Transcript 3rd Law notes

For every action force
there is an equal and
opposite reaction force!!
For every action force
there is an equal and
opposite reaction force!!
For every action force
there is an equal and
opposite reaction force!!
•These Two forces are known as an
action/reaction pair.
•Gravity acts on an object and the
object acts with equal force.
Newton’s Third Law

Action-Reaction Pairs

The hammer exerts a force
on the nail to the right.

The nail exerts an equal but
opposite force on the
hammer to the left.
Acting and Reacting Forces
• Use the words by and on to study
action/reaction forces below as they
relate to the hand and the bar:
Action
The action force is exerted by
bar
the _____
hands on the _____.
The reaction force is exerted
bar on the _____.
hands
by the _____
Reaction
A 60-kg athlete exerts a force on a 10-kg
skateboard. If she receives an acceleration
of 4 m/s2, what is the acceleration of the
skateboard?
Force on runner = -(Force on board)
mr ar = -mb ab
Force on
Board
(60 kg)(4 m/s2) = -(10 kg) ab
Force on
Runner
(60 kg)(4 m/s)
2
a
 24 m/s
-(10 kg)
a = - 24 m/s2
Applying Newton’s Law
• Read, draw, and label problem.
• Draw free-body diagram for each body.
• Choose x or y-axis along motion and choose
direction of motion as positive.
• Write Newton’s law for both axes:
SFx = m ax
SFy = m ay
• Solve for unknown quantities.
What is the tension T in the rope below if
the block accelerates upward at 4 m/s2?
(Draw sketch and free-body.)
T
a
10 kg
a = +4 m/s2
T
mg
+
SFx = m ax = 0 (No Motion)
SFy = m ay = m a
T - mg = m a
mg = (10 kg)(9.8 m/s) = 98 N
m a= (10 kg)(4 m/s) = 40 N
T - 98 N = 40 N
T = 138 N
Two-Body Problem: Find tension in the connecting
rope if there is no friction on the surfaces.
12 N
2 kg
4 kg
Find acceleration of
system and tension
in connecting cord.
First apply F = ma to entire system (both masses).
n
SFx = (m2 + m4) a
12 N
12 N = (6 kg) a
(m2 + m4)g
a=
12 N
6 kg
a = 2 m/s2
The two-body problem.
12 N
2 kg
4 kg
Now find tension T
in connecting cord.
Apply F = m a to the 2 kg mass where a = 2 m/s2.
n
T
SFx = m2 a
T = (2 kg)(2 m/s2)
m2 g
T=4N
The two-body problem.
2 kg
4 kg
12 N Same answer for T
results from focusing
on 4-kg by itself.
Apply F = m a to the 4 kg mass where a = 2 m/s2.
T
n
12 N
m2 g
SFx = m4 a
12 N - T = (4 kg)(2 m/s2)
T=4N
Find acceleration of system and tension in
cord for the arrangement shown.
First apply F = m a to entire
system along the line of motion.
2 kg
SFx = (m2 + m4) a
n
4 kg
T
m4g = (m2 + m4) a
+a
T
m2 g
m4 g
Note m2g is balanced by n.
a=
m4g
m2 + m4
=
(4 kg)(9.8 m/s2)
2 kg + 4 kg
a = 6.53 m/s2
Now find the tension T given that the
acceleration is a = 6.53 m/s2.
To find T, apply F = m a to just
the 2 kg mass, ignoring 4 kg.
2 kg
n
4 kg
T
+a
T
m2 g
m4 g
F
x
 m2 a or T  m2a
T = (2 kg)(6.53 m/s2)
T = 13.1 N
Same answer if using 4 kg.
m4g - T = m4 a
T = m4(g - a) = 13.1 N
Summary
Newton’s First Law: An object at rest or an object
in motion at constant speed will remain at rest or
at constant speed in the absence of a resultant
force.
Newton’s Second Law: A resultant force produces
an acceleration in the direction of the force that
is directly proportional to the force and inversely
proportional to the mass.
Newton’s Third Law: For every action force,
there must be an equal and opposite reaction
force. Forces occur in pairs.