Download Lecture-06-09

Lecture 5
Newton’s Laws of Motion
(sections 5.5-5.7)
Exam #1 - next Thursday!
-
21 multiple-choice problems
- A calculator will be needed.
- CHECK YOUR BATTERIES!
- NO equations or information may be stored in
your calculator. This is part of your pledge on the
exam.
- Scratch paper will be provided, to be turned in at
the end of the exam.
- A sign-in sheets will be used, photographs of the
class will be taken, all test papers will be collected
A 71-kg parent and a 19-kg child meet at the center of an ice rink. They
place their hands together and push.
(a) Is the force experienced by the child more than, less than, or the
same as the force experienced by the parent?
(b) Is the acceleration of the child more than, less than, or the same as
the acceleration of the parent? Explain.
(c) If the acceleration of the child is 2.6 m/s2 in magnitude, what is the
magnitude of the parent’s acceleration?
3
4
On vacation, your 1300-kg car pulls a 540-kg trailer away from a
stoplight with an acceleration of 1.9 m/s2
(a) What is the net force exerted by the car on the trailer?
(b) What force does the trailer exert on the car?
(c) What is the net force acting on the car?
5
6
Collision Course I
a) the car
A small car collides with
a large truck. Which
experiences the greater
impact force?
b) the truck
c) both the same
d) it depends on the velocity of each
e) it depends on the mass of each
Collision Course I
a) the car
A small car collides with
a large truck. Which
experiences the greater
impact force?
b) the truck
c) both the same
d) it depends on the velocity of each
e) it depends on the mass of each
According to Newton’s Third Law, both vehicles experience
the same magnitude of force.
Force and Two Masses
a) ¾a1
A force F acts on mass m1 giving acceleration a1.
b) 3/2a1
The same force acts on a different mass m2
giving acceleration a2 = 2a1. If m1 and m2 are
c) ½a1
glued together and the same force F acts on this
d) 4/3a1
combination, what is the resulting acceleration?
e) 2/3a1
F
F
F
m1
a1
F = m1 a 1
a2 = 2a1
m2
F = m2 a 2
m2
m1
a3
a3 = ?
Force and Two Masses
a) ¾a1
A force F acts on mass m1 giving acceleration a1.
b) 3/2a1
The same force acts on a different mass m2
giving acceleration a2 = 2a1. If m1 and m2 are
c) ½a1
glued together and the same force F acts on this
d) 4/3a1
combination, what is the resulting acceleration?
e) 2/3a1
F
F
m1
a1
F = m1 a1
a2 = 2a1
m2
F = m2 a2 = (1/2 m1 )(2a1 )
Mass m2 must be ( m1) because its
acceleration was 2a1 with the same
force. Adding the two masses
together gives ( )m1, leading to an
F
m2
m1
a3
F = (3/2)m1 a3 => a3 = (2/3) a1
acceleration of ( )a1 for the same
applied force.
Newton’s Laws
(I)
In order to change the velocity of an object –
magnitude or direction – a net force is required.
(II)
(III)
Newton’s third law: If object 1 exerts a force on
object 2, then object 2 exerts a force – on object 1.
An archer shoots a 0.022-kg arrow at a target with a speed of 57 m/s.
When it hits the target, it penetrates to a depth of 0.085 m.
(a) What was the average force exerted by the target on the arrow?
(b) If the mass of the arrow is doubled, and the force exerted by the
target on the arrow remains the same, by what multiplicative factor
does the penetration depth change? Explain.
12
13
When you lift a bowling ball with a force of 82 N, the ball accelerates upward
with an acceleration a. If you lift with a force of 92 N, the ball’s acceleration
is 2a. Find (a) the weight of the bowling ball, and (b) the acceleration a.
When you lift a bowling ball with a force of 82 N, the ball accelerates upward
with an acceleration a. If you lift with a force of 92 N, the ball’s acceleration
is 2a. Find (a) the weight of the bowling ball, and (b) the acceleration a.
a)
b)
Contact Force
Two blocks of masses 2m and m
a) 2F
are in contact on a horizontal
b) F
frictionless surface. If a force F0
c) ½ F
is applied to mass 2m, what is the
d) 1/3 F
force on mass m ?
e) ¼ F
F
2m
16
m
Contact Force
Two blocks of masses 2m and m
a) 2F
are in contact on a horizontal
b) F
frictionless surface. If a force F0
c) ½ F
is applied to mass 2m, what is the
d) 1/3 F
force on mass m ?
e) ¼ F
The force F0 leads to a specific
acceleration of the entire system. In
order for mass m to accelerate at the
F
2m
same rate, the force on it must be
smaller! For the two blocks together,
F0 = (3m)a. Since a is the same for both
blocks, Fm = ma
17
m
Two boxes sit side-by-side on a smooth horizontal surface. The lighter box has a mass
of 5.2 kg, the heavier box has a mass of 7.4 kg.
(a) Find the contact force between these boxes when a horizontal force of 5.0 N is
applied to the light box.
(b) If the 5.0-N force is applied to the heavy box instead, is the contact force between
the boxes the same as, greater than, or less than the contact force in part (a)? Explain.
(c) Verify your answer to part (b) by calculating the contact force in this case.
The boxes will remain in contact, so must have the same acceleration. Use
Newton’s second law to determine the magnitude of the contact forces
(a)
Mm
(b)The lighter box will require less net force than
the heavier box did (for the same acceleration). If
the external force is instead applied to the heavier
box, the contact force must be less than it was
before:
m
mM
Forces in Two Dimensions
The easiest way to handle forces in two
dimensions is to treat each dimension separately,
as we did for kinematics.
Normal Forces
The normal force is
the force exerted by
a surface on an
object, to keep an
object above the
surface.
The normal force
is always
perpendicular to
the surface.
The normal force may be equal
to, greater than, or less than
the weight.
No vertical motion, so
Apparent weight:
Weight
Your perception of your weight is based on the
contact forces between your body and your
surroundings.
If your surroundings
are accelerating, your
apparent weight may
be more or less than
your actual weight.
In this case the “apparent weight” is the normal force, and is
equal to the sum of the gravitational attaction (actual weight)
and the force required to accelerate the body, as specified
Normal Force
Below you see two cases: a
physics student pulling or
pushing a sled with a force F
that is applied at an angle . In
which case is the normal force
greater?
a) case 1
b) case 2
c) it’s the same for both
d) depends on the magnitude of
the force F
e) depends on the ice surface
Case 1
Case 2
Normal Force
Below you see two cases: a
physics student pulling or
pushing a sled with a force F
that is applied at an angle . In
which case is the normal force
greater?
a) case 1
b) case 2
c) it’s the same for both
d) depends on the magnitude of
the force F
5) depends on the ice surface
Case 1
In case 1, the force F is pushing down (in
addition to mg), so the normal force needs
to be larger. In case 2, the force F is
pulling up, against gravity, so the normal
force is lessened.
Case 2
On an Incline
Consider two identical
a) case A
blocks, one resting on a
b) case B
flat surface and the other
c) both the same (N = mg)
resting on an incline. For
which case is the normal
force greater?
A
d) both the same (0 < N < mg)
e) both the same (N = 0)
B
On an Incline
Consider two identical
a) case A
blocks, one resting on a
b) case B
flat surface and the other
c) both the same (N = mg)
resting on an incline. For
which case is the normal
force greater?
d) both the same (0 < N < mg)
e) both the same (N = 0)
In case A, we know that N = W.
y
In case B, due to the angle of
N
the incline, N < W. In fact, we
f
can see that N = W cos().
W
A
B

 Wy
x
Translation Equilibrium
“translational equilibrium” = fancy term for not accelerating
= the net force on an object is zero
example: book on a table
example: book on a table in an elevator
at constant velocity
A weight on a string...
if I pull the bottom string
down, which string will
break first?
Ftop
a) top string
b) bottom string
Fbot
W
c) there is not enough information to answer
this question
How quickly is the string pulled? A sudden, strong
tug is resisted by the inertia of the mass, protecting
the top string. A gradual pull forces the top string
to keep the system in equilibrium.
Before practicing his routine on the rings, a 67-kg gymnast stands
motionless, with one hand grasping each ring and his feet touching the
ground. Both arms slope upward at an angle of 24° above the
horizontal.
(a) If the force exerted by the rings on each arm has a magnitude of 290
N, and is directed along the length of the arm, what is the magnitude
of the force exerted by the floor on his feet?
(b) If the angle his arms make with the horizontal is greater that 24°,
and everything else remains the same, is the force exerted by the
floor on his feet greater than, less than, or the same as the value
found in part (a)? Explain.
Before practicing his routine on the rings, a 67-kg gymnast stands
motionless, with one hand grasping each ring and his feet touching the
ground. Both arms slope upward at an angle of 24° above the
horizontal.
(a) If the force exerted by the rings on each arm has a magnitude of 290
N, and is directed along the length of the arm, what is the magnitude
of the force exerted by the floor on his feet?
(b) If the angle his arms make with the horizontal is greater that 24°,
and everything else remains the same, is the force exerted by the
floor on his feet greater than, less than, or the same as the value
found in part (a)? Explain.
N
a)
b) if the angle is larger and everything else remains the same,
the applied forces are more vertical. With more upward
force from the arms, LESS normal force is required for zero
acceleration
Lecture 6
Applications of
Newton’s Laws
(Chapter 6)
Going Up II
A block of mass m rests on the
floor of an elevator that is
accelerating upward. What is
a) N > mg
b) N = mg
c) N < mg (but not zero)
the relationship between the
d) N = 0
force due to gravity and the
e) depends on the size of the
elevator
normal force on the block?
m
a
3
Going Up II
A block of mass m rests on the
floor of an elevator that is
accelerating upward. What is
a) N > mg
b) N = mg
c) N < mg (but not zero)
the relationship between the
d) N = 0
force due to gravity and the
e) depends on the size of the
elevator
normal force on the block?
The block is accelerating upward, so it
must have a net upward force. The
forces on it are N (up) and mg (down),
so N must be greater than mg in order
to give the net upward force!
Follow-up: What is the normal force if
the elevator is in free fall downward?
N
m
a>0
mg
F = N – mg = ma > 0
 N > mg
3
Elevate Me
You are holding your 2.0 kg
physics text book while
standing on an elevator.
Strangely, the book feels as if
it weighs exactly 2.5 kg. From
this, you conclude that the
elevator is:
a) in freefall
b) moving upwards with a constant
velocity of 4.9 m/s
c) moving down with a constant velocity
of 4.9 m/s
d) experiencing a constant acceleration
of about 2.5 m/s2 upward
e) experiencing a constant acceleration
of about 2.5 m/s2 downward
3
35
Elevate Me
You are holding your 2.0 kg
physics text book while
standing on an elevator.
Strangely, the book feels as if
it weighs exactly 2.5 kg. From
this, you conclude that the
elevator is:
a) in freefall
b) moving upwards with a constant
velocity of 4.9 m/s
c) moving down with a constant velocity
of 4.9 m/s
d) experiencing a constant acceleration
of about 2.5 m/s2 upward
e) experiencing a constant acceleration
of about 2.5 m/s2 downward
Use Newton’s 2nd law! the apparent weight:
and the sum of forces:
give a positive acceleration ay
Frictional Forces
Friction has its basis in surfaces that are not
completely smooth:
Kinetic friction
Kinetic friction: the friction experienced by surfaces
sliding against one another
The frictional force is proportional to the contact force
between the two surfaces (normal force):
The constant
is called the coefficient of
kinetic friction.
fk always points in the direction
opposing motion of two surfaces
Frictional Forces
fk
fk
Naturally, for any frictional
force on a body, there is an
opposing reaction force on
the other body
Frictional Forces
fk
fk
when moving, one bumps
“skip” over each other
fs
fs
when relative motion
stops, surfaces settle into
one another
static friction
Static Friction
The static frictional force tries to keep an object from
starting to move when other forces are applied.
The static frictional force has a maximum value, but
may take on any value from zero to the maximum...
depending on what is needed to keep
the sum of forces to zero.
The maximum static frictional
force is also proportional to
the contact force
Static Friction
A block sits on a flat table. What is
the force of static friction?
a) zero
b) infinite
c) you need to tell me stuff, like the
mass of the block, μs, and what
planet this is happening on
Characteristics of Frictional Forces
• Frictional forces always oppose relative motion
•Static and kinetic frictional forces are independent of
the area of contact between objects
• Kinetic frictional force is also independent of the
relative speed of the surfaces.
• Coefficients of friction are
independent of the mass
of objects, but in (most)
cases forces are not:
(twice the mass
= twice the weight
= twice the normal force
= twice the frictional force)
Coefficients of Friction
Q: what units?
Measuring static coefficient of friction
If the block doesn’t move, a=0.
Given the “critical angle” at
which the block starts to slip,
what is μs?
y
x
N
fs
Wx
at the critical point

W

Wy
Acceleration of a block on an incline
If the object is sliding down -
v
y
x
N
fk
Wx

W

Wy
Acceleration of a block on an incline
If the object is sliding up -
v
y
x
N
Wx
fk

W

Wy
What will happen
when it stops?