for air resistance?
B
x
A
x
t
C
t
x
D
x
t
t
The final equation shows how acceleration approaches zero.
Which of these is the graph that best represents the general
acceleration versus time for air resistance?
31
a
a
A
B
t
C
a
t
D
a
t
t