Transcript Document

Topic 2: Mechanics
2.2 Forces and dynamics
Mechanics is the branch of physics which concerns
itself with forces, and how they affect a body's
motion.
Kinematics is the sub-branch of mechanics which
studies only a body's motion without regard to
causes.
Dynamics is the sub-branch of mechanics which
studies the forces which cause a body's motion.
The two pillars of
mechanics
Galileo
Kinematics
Topic 2.1
Newton
Dynamics
Topic 2.2
Topic 2: Mechanics
2.2 Forces and dynamics
2.2.1 Calculate the weight of a body using the
expression W = mg.
2.2.2 Identify the forces acting on an object and
draw free-body diagrams representing those
forces.
2.2.3 Determine the resultant force in different
situations.
2.2.4 State Newton’s first law.
2.2.5 Describe examples of Newton’s first law.
2.2.6 State the condition for translational
equilibrium.
2.2.7 Solve problems involving translational
equilibrium.
Topic 2: Mechanics
2.2 Forces and dynamics
Calculate the weight of a body using the
expression W = mg. Identify the forces acting on
an object and draw free-body diagrams
representing those forces.
A force is a push or a pull measured in newtons.
One force we are very familiar with is the force
of gravity, AKA the weight.
The very concepts of push and pull imply
direction. The direction of the weight is down
toward the center of the earth.
If you have a weight of 90 newtons (or 90 N),
your weight can be expressed as a vector, 90 N,
down.
We will show later that weight has the formula
weight
where g = 10 m s-2
W = mg
and m is the mass in kg
Topic 2: Mechanics
2.2 Forces and dynamics
Calculate the weight of a body using the
expression W = mg. Identify the forces acting on
an object and draw free-body diagrams
representing those forces.
EXAMPLE: Calculate the weight of a 25-kg
object.
SOLUTION:
Since m = 25 kg and g = 10 m s-2,
W = mg = (25)(10) = 250 N (or 250 n).
Note that W inherits its direction from
the fact that g points downward.
We sketch the mass as a dot, and the
weight as a vector in a free body diagram:
Free body
diagram
mass
force
W = mg
weight
where g = 10 m s-2
and m is the mass in kg
W
Topic 2: Mechanics
2.2 Forces and dynamics
Identify the forces acting on an object and draw
free-body diagrams representing those forces.
Certainly there are other forces besides
weight that you are familiar with.
For example, when you set a mass on a
tabletop, even though it stops moving, it
still has a weight. The implication is that
the tabletop applies a counterforce to the
weight, called a normal force.
Note that the weight and the normal forces
are the same length – they balance.
The normal force is called a surface
contact force.
N
W
Topic 2: Mechanics
2.2 Forces and dynamics
Identify the forces acting on an object and draw
free-body diagrams representing those forces.
Tension can only be a pull and never a push.
Friction tries to oppose the motion.
Friction is parallel to the contact surface.
Normal is perpendicular to the contact surface.
Friction and normal are mutually perpendicular.
Friction and normal are surface contact forces.
N
T
f
the tension
Contact surface
W
Topic 2: Mechanics
2.2 Forces and dynamics
Identify the forces acting on an object and draw
free-body diagrams representing those forces.
Weight is drawn from the center of an object.
Normal is always drawn from the contact surface.
Friction is drawn along the contact surface.
Tension is drawn at whatever angle is given.
N
T
f
W
Topic 2: Mechanics
2.2 Forces and dynamics
W
f
W
Free body diagram
Identify the forces acting on an object and draw
free-body diagrams representing those forces.
EXAMPLE: An object has a tension acting on it at
30° as shown. Sketch in the forces, and draw a
free body diagram.
N
SOLUTION:
Weight from center, down.
T
Normal from surface, up.
30°
f
Friction from surface, N
T
parallel.
30°
Topic 2: Mechanics
2.2 Forces and dynamics
Determine the resultant force in different
situations.
The resultant (or net) force is just the vector
sum of all of the forces acting on a body.
EXAMPLE: An object has mass of 25 kg. A tension
of 50 n and a friction force of 30 n are acting
on it as shown. What is the resultant force?
SOLUTION:
Since the weight and the normal
N
forces cancel out in the ydirection, we only need to worry
about the forces in the x50 n
direction.
f
The net force is thus
30 n
50 – 30 = 20 n (+x-dir)
W
T
Topic 2: Mechanics
2.2 Forces and dynamics
Determine the resultant force in different
situations.
The resultant (or net) force is just the vector
sum of all of the forces acting on a body.
30 n
Fy,net = Fy
Fx,net = Fx
Fnet = F
net force
EXAMPLE: An object has exactly two forces F1 = 50
n and F2 = 30 n applied simultaneously to it. What
is the resultant force’s magnitude?
SOLUTION:
F2
Fnet = F = F1 + F2 so we
simply graphically add the
two vectors:
The magnitude is just given
by Fnet2 = 502 + 302 so that
50 n F1
Fnet = 58 n.
Topic 2: Mechanics
2.2 Forces and dynamics
Determine the resultant force in different
situations.
The resultant (or net) force is just the vector
sum of all of the forces acting on a body.
F2
30 n
Fy,net = Fy
Fx,net = Fx
Fnet = F
net force
EXAMPLE: An object has exactly two forces F1 = 50
n and F2 = 30 n applied simultaneously to it as
shown. What is the resultant force’s direction?
SOLUTION:
Direction is measured from the x-axis
traditionally.
Opposite and adjacent are given

directly, so use tangent.
50 n F1
tan  = opp/adj = 30/50 = 0.6 so that
 = tan-1(0.6) = 31°.
Topic 2: Mechanics
2.2 Forces and dynamics
Determine the resultant force in different
situations.
The resultant (or net) force is just the vector
sum of all of the forces acting on a body.
30 n
Fy,net = Fy
Fx,net = Fx
Fnet = F
net force
EXAMPLE: An object has exactly two forces F1 = 50
n and F2 = 30 n applied simultaneously to it. What
is the resultant force’s magnitude?
SOLUTION:
F2
Begin by resolving F1 into
its x- and y-components.
Then Fnet,x = 44 n and
50sin28
Fnet,y = 23 + 30 = 53 n.
23 n
28°
Fnet2 = Fnet,x2 + Fnet,y2 so that
50cos28
Fnet2 = 442 + 532, Fnet = 70 n.
44 n
Topic 2: Mechanics
2.2 Forces and dynamics
State Newton’s first law.
Newton’s first law is related to certain studies
made by Galileo Galilee which contradicted
Aristotelian tenets.
Aristotle basically said
“The natural state of motion of all objects
(except the heavenly ones) is one of rest.”
A child will learn that if you stop pushing a
cart, the cart will eventually stop moving.
This simple observation will lead the child to
come up with a force law that looks something
like this:
“In order for a body to be in motion, there must
be a force acting on it.”
As we will show on the next slide, both of these
statements is false!
Topic 2: Mechanics
2.2 Forces and dynamics
Inertia will
only
change if
there is a
force.
State Newton’s first law.
Here’s how Galileo (1564-1642) thought:
If I give a cart a push on a smooth, level
surface, it will eventually stop.
What can I do to increase the distance without
pushing it harder?
If I can minimize the friction, it’ll go farther.
In fact, he reasoned, if I eliminate the friction
altogether the cart will roll forever!
Galileo called the tendency of an object to not
change its state of motion inertia.
A body’s inertia will
only change if there
Topic 2: Mechanics
is a net force
2.2 Forces and dynamics applied to it.
State Newton’s first law. State the condition for
translational equilibrium.
Newton’s first law is drawn from his concept of
net force and Galileo’s concept of inertia.
Essentially, Newton’s first law says that the
velocity of an object will not change if there is
no net force acting on it.
v = 0
In his words...
Every body continues in its state of rest, or
of uniform motion in a straight line, unless
it is compelled to change that state by
forces impressed thereon.
v = CONST
F
In symbols...
If F = 0, then v = CONST.
Newton’s first law
The above equation is known as the condition for
translational equilibrium.
Topic 2: Mechanics
2.2 Forces and dynamics
Describe examples of Newton’s first law.
As a memorable demonstration of inertia –
matter’s tendency to not change its state of
motion (or its state of rest) - consider this:
A water balloon is cut very rapidly with a knife.
For an instant the water remains at rest!
Don’t try this at home, kids.
30°
Topic 2: Mechanics
2.2 Forces and dynamics
T1
45° T
2
T3
m
Solve translational equilibrium problems.
EXAMPLE: An object of mass m is hanging
via three cords as shown. Find the tension
in each of the three cords, in terms of m.
SOLUTION:
Give each tension a name to organize your
effort.
Draw a free body diagram of the mass and
the knot.
T3 is the easiest force to find. Why?
m is not moving so its FBD tells us that
Fy = 0 or T3 – mg = 0 or T3 = mg .
T3
mg
FBD, m
T2
T1
30°
45°
T3
FBD, knot
30°
Topic 2: Mechanics
2.2 Forces and dynamics
T1
45° T
2
T3
Solve translational equilibrium problems.
EXAMPLE: An object of mass m is hanging
via three cords as shown. Find the tension
in each of the three cords, in terms of m.
T3
SOLUTION:
T3 = mg
Now we break T1 and T2 down to components.
Looking at FBD knot we see that
T1x = T1 cos 30° = 0.866T1
T1y = T1 sin 30° = 0.500T1
T2x = T2 cos 45° = 0.707T2
T1
T2y = T2 sin 45° = 0.707T2
30°
mg
FBD, m
T2
45°
T3
FBD, knot
30°
Topic 2: Mechanics
2.2 Forces and dynamics
T1
45° T
2
T3
Solve translational equilibrium problems.
EXAMPLE: An object of mass m is hanging
via three cords as shown. Find the tension
in each of the three cords, in terms of m.
SOLUTION:
T3 = mg
Putting this into the FBD of knot we get:
∑Fx = 0
0.707T2 - 0.866T1 = 0
T2 = 1.225T1
∑Fy = 0
0.707T2 + 0.500T1 - T3 = 0
0.707(1.225T1) + 0.500T1 = T3
T1 = mg/1.366
T2 = 1.225(mg/1.366)
T2 = 0.897mg
T3
mg
FBD, m
T2
T1
30°
45°
T3
FBD, knot
Topic 2: Mechanics
2.2 Forces and dynamics
Solve translational equilibrium problems.
EXAMPLE: A 25-kg mass is hanging via three cords
as shown. Find the tension in each of the three
cords, in newtons.
30°
T1
SOLUTION:
Since all of the angles are the same use the
formulas we just derived:
T3 = mg = 25(10) = 250 n
T1 = mg/1.366 = 25(10)/1.366 = 180 n
T2 = 0.897mg = 0.897(25)(10) = 220 n
45° T
T3
FYI
This is an example of using Newton’s first law
with v = 0. The next example shows how to use
Newton’s first when v is constant, but not zero.
2
Topic 2: Mechanics
2.2 Forces and dynamics
Solve translational equilibrium problems.
EXAMPLE: A 1000-kg airplane is flying at a
constant velocity of 125 m s-1. Label and
determine the value of the weight W, the lift L,
the drag D and the thrust F if the drag is 25000
L
n.
F
D
SOLUTION:
W
Since the velocity is constant, Newton’s first
law applies. Thus Fx = 0 and Fy = 0.
W = mg = 1000(10) = 10000 n and points down.
L = W = 10000 n and points up since Fy = 0.
D tries to impede the aircraft and points left.
F = D = 25000 n and points right since Fx = 0.
Topic 2: Mechanics
2.2 Forces and dynamics
2.2.8 State Newton’s second law of motion.
2.2.9 Solve problems using Newton’s second law.
Topic 2: Mechanics
2.2 Forces and dynamics
a = F/m
State Newton’s second law of motion.
Newton reasoned: “If the sum of the forces is
not zero, the velocity will change.”
But we know, and he also did, that a change in
velocity is an acceleration.
So Newton then asked himself: “How is the sum of
the forces related to the acceleration.”
Here is what Newton said: “The acceleration of an
object is proportional to the net force acting on
it, and inversely proportional to its mass.”
In other words, the bigger the force the bigger
the acceleration, and the bigger the mass the
smaller the acceleration.
In formula form
(or F = ma )
Fnet = ma
Newton’s second law
Topic 2: Mechanics
2.2 Forces and dynamics
State Newton’s second law of motion.
(or F = ma )
Fnet = ma
Newton’s second law
Looking at the form F = ma note that if a = 0
then F = 0.
But if a = 0, v = CONST.
Thus Newton’s first law is just a special case of
his second.
Topic 2: Mechanics
2.2 Forces and dynamics
Solve problems using Newton’s second law.
(or F = ma )
Fnet = ma
Newton’s second law
EXAMPLE: An object has mass of 25 kg. A tension
of 50 n and a friction force of 30 n are acting
on it as shown. What is its acceleration?
N
50 n
f
30 n
SOLUTION:
The vertical forces W and N cancel out. W
The net force is thus 50 – 30 = 20 n (+x-dir).
From Fnet = ma we get 20 = 25a so that
a = 20 / 25 = 0.8 m s-2.
T
Topic 2: Mechanics
2.2 Forces and dynamics
Solve problems using Newton’s second law.
(or F = ma )
Fnet = ma
Newton’s second law
PRACTICE: Use F = ma to show that the formula for
weight is correct.
F = ma.
But F is the weight W.
And a is the freefall acceleration g.
Thus F = ma becomes W = mg.
Topic 2: Mechanics
2.2 Forces and dynamics
Solve problems using Newton’s second law.
(or F = ma )
Fnet = ma
Newton’s second law
EXAMPLE: A 1000-kg airplane is flying in
perfectly level flight. The drag D is 25000 n and
the thrust F is 40000 n. Find its acceleration.
L
D
SOLUTION:
W
Since the flight is level, Fy = 0.
Fx = F – D = 40000 – 25000 = 15000 n = Fnet.
From Fnet = ma we get
15000 = 1000a
a = 15000 / 1000 = 15 m s-2.
F
Topic 2: Mechanics
2.2 Forces and dynamics
Solve problems using Newton’s second law.
(or F = ma )
Fnet = ma
Newton’s second law
30 n
EXAMPLE: A 25-kg object has exactly two forces F1
= 40. n and F2 = 30. n applied simultaneously to
it. What is the object’s acceleration?
SOLUTION:
F2
Begin by resolving F1 into its
x- and y-components.
Then Fnet,x = 36 n and
40sin25
Fnet,y = 17 + 30 = 47 n.
17 n
25°
Fnet2 = Fnet,x2 + Fnet,y2 so that
40cos25
Fnet2 = 362 + 472, Fnet = 59 n.
36 n
Then from Fnet = ma we get
59 = 25a
a = 59 / 25 = 2.4 m s-2.
Topic 2: Mechanics
2.2 Forces and dynamics
Solve problems using Newton’s second law.
(or F = ma )
Fnet = ma
Newton’s second law
6.0 m
EXAMPLE: A 25-kg object resting
N
on a frictionless incline is
released, as shown. What is its
acceleration?
60
30°
SOLUTION:
mg cos 30
Begin with a FBD.
mg
mg sin 30
Break down the weight into its components.
Since N and mg cos 30° are perpendicular to the
path of the crate they do NOT contribute to its
acceleration.

Fnet = ma
mg sin 30° = ma
a = 10 sin 30° = 5.0 m s-2.
Topic 2: Mechanics
2.2 Forces and dynamics
6.0 m
Solve problems using Newton’s second law.
(or F = ma )
Fnet = ma
Newton’s second law
u = 0
EXAMPLE: A 25-kg object resting
on a frictionless incline is
released, as shown. What is its
speed at the bottom?
v = ?
30°
SOLUTION:
We found that its acceleration is 5.0 m s-2.
We will use the timeless equation to find v so
we need to know what s is.
From trigonometry, we have opposite and we want
hypotenuse so we use sin  = opp/hyp. Thus
s = hyp = opp/sin  = 6 / sin 30° = 12 m.
v2 = u2 + 2as = 02 + 2(5)(12)
v = 11 m s-1.
Topic 2: Mechanics
2.2 Forces and dynamics
2.2.10 Define linear momentum and impulse.
2.2.11 Determine the impulse due to a timevarying force by interpreting a force-time
graph.
2.2.12 State the law of conservation of linear
momentum.
2.2.13 Solve momentum and impulse problems.
Topic 2: Mechanics
2.2 Forces and dynamics
Define linear momentum and impulse.
Linear momentum, p, is defined to be the product
of an object’s mass m with its velocity v.
p = mv
linear momentum
Its units are obtained directly from the formula
and are kg m s-1.
EXAMPLE: What is the linear
momentum of a 4.0-gram NATO SS
109 bullet traveling at 950 m/s?
SOLUTION:
Convert grams to kg (jump 3
decimal places left) to get m =
.004 kg.
Then p = mv = (.004)(950) = 3.8
kg m s-1.
Topic 2: Mechanics
2.2 Forces and dynamics
Define linear momentum and impulse.
p = mv
linear momentum
From Fnet = ma we can get
Fnet = m∆v/∆t
Fnet = ∆p/∆t
This last is just Newton’s second law in terms
of change in momentum rather than mass and
acceleration.
Fnet = ∆p/∆t
Newton’s second law (p-form)
EXAMPLE: A 6-kg object has its speed increase
from 5 m s-1 to 25 m s-1 in 30 s. What is the net
force acting on it?
SOLUTION:
Fnet = ∆p/∆t = m(v – u)/∆t
= 6(25 – 5)/30 = 4 n.
Topic 2: Mechanics
2.2 Forces and dynamics
Define linear momentum and impulse.
Fnet = ∆p/∆t
Newton’s second law (p-form)
If we manipulate Newton’s second law (p-form) to
isolate the change in momentum we get:
Fnet = ∆p/∆t
Fnet∆t = ∆p
We call the force times the time the impulse J.
J = Fnet∆t = ∆p
impulse
EXAMPLE: A baseball has an average force of 12000
n applied to it for 25 ms. What is the impulse
imparted to the ball from the bat?
SOLUTION:
FYI
J = Fnet∆t
The units for impulse
are the units of force
= 12000(2510-3)
(n) times time (s).
= 300 n s.
Topic 2: Mechanics
2.2 Forces and dynamics
Determine the impulse due to a time-varying force
by interpreting a force-time graph.
J = Fnet∆t = ∆p
impulse
The impulse is the area under an F vs. t graph.
EXAMPLE: A bat striking a ball imparts a force to
it as shown in the graph. Find the impulse.
Force F/n
SOLUTION:
Break the graph into simple areas of rectangles
9
and triangles.
A1 = (1/2)(3)(9) = 13.5 n s
6
A2 = (4)(9) = 36 n s
3
A3 = (1/2)(3)(9) = 13.5 n s
Atot = A1 + A2 + A3
0
5
Atot = 13.5 + 36 + 13.5 = 63 n s. 0
Time t/s
10
Topic 2: Mechanics
2.2 Forces and dynamics
State the law of conservation of linear momentum.
Recall Newton’s second law (p-form):
Fnet = ∆p/∆t
Newton’s second law (p-form)
If the net force acting on an object is zero, we
have
Fnet = ∆p/∆t
0 = ∆p/∆t
0 = ∆p
In words, if the net force is zero, then the
momentum does not change – p is constant.
In symbols
If Fnet = 0 then p = CONST
conservation of linear
momentum
Topic 2: Mechanics
2.2 Forces and dynamics
2.2.14 State Newton’s third law.
2.2.15 Discuss examples of Newton’s third law.
Students should understand that when two
bodies interact, the forces they exert on each
other are equal and opposite.
Topic 2: Mechanics
2.2 Forces and dynamics
State Newton’s third law. Discuss examples of
Newton’s third law.
In words
“For every action force there is an equal and
opposite reaction force.”
In symbols
FAB = -FBA
Newton’s third law
FAB is the force on body A by body B.
FBA is the force on body B by body A.
In the big picture, if every force in the
universe has a reaction force that is equal and
opposite, the net force in the whole universe is
zero!
So why are there accelerations all around us?
Topic 2: Mechanics
2.2 Forces and dynamics
State Newton’s third law. Discuss examples of
Newton’s third law.
EXAMPLE: Consider a door. When you
push on a door with 10 n, because
of Newton’s third law the door
FBA
FAB
pushes on your hand with the same the
B
10 n, but in the opposite
door’s
your
direction. Why does the door move, reaction
action
and you don’t?
A
A
SOLUTION:
Even though the forces are equal
and opposite, they are acting on different
bodies.
Each body acts in response only to the force
acting on it.
The door can’t resist FAB, but you CAN resist FBA.
Topic 2: Mechanics
2.2 Forces and dynamics
FBE
State Newton’s third law. Discuss examples of
NBT
Newton’s third law.
EXAMPLE: Consider a baseball
resting on a tabletop. Discuss all
of the forces acting on the
NTB
baseball, and their reactions.
SOLUTION:
Acting on the ball is its weight
FEB
FBE and the normal force on the
ball caused by the table NBT.
The reactions are FEB and NTB.
Note that FBE (the weight force) and NBT (the
normal force) are acting on the ball.
NBT (the normal force) acts on the table.
FEB (the weight force) acts on the earth.
Topic 2: Mechanics
2.2 Forces and dynamics
State Newton’s third law. Discuss examples of
Newton’s third law.
We define a system as a collection of more than
one body, mutually interacting with each other.
EXAMPLE: Three billiard balls interacting on a
pool table constitute a system.
The action-reaction force pairs between the
balls are called internal forces.
For any system all internal forces cancel!
Topic 2: Mechanics
2.2 Forces and dynamics
State Newton’s third law. Discuss examples of
Newton’s third law.
We define a system as a collection of more than
one body, mutually interacting with each other.
EXAMPLE: Three billiard balls interacting on a
pool table constitute a system. Label and discuss
all of the internal forces.
The internal force pairs only exist while the
balls are in contact with one another.
Note that a blue and a red force act on the
white ball. It responds only to those two forces.
Note that a single white force acts on the red
ball. It responds only to that single force.
Note that a single white force acts on the blue
ball. It responds only to that single force.
Topic 2: Mechanics
2.2 Forces and dynamics
State Newton’s third law. Discuss examples of
Newton’s third law.
We define a system as a collection of more than
one body, mutually interacting with each other.
EXAMPLE: Three billiard balls interacting on a
pool table constitute a system. Describe the
external forces.
External forces are the forces that the balls
feel from external origins.
For billiard balls, these forces are the balls’
weights, the cushion forces, and the queue stick
forces.
Topic 2: Mechanics
2.2 Forces and dynamics
State the law of conservation of linear momentum.
In light of Newton’s third law and a knowledge of
systems, the conservation of linear momentum can
be refined for a system of particles.
If Fnet = 0 then p = CONST
conservation of linear
momentum
Since in a system all of the internal forces sum
up to zero, Fnet can only be the sum of the
external forces. Thus
If Fnet,ext = 0 then p = CONST
conservation of linear
momentum - system
In other words, internal forces cannot change the
momentum of a system of particles – ever!
Topic 2: Mechanics
2.2 Forces and dynamics
Solve momentum and impulse problems.
If Fnet,ext = 0 then p = CONST
conservation of linear
momentum - system
EXAMPLE: A 12-kg block of ice at rest has a fire
cracker inside a hole drilled in its center. When
it explodes, the block breaks into 2 pieces, one
of which travels at +16 m s-1 in the x-direction.
What is the velocity of the other 8.0 kg piece?
SOLUTION:
8 4
Make before and after sketches.
The initial velocity of the two is 0.
4
8
16
From conservation of momentum we havev
p = CONST which means p0 = pf. Since p = mv,
(8+4)(0) = 8v + 4(16)
so that v = -8 m s-1.
Topic 2: Mechanics
2.2 Forces and dynamics
Solve momentum and impulse problems.
If Fnet,ext = 0 then p = CONST
conservation of linear
momentum - system
EXAMPLE: A 730-kg Smart Car traveling at 25 m s-1
(x-dir) collides with a stationary 1800-kg Dodge
Charger. The two vehicles stick together. Find
their velocity immediately after the collision.
v
25
0
SOLUTION:
730
Make before and after sketches. 730
1800
+1800
p0 = pf so that (730)(25) = (730 + 1800)v
18250 = 2530v
v = 18250/2530 = 7.2 m s-1.