Transcript Lecture 5

Kinematics in One Dimension
MECHANICS comes in two parts:
kinematics: motion (displacement, time, velocity)
x, t, v, a
dynamics: motion and forces
x, t, v, a, p, F
v  v0  a t
a  const
 v  v0 
v 
t

0
0

 2 
2
1
x  x0  v0 t  2 a t
v  v  2a  x  x0 
2
2
0
Dynamics: Laws of Motion
Recall the two parts of MECHANICS - kinematics
dynamics
Introduce the motivation for motion as a derived concept
called FORCE.
F = ma
Have you seen a force lately?
A push or a pull; motion not required
What is the effect of a force acting on a free body?
F
g
m
accelerated motion
Newton’s First Law of Motion
Every body continues in its state of rest or uniform
speed in a straight line unless acted on by a nonzero
net force.
Preceding defines INERTIA
Preceding defines MASS
(m  W)
Best when observed in the absence of friction.
Newton’s Second Law of Motion
The acceleration of an object is directly proportional
to the net force acting on it and is inversely
proportional to its mass. The direction of the
acceleration is in the direction of the net force
acting on the object.
a=
F
m
F = m a
dp
(F=
)
dt
Newton’s Second Law of Motion
F = m a
F
x
 m ax
F
y
 m ay
Units for FORCE  kg m/s = N (Newton)
2
g cm/s 2 = dyne (dyne)
slug ft/s 2 = lb (pound)
Weight, and a Mystery in a Box.
You have received a gift box, of mass 10.0 kg, with
a mystery object inside. You place it on a table.
What is the weight of the box and the normal force
on it (from where)?
If a force of 40.0 N is added to the box, what is the
normal force?
If a force of 40.0 N is subtracted from the box, what
is the normal force?
Weight, and a Mystery in a Box.
W = mg
= 10.0 kg  9.80 m/s2
= 98.0 N
FN = mg
Weight, and a Mystery in a Box
FN - W - 40.0 N = 0
FN = mg + 40.0 N
= 98.0 N + 40.0 N
= 138 N
Weight, and a Mystery in a Box
FN - W + 40.0 N = 0
FN = mg - 40.0 N
= 98.0 N - 40.0 N
= 58.0 N
Force, and the Mystery in a Box
The box is now pulled across a frictionless (!)
table with a force of 40.0 N at an angle of 30.0.
What is its acceleration?
What is the normal force exerted by the table?
Checking the normal force to see if the box can fly...
FN  W  FPy  0
FN  mg  FPy
 98.0 N  40.0 N  sin(30.0)
 78.0 N
Slip, sliding away…
m ax  FPx  0
FPx
ax 
m
40.0 N  cos(30.0)

10.0 kg
 3.46 m/s
2
Still going down the slippery slope...
The mystery box slides freely down a ramp
6.0 m long inclined at 9.5. How long does
it take the box to reach the bottom? Would
this change if the box’s mass were doubled?
F = m a
Y direction Forces
FN - Wy = 0
X direction Forces
Wx = m a
ma  Wx  mg sin( )
a  g sin( )
 9.80 m/s 2  sin(9.5)
 1.62 m/s 2
x  12 at 2
FN = Wy
 mg cos( )
 98.0 N  cos(9.5)
 97.7 N
2x
t
a
2(6.00 m)

1.62 m/s 2
 2.7 s
One 3.5-kg paint bucket is hanging by
a cord from another 3.5-kg paint bucket
which is also hanging by a cord.
What is the tension in the cords?
If the two buckets are pulled upward
with an acceleration of 1.60 m/s2, what
is the new tension in the cords?
 T1 - W1 - W2 = 0
T1 = mg + mg
= 2 ( 3.5 kg  9.80 m/s2 )
= 68.6 N
 T2 - W2 = 0
T2 = mg
= 34.3 N
 T1 - W1 - W2 = (m + m) a
T1 = 2 mg + 2m a
= 2 ( 34.3 N ) +
2 ( 3.5 kg)  1.60 m/s2
= 79.8 N
 T2 - W2 = m a
T2 = mg + m a
= 39.9 N
The Bane of Galileo: Friction
Friction is everywhere! There is little wonder
why it played such a prominent part in
MECHANICS for the ancients (e.g., Aristotle).
Galileo recognized friction as separate from
motion, so the equations of kinematics could be
discovered (using geometry).
Friction - treated as a force (though not a vector)
- always opposite to the motion
- sometimes related to the motion,
sometimes not so related
Friction
Static friction is always greater than moving
(kinetic) friction.
• Calculating frictional force, Ffr
• The force of friction, Ffr or f, is equal to the
normal force times the coefficient of friction.
fs ≤µs·N (Static) fk = µk·N (kinetic)
• µ is like the % of the force pushing the two
substances together which results in friction.
Kinetic friction
Ffr = k FN
Static friction
Ffr  s FN
where
s  k
• Static friction and kinetic friction plotted for
a pull versus time.
• In lab you will plot Frictional force vs the Mass and
the slope will be mµ.
A 10.0-kg box rests on a surface and experiences a force
of 40.0 N at an angle of 30. The coefficient of static
friction is 0.40 and the coefficient of kinetic friction is
0.30.
Will the box move?
If it moves, how fast will it move?
FN - mg + FPy = 0
FPx ? s FN
FPx - Ffr = m a
FN - mg + FPy = 0
FN = mg  FPy
 98.0 N  40 N  sin(30)
 78.0 N
FPx  FP  cos( )
 40.0 N  cos(30)
 34.6 N
Fsf   s FN
 0.40  78.0 N
 31.2 N
Since FPx  Fsf , the box can move!
FPx - Ffr = m a
FPx  Ffr
a
m
FP cos( )   s FN

m
40.0 N  cos(30)  0.30  78.0 N

10.0 kg
 1.1 m/s 2
A block is placed at the top of an inclined plane. The
incline is 30.0 and 9.3 m long, with a coefficient of
kinetic friction of 0.17.
What is the acceleration of the block?
What speed does the block have when it reaches the
bottom?
FN - Wy = 0
FN = mg cos()
Wx - Ffr = m a
m a = mg sin() - s FN
FN = mg cos()
ma = mg sin() - k FN
Wx  Ffr
a
m
mg sin( )   k mg cos( )

m
 g  sin( )   k cos( ) 
 9.80 m/s 2   sin(30)  0.17 cos(30) 
 3.46 m/s 2
How fast does it go?
v 2  v02  2ax
v  2ax
 2  3.46 m/s 2  9.3 m
 8.0 m/s
Summary:
When solving problems:
1) Draw a diagram with the forces.
2) Resolve the components into x and y
components.
3) Remember Newton’s Second Law
4) Write Newton’s 2nd law for the x and y
components.
Next time: Gravity and Centripetal force