Transcript PPT

Newton’s Laws
Straight Line Motion
A person pulls on a crate that weighs 600 N
across a flat frictionless surface with a force
of 30N. What is the motion of the crate?

N

F
1. Redraw the object of interest and label all forces
(contact and field).
r
v
FG  mg
F
m  G  61.2kg
g
1st
2. Apply Newton’s Law to vertical case – since
object has no motion in the vertical sense the forces
in the vertical direction MUST add up to be zero.
r
r
FG  N  0
r
r
 N  FG

Horizontally

N
There is only one force in the
horizontal direction so it
must also be the net force
 
F

F

m
a
n et

F
r
v
FG  mg
F
m  G  61.2kg
g
So…
 

FF
m
a
net
30
61
.2
a
m
a0
.49
s2

In the
same
direction
as
the
force
Force at an Angle
30o

N

F
30o
A person pulls on a crate that weighs
600 N across a flat frictionless
surface with a force of 30N. What is
the motion of the crate?
Resolve Forces that are at an angle to coordinate system


F

F
cos
30

30
cos
30

26
N
horiz


F

F
sin
30

30
sin
30

15
N
vert
r
v
FG  mg
Vertically:
r r
r
FG
N  Fvert  FG  0
m   61.2kg
g
r
N 15  600  0
r
N  585N

Horizontally:


F Horiz  Fnet
26  ma
26  61 . 2 a
a  0 . 425 m s 2
Motion on a Ramp
The person is pulling the same crate up a 25o ramp with a force of 300N. What is the
motion of the crate?
Draw a free body diagram that labels
direction of forces (contact and field)
We can anticipate the direction of the motion of the crate
and simplify things by selecting a coordinate system in that
direction.
+y
25o
Notice that the weight vector is at an angle WRT our
coordinate system. We’ll need to determine its
components.

N
r
v
FG  mg

F
25o
+x
Ramp Problem
r
r
FG,y  FG cos25  600cos25  544N
r
r
FG,x  FG sin 25  600sin 25  254N
2. Apply Newton’s 1st Law to vertical case –
since the object has no motion in the vertical
sense the forces in the vertical direction
MUST add up to be zero.
r
r
FG,y  N  0
r
r
 N  FG,y  544 N
3. There are two forces in the
horizontal direction so we must find
the net force
+y

N
r
v
FG  mg

F
25o
r
FG,y
r
FG,x
r
r r 
r
FG,x  F  Fnet  ma

254  300  46  61.2a

m
a  0.75 s2
directed up the ramp
+x
Motion on a Ramp
The person is pulling the same crate up a 25o ramp when the rope breaks and the box
begins to slide back down the ramp. What is the motion of the crate?
Draw a free body diagram that labels
direction of forces (contact and field)
+y

N
+x
25o
Notice that the weight vector is at an angle
WRT our coordinate system. We’ll need to
determine its components.

r
v
FG  mg
25o
Ramp Example cont…
r
r
FG,y  FG cos25  600cos25  544N
r
r
FG,x  FG sin 25  600sin 25  254N
2. Apply Newton’s 1st Law to vertical case –
since the object has no motion in the vertical
sense the forces in the vertical direction
MUST add up to be zero.
r
r
FG,y  N  0
r
r
 N  FG,y  544 N
3. There is only one force in the

horizontal direction so it must also
be the net force
+y

N
r
v
FG  mg
+x
25o
r
FG,y
r
FG,x
r
r 
r
FG,x  Fnet  ma
254  61.2a
 m
a  4.15 s2
directed down the ramp
Atwood’s Machine
Two masses hang over a pulley by means of a
massless rope. Determine the motion of the system.
10 kg
20 kg
Conceptual approach:
• Clearly the 20 kg object is falling and the 10 kg object is
rising.
• If the 20kg object were only a 10 kg object then the
system would be in equilibrium. So the effect is that
there is an extra 10kg hanging on the left side.
• This provides a net force of 10(9.8) = 98 N
• This 98 N force acts on both masses (they both move due
to this extra mass)


Fnet  ma
98 (1020)a
a 3.26ms2
Atwood’s Machine cont…
What is the tension in the rope?
Conceptual approach:
•The entire system is accelerating at the rate of 3.26m/s2
•On the 10kg object, the tension force provided by the
string must be larger than the 10kg object’s weight.
•Newton’s 2nd Law says:
10 kg
20 kg

r
r r
Fnet  T  FG
r r r
ma  T  FG
r
10(3.26)  T 10(9.8)
T  130.6N (up)
Friction

N
r
f fric
Friction as a Contact

Force
F
r
v
FG  mg

Friction arises from the
roughness between two
surfaces and how hard the two
surface are pressed together
Friction ALWAYS opposes motion.
As the two surfaces are pushed closer together, more of the
nooks and crannies come into contact and the friction increases.
Friction Defined Mathematically
The force of friction depends upon two things:
1. The inherent roughness between the two surfaces (m)
2. The force which presses the two surfaces together
Ordinarily, you might be tempted to use the weight of the object as the force
which presses the two surfaces together – and in many case you would be right.
But not always!



N
r
f fric

F
r
v
FG  mg
Here the weight is equal to
the Normal Force

N
f fric
r
v
FG  mg
In this case the force that pushes the two surfaces
together is only one component of the weight. But is
still
 equal to the Normal Force
r
r
f fric ~ mN
2 Cases -Static and Kinetic Friction
f
fs,max
F
InKinetic
the
Region
In Static
the
Region
f
m
N
fs
m
N
k
k
s
where
m
is
the
coefficien
t
where
m
is
the
coefficien
t
s
k
of
static
friction
of
static
friction
In both cases ms,k is the ratio of two forces fs,k/N and is dimensionless
Straight Line Motion - Revisited
A person pulls on a crate that weighs 600 N across a
flat surface with a force of 30N. The coefficient of
kinetic friction between the crate and the surface is
0.28. The coefficient of static friction between the
surfaces is 0.35. What is the motion of the crate?
1. Redraw the object of interest and label all forces
(contact and field). (pick “up” to be positive and
the direction of F to be positive)
1st
2. Apply Newton’s Law to vertical case – since
object has no motion in the vertical sense the forces
in the vertical direction MUST add up to be zero.
r
r
FG  N  0
r
r
 N  FG  (600N)

N
r
f k  mk N

F
r
v
FG  mg
F
m  G  61.2kg
g


Horizontally
There are two forces in the
horizontal direction. So, we must
determine the net force
  
F

f

F

m
a
s
net

fs m
N
30  0.35(600)  61.2a
a  2.94
m
s2

F
r
v
FG  mg
F
m  G  61.2kg
g
So…
r r
r
r
F  f s  Fnet  ma
r
r
r
F  msN  Fnet  ma

N

Minimum Force to Move Crate
If the crate is initially at rest then the person must overcome
the static friction force.
m
f

N
maximum
value
occurs
when

beco

s
s
f

0
.
35
(
600
)

210
N
s
If the person pulls with a force of exactly 210N then the box is right on
the verge of moving but doesn’t move. He must pull with a force
slightly higher than 210N.
Once the crate begins to move, he need only match the force of kinetic
friction in order to pull the crate at constant speed.
m
f

N

0.28(60

168
N
k
k
If he pulls with a force larger than 168N, then the crate is accelerating instead
of traveling at constant speed in agreement with Newton’s 2nd Law.
Bobsled
25 m/s
In the horizontal sense:
r
r
f k  Fnet
mk N  ma
mk N mk mg
a

m
m
a  0.05 * 9.8  0.49 m s2
A bobsled of mass 180 kg is traveling
at 50 mph (~25 m/s) along an icy
track. The coefficient of kinetic
friction between the sled runners and
the ice is 0.05. If the sled were to
coast to a stop at the bottom of the
run, how far would it coast?
In the vertical sense we should be able to
see that the Normal force is equal in
magnitude to the Weight of the sled

N
fk
m
N
k
r
r
FG  mg
A bobsled of mass 180 kg is traveling
at 50 mph (~25 m/s) along an icy
track. The coefficient of kinetic
friction between the sled runners and
the ice is 0.05. If the sled were to
coast to a stop at the bottom of the
run, how far would it coast?
Bobsled
25 m/s
In the horizontal sense:
r
r
f k  Fnet
mk N  ma
mk N mk mg
a

m
m
v 2f  v i2  2a(d f  di )
0  25 2  2(0.49)(d f  0)
d f  638m
a  0.05 * 9.8  0.49 m s2

Bobsled
25 m/s
What must the coefficient of friction
be if the sled had to be stopped
within 50m?
v 2f  v i2  2a(d f  di )
a
v 2f  v i2
2(d f  di )
25 2  0 2

2(50  0)
 6.25 m s2
a  mk g
a 6.25
mk  
 0.638
g 9.8
Block on the wall
An 8 N horizontal force F pushes a block weighing
12.0 N against a vertical wall . The coefficient of
static friction between the wall and the block is
0.69, and the coefficient of kinetic friction is 0.49.
Assume that the block is not moving initially.

N


fk
m
kN
r
f s  msN

F
8
N
r
v
FG  mg  12N
m 
12
 1.22kg
9.8
Horizontally the forces add up to be
zero (no motion in this direction)
Therefore, N and F have the same
magnitude but opposite directions

f
m
N

.
49
(
8
)

3
.
92
N
k
k

f
m
N

.
69
(
8
)

5
.
52
N
s
s
Since the weight is larger than the static
friction force then the block is sliding
down the wall (kinetic).
Block on the wall
An 8 N horizontal force F pushes a block weighing
12.0 N against a vertical wall . The coefficient of
static friction between the wall and the block is
0.69, and the coefficient of kinetic friction is 0.49.
Assume that the block is not moving initially.
m

f

N

3
.
92
N
k
k
r
f s  msN


N
F
8
N

r
v
FG  mg  12N
m 
12
 1.22kg
9.8
Vertically the forces don’t add up to be
zero (acceleration)
r r
r
f k  FG  Fnet
mk N  mg  ma
mk N  mg
a
m
.49(8) 1.22(9.8)

1.22
 6.59 m s2
Block on the wall (revisited)
How much force must the person exert against the
same block to hold it in place on the wall? The
coefficient of static friction between the wall and
the block is 0.69, and the coefficient of kinetic
friction is 0.49. Assume that the block is not moving
initially.
r
f s 12N

N


F?
r
v
FG  mg  12N
m 
12
 1.22kg
9.8
Vertically the forces have to add up to
be zero.
r
f s  ms N
fs
m N
s
12
0.69
N
17.39  N
r r
NF 0
r
17.39  F  0
r
F  17.39