Kinematics in One Dimension

copygrouperMechanics

Nov 13, 2013 (3 years and 11 months ago)

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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




0
0
0
2
1
0 0
2
2 2
0 0

0
2
2
v v at a const
v v
v t
x x v t at
v v a x x
  

 
 
 
 
  
  
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.

F
a =
p
F = a ( F = )
m
d
m
dt


Newton’s Second Law of Motion

2
2
2
F = a

Units for FORCE kg m/s = N (Newton)
g c
m/s = dyne (dyne)
slu
g ft/s = lb (poun
x x y y
m
F ma F ma
 


 
d)
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/s
2


= 98.0 N


F
N

= mg

Weight, and a Mystery in a Box

F
N

-

W
-

40.0 N = 0




F
N

= mg + 40.0 N


= 98.0 N + 40.0 N


= 138 N

Weight, and a Mystery in a Box

F
N

-

W + 40.0 N = 0



F
N

= 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...

98.0 N 40.0 N sin(30.0 )
78.0 N
N Py
F mg F
 
   

0
N Py
F W F
  
Slip, sliding away…

0
x Px
ma F
 
2
40.0 N cos(30.0 )
10.0 kg
3.46 m/s
Px
x
F
a
m

 


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 F
N

-

W
y

= 0


X direction Forces W
x

=
m a

N y
F = W
cos( )
98.0 N cos(9.5 )
97.7 N
mg


  

x
2
2
2
1
2
2
W sin( )
sin( )
9.80 m/s sin(9.5 )
1.62 m/s
2
2(6.00 m)
1.62 m/s
2.7 s
ma mg
a g
x at
x
t
a


 

  





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/s
2
, what

is the new tension in the cords?



T
1

-

W
1

-

W
2

= 0



T
1

=
mg

+
mg


= 2 ( 3.5 kg


9.80 m/s
2

)


= 68.6 N





T
2

-

W
2

= 0



T
2

=
mg


= 34.3 N



T
1

-

W
1

-

W
2

= (
m + m
)
a



T
1

= 2
mg

+ 2
m a


= 2 ( 34.3 N

) +


2 ( 3.5 kg)


1.60 m/s
2


= 79.8 N





T
2

-

W
2

=
m a



T
2

=
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, F
fr



The force of friction, F
fr

or
f
, is equal to the

normal force times the coefficient of friction.


f
s


µ
s
∙N (Static)
f
k

= µ
k
∙N (kinetic)




µ is like the % of the force pushing the two

substances together which results in friction.

Kinetic friction


F
fr

=

k

F
N




Static friction


F
fr




s

F
N



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 m
µ.

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?

F
N

-

mg

+ F
Py

= 0

F
Px

-

F
fr

=
m a

F
Px

?

s

F
N

F
N

-

mg

+ F
Py

= 0

N Py
sf N
F = F
98.0 N 40 N sin(30 )
78.0 N
F F
0.40 78.0 N
31.2 N
s
mg


   


 

Px P
F F cos( )
40.0 N cos(30 )
34.6 N

 
  

Since F
Px



F
sf

, the box can move!

F
Px

-

F
fr

=
m a

Px fr
P N
2
F F
F cos( ) F
40.0 N cos(30 ) 0.30 78.0 N
10.0 kg
1.1 m/s
s
a
m
m
 




   



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?


F
N

-

W
y

= 0 F
N

=
mg

cos(

)



W
x

-

F
fr

=
m a

m a

=
mg

sin(

)
-


s

F
N

F
N

=
mg

cos(

)



ma

=
mg

sin(

)
-


k

F
N





x fr
2
2
W F
sin( ) cos( )
sin( ) cos( )
9.80 m/s sin(30 ) 0.17cos(30 )
3.46 m/s
k
k
a
m
mg mg
m
g
  
  




 
    

How fast does it go?

2 2
0
2
2
2
2 3.46 m/s 9.3 m
8.0 m/s
v v ax
v ax
 

  

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 2
nd

law for the x and y
components
.

Next time: Gravity and Centripetal force