Transcript PHYSICS SAE 4

Physics Subject
Area Test
MECHANICS:
DYNAMICS
Dynamics: Newton’s Laws of Motion
Newton’s First Law
Force
Dynamics – connection between force and motion
Force – any kind of push or pull
required to cause a change in motion
(acceleration)
measured in Newtons (N)
Dynamics: Newton’s Laws of Motion
Newton’s First Law of Motion
Newton’s First Law of Motion
First Law – Every object continues in its state of rest, or of
uniform velocity in a straight line, as long as no net force acts on
it.
First Law – (Common) An object at rest remains at rest, and a
object in motion, remains in motion unless acted upon by an
outside force.
Newton’s First Law of Motion
Newton’s Laws are only valid in an
Inertial Frame of Reference
For example, if your frame of
reference is an accelerating car – a
cup in that car will slide with no
apparent force being applied
Newton’s First Law of Motion
An inertial frame of reference is one where if the first law is
valid
Inertia – resistance to change in motion
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Dynamics: Newton’s Laws of Motion
Mass
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Mass
Mass – a measurement of inertia
A larger mass requires more force to accelerate it
Weight – is a force, the force of gravity on a specific mass
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Dynamics: Newton’s Laws of Motion
Newton’s Second Law
Newton’s Second Law
Second Law – acceleration is directly proportional to the net force
acting on it, and inversely proportional to its mass.
-the direction is in the direction of the net force
Easier to see as an equation
more commonly written
a 
F
m
F  ma
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Newton’s Second Law
F – the vector sum of the forces
In one dimension this is simply adding or subtracting forces.
Free Body Diagram
The most important step in
solving problems involving
Newton’s Laws is to draw the
free body diagram
 Be sure to include only the
forces acting on the object
of interest
 Include any field forces
acting on the object
 Do not assume the normal
force equals the weight

F table on book
F Earth on book
Objects in Equilibrium
Objects that are either at rest or moving with
constant velocity are said to be in equilibrium
 Acceleration of an object can be modeled as

zero: a  0
 Mathematically, the net force acting on the
object is zero
F 0
 Equivalent to the set of component equations
given by


F
x
0
F
y
0
A lamp is suspended from a
chain of negligible mass
 The forces acting on the
lamp are




the downward force of gravity
the upward tension in the
chain
Applying equilibrium gives
F
y
 0  T  Fg  0  T  Fg

A traffic light weighing 100 N hangs from a vertical
cable tied to two other cables that are fastened to a
support. The upper cables make angles of 37 ° and
53° with the horizontal. Find the tension in each of
the three cables.

Conceptualize the traffic light



Assume cables don’t break
Nothing is moving
Categorize as an equilibrium problem


No movement, so acceleration is zero
Model as an object in equilibrium
F
x
0
F
y
0

Need 2 free-body diagrams
Apply equilibrium equation to light


T3  F g  100 N
Apply equilibrium equations to
knot

F
F y  0  T3  F g  0
x
 Fy

 T1x  T2 x  T1 cos 37  T2 cos 53  0
 T1y  T2 y  T3 y
 T1 sin 37  T2 sin 53  100 N  0
cos 37 
T2  T1 
  1.33 T1
cos
53


T1  60 N
T2  1.33 T1  80 N
F
y
 0  T3  F g  0
T3  F g  100 N

Suppose a block with a
mass of 2.50 kg is
resting on a ramp. If
the coefficient of static
friction between the
block and ramp is 0.350,
what maximum angle
can the ramp make
with the horizontal
before the block starts
to slip down?

F
F
Newton 2nd law:
x
 mg sin    s N  0
y
 N  mg cos   0

Then
F

So
y
N  mg cos 
 mg sin    s mg cos   0
tan    s  0 . 350
  tan
1
( 0 . 350 )  19 . 3


A block of mass m1 on a rough, horizontal surface is
connected to a ball of mass m2 by a lightweight cord
over a lightweight, frictionless pulley as shown in figure.
A force of magnitude F at an angle θ with the horizontal is
applied to the block as shown and the block slides to the right.
The coefficient of kinetic friction between the block and
surface is μk. Find the magnitude of acceleration of the two
objects.
We all remember the fun
see-saw of our youth.
But what happens if . . .
Moral
Both the masses and their positions affect
whether or not the “see saw” balances.
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M1
M2
d1
Need:
M1 d1 = M2 d2
d2
The great Greek mathematician
Archimedes said, “give me a
place to stand and I will move the
Earth,” meaning that if he had a
lever long enough he could lift the
Earth by his own effort.
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We can think of leaving the masses in place and moving the fulcrum.
It would have to be a pretty long
see-saw in order to balance the
school bus and the race car,
though!
M1
M2
d1
d2
(We still) need:
M1 d1 = M2 d2