Transcript Newton's Laws

NEWTON’S LAWS & FORCES
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.
Newton: The Unit of Force
One newton is that resultant force which imparts
an acceleration of 1 m/s2 to a mass of 1 kg.
F (N) = m (kg) a (m/s2)
What resultant force will give a 3 kg mass an
acceleration of 4 m/s2? F = m a
F=?
3 kg
F  (3 kg)(4 m/s2 )
a = 4 m/s2
F = 12 N
Acceleration and Force With
Zero Friction Forces
Pushing the cart with twice the force
produces twice the acceleration. Three
times the force triples the acceleration.
Example 2. A net force of 4.2 x 104 N acts on
a 3.2 x 104 kg airplane during takeoff. What
is the force on the plane’s 75-kg pilot?
First we find the
acceleration a of
plane.
F = 4.2 x 104 N
F 4.2 x 104 N
a 
m 3.2 x 104 kg
+
F = ma
m = 3.2 x 104 kg
a = 1.31 m/s2
To find F on 75-kg pilot, assume same acceleration:
F = ma = (75 kg)(1.31 m/s2);
F = 98.4 N
Problem Solving Strategy
(For the Simpler Problems.)
• Read problem; draw and label sketch.
• List all given quantities and state what is to
be found.
• Make sure all given units are consistent with
Newton’s second law of motion (F = m a).
• Determine two of the three parameters in
Newton’s law, then solve for the unknown.
Example 3. A 54 g tennis ball is in contact
with the racket for a distance of 40 cm as it
leaves with a velocity of 48 m/s. What is the
average force on the ball?
First, draw sketch and list
given quantities:
Given: vo = 0; vf = 48 m/s
d = 40 cm; m = 54 g
a=?
Consistent units require converting grams
to kilograms and centimeters to meters:
Given: vo = 0; vf = 48 m/s d = 0.40 m;
Cont. . .
m = 0.0540 kg; a = ?
Example 3 (Cont). A 54-gm tennis ball is in
contact with the racket for a distance of 40
cm as it leaves with a velocity of 48 m/s.
What is the average force on the ball?
Knowing that F = m a, we need
first to find acceleration a:
0
2ax  v  v ; a 
2
f
(48 m/s)2
a
;
2(0.40 m)
2
0
v
2
f
2x
a  2880 m/s 2
F= (0.054 kg)(2880 m/s2);
F = ma
F = 156 N
Weight and Mass
• Weight is the force due to gravity. It is
directed downward and it varies from
location to location.
• Mass is a universal constant which is a
measure of the inertia of a body.
F = m a so that: Fw = mg and m =
Fw
g
Weight and Mass: Examples
• What is the weight of a 10-kg block?
10 kg
m
9.8 m/s2 Fw
Fw = mg = (10 kg)(9.8 m/s2)
Fw = 98 N
Newton’s Third Law (Reviewed):
• Third Law: For every action force, there
must be an equal and opposite reaction
force. Forces occur in pairs.
Action
Reaction
Reaction
Action
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
Example 6: 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 Second 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.
Example 7: A cart and driver have a mass of
120 kg. What force F is required to give an
acceleration of 6 m/s2 with no friction?
1. Read problem and draw a sketch.
Diagram for Cart:
FN
FW
Fapp
+
x
2. Draw a vector force diagram and label forces.
3. Choose x-axis along motion and indicate the
right direction as positive (+).
Example 7 (Cont.) What force F is required
to give an acceleration of 6 m/s2?
4. Write Newton's Law equation for both axes.
m = 120 kg
Diagram for cart:
FN
F
FW
ay = 0
SFy = 0; FN - Fw= 0
The normal force FN
is equal to weight FW
+
x
SFx = max; F = ma
F = (120 kg)(6 m/s2)
F = 720 N
Example 8: 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 info)
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