Transcript Newton`s Second Law (PowerPoint)

Newton’s Second Law
Why do objects of different mass and therefore different
weights fall at the same rate (g) in the absence of air
resistance?
g
g
The more mass an object has the greater its inertia or resistance to a
change in motion.
The more mass an object has the greater the force of gravity acting on
that mass.
We know that an object’s ability to change its motion is related to
both inertia and unbalanced force by…….
Fg = m g
M
m
g
g
g = Fg = Fg
m
M
Fg
Fg
The bigger mass has more
inertia but proportionally
more force acting upon it
so will fall at the same rate.
Putting it all together……
The acceleration of an object is directly proportional to the
net force acting on it and inversely proportional to its
mass.
a = FNET / m
or
FNET = m a
The direction of the acceleration is in the direction of the
applied force.
The net force in this case is:
275 N + 395 N – 560 N = +110 N
and is directed along the + x axis of the coordinate system.
If the mass of the car is 1850 kg then, by
Newton’s second law, the acceleration is
FNet
110 N
2
a

 0.059 m s
1850 kg
m
Inertial Mass
Inertial mass is the property of an object that
determines how it responds to a given force
Different masses have different accelerations
when a force acts on them
Inertial mass, M(i) = F / a
Gravitational Mass
Gravitational mass is the property of an object
that determines how much gravitational force it
feels near another object
Different masses have different gravitational
forces acting between them
Gravitational force  M(g)
Gravitational Mass and Inertial Mass
Inertial mass and gravitational mass are very
different. The surprise is that they turn out to be
equivalent.
In other words, an object’s gravitational mass is
equal to its inertial mass.
The fact that different objects have the same
value for free-fall acceleration shows this.
Compound Systems
A system consists of two or more interacting objects.
A compound system is composed of two or more objects that are
restrained so that they have to move together.
Forces are external if they act on the system and internal if they act
within the system
Rules:
1. Treat the two connected objects as one object in order to determine
the change in motion (acceleration) of the system and/or any
external forces acting on the system
2. Look at each individual object separately (FBD) in order to
determine any internal forces
A 4000 kg airplane accelerates down the runway pulling a 1000 kg glider with a
towing cable. The forward thrust provided by the planes propellers is 20000 N.
i) What is the acceleration of the glider down the runway
ii) What is the tension in the towing cable
-
+
4000kg
1000kg
FT
i) mp = 4000kg mg = 10000kg FE = 20000 N a = ?
FT
20000N
ii) Look at just the glider
Fnet = mT a
a = 4 m/s2
FE = mT a
Fnet = mg a
a = FE / mT
FT = mg a = (1000kg) (4 m/s2)
a = 20000N / (4000kg+1000kg)
a = 4 m/s2
FT = ?
= 4000N
Check: Look at just the airplane
Fnet = mp a
FE - FT = mp a
FT = FE - mp a
FT = 20000N - (4000kg)(4m/s2) = 4000N
Write an expression for the acceleration of the cart/weight system shown below
and the tension in the string
Mc
FT
FT
mH
mH g
Fnet = mH g
Looking just at the cart
a = Fnet / mT
a = mH g / (Mc + mH)
FT = MC a
If MC  mH
Looking just at the hanging mass
a = mH g / (mH) = g
If mH  MC
a = mH g / (Mc)
Fnet,C = MC a
Fnet,H = mH a
mH g - FT = mH a
= 0
FT = mH g - mH a = mH (g - a)
A 500 g cart on a horizontal surface is being pulled by a 200 g hanging mass.
What is its acceleration and tension in the connecting string?
mH = 0.2 kg
Mc = 0.5 kg g = 9.8 m/s2
a = mH g / (Mc + mH)
a = (0.2kg)(9.8 m/s2) / (0.5kg+0.2kg)
a = 2.8 m/s2
Looking just at the cart
Looking just at the hanging mass
Fnet,C = MC a
FT = mH (g - a)
FT = MC a
FT = (0.2kg) (9.8N/kg - 2.8N/kg)
FT = (0.5kg)(2.8N/kg) = 1.4 N
FT = 1.4 N