Transcript 5.5 The Gravitational Force and Weight

5.5 The Gravitational Force
and Weight
The Gravitational Force
and Weight
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The gravitational force:
Fg ≡ Force that The Earth exerts on an object
This force is directed toward the center of the
earth.
Its magnitude is called THE WEIGHT of the object
Weight ≡ |Fg| ≡ mg
(5.6)
Because it is dependent on g, the weight varies
with location
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g, and therefore the weight, is less at higher altitudes
Weight is not an inherent property of the object
Gravitational Force, 2
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Object in FREE FALL. Newton’s 2nd Law:
∑F = m a
If no other forces are acting, only Fg acts (in
vertical direction). ∑Fy = m ay
Or: Fg = mg (down, of course)
(5.6)
Where: g = 9.8 m/s2
SI Units: Newton (just like any force!).
2
 If m = 1 kg  Fg = (1kg)(9.8m/s ) = 9.8N
Gravitational Force, final
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REMARKS!!!
Fg Depends on g, then it varies with geographic
location.
Fg Decreases from Sea level to a higher altitude.
You want lose Weight without diet  climb a
high mountain.
m in Equation 5.6 is called the gravitational mass
The kilogram is not a unit of Weight is a unit of
Mass.
Mass and Weight are two different quantities
Gravitational & Inertial Mass
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In Newton’s Laws, the mass is the inertial
mass and measures the resistance to a
change in the object’s motion
In the gravitational force, the gravitational
mass is determining the gravitational
attraction between the object and the Earth
Experiments show that gravitational mass
and inertial mass have the same value
5.6 Newton’s Third Law
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If two objects interact, the force F12 exerted
by object 1 on object 2 is equal in magnitude
and opposite in direction to the force F21
exerted by object 2 on object 1
F12 = ̶ F21 (5.7)
Note on notation: FAB is the force exerted by
A on B
Newton’s Third Law, 2
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Forces always occur in pairs
A single isolated force cannot exist
The action force is equal in magnitude
to the reaction force and opposite in
direction
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One of the forces is the action force, the other
is the reaction force
It doesn’t matter which is considered the action
and which the reaction
Newton’s Third Law, 3
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The action and reaction forces must:
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act on different objects.
be of the same type
Forces exerted BY a body DO NOT (directly)
influence its motion!!
Forces exerted ON a body (BY some other
body) DO influence its motion!!
When discussing forces, use the words “BY”
and “ON” carefully.
Example: 5.1 Action-Reaction
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The force F12 exerted
BY object 1 ON
object 2 is equal in
magnitude and
opposite in direction
to F21 exerted BY
object 2 ON object 1
F12 = – F21
Example: 5.2 Action-Reaction
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The force Fhn exerted
BY the hammer ON
the nail is equal in
magnitude and
opposite in direction
to Fnh exerted BY the
nail ON the hammer
Fhn = – Fnh
Example: 5.3 Action-Reaction
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We can walk forward
because when one
foot pushes
backward against
the ground, the
ground pushes
forward on the foot.
– FPG ≡ FGP
Example: 5.4 Action-Reaction
(Normal Force)
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Does the force of gravity stop? OF
COURSE NOT
But, object does not move:
2nd Law  ∑F = m a = 0
 There must be some other
force acting besides gravity
(weight) to have ∑F = 0.
The Normal Force : FN = n
 Normal is math term for
Perpendicular ()
FN is  to the surface & opposite
to the weight (in this simple
case only)
Example: 5.4 Normal Force, 2
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The normal force: n (table
on monitor) is the reaction of
the force the monitor exerts
ON the table
The action (Fg, Earth on
monitor) force is equal in
magnitude and opposite in
direction to the reaction
force, the force the monitor
exerts ON the Earth
Caution: The normal force
is not always = & opposite
to the weight!! As we’ll see!
Free Body Diagram
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In a free body diagram,
you want the forces
acting ON a particular
object
The normal force and
the force of gravity
are the forces that act
ON the monitor
Normal Force
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Where does the Normal Force come
from?
From the other body!!!
Does the normal force ALWAYS equal
to the weight ?
NO!!!
Weight and Normal Force are not ActionReaction Pairs!!!
Example 5.5 Normal Force
m = 10 kg
Weight: Fg = mg = 98.0N
The normal
force is equal
to the weight!!
Only this case
FN = n = mg = 98.0N
Example 5.6 Normal Force, 2
m = 10 kg
Weight: Fg = mg = 98.0N
Pushing Force = 40.0N
The normal force is NOT
ALWAYS equal to the
weight!!
FN = n = 40.0N mg= 138.0N
Example 5.7 Normal Force, final
m = 10 kg
Weight: Fg = mg = 98.0N
Puling Force = 40.0N
The normal force is NOT
ALWAYS equal to the weight!!
FN = n = 98.0N – 40.0N= 58.0N
Example 5.8 Accelerating the box
m = 10 kg  Fg = 98.0N
From Newton’s 2nd Law:
∑F = ma
FP – mg = m a 
m a = 2.0N
The box accelerates
upwards because
FP > m g
Example 5.9 Weight Loss
(Example 5.2 Text Book)
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Apparent weight loss. The lady weights
65kg = 640N, the elevator descends with
a = 0.2m/s2.
 What does the scale read (FN)?
From Newton’s 2nd law: ∑F = ma
FN – mg = – m a  FN = mg – m a
FN = 640N – 13N = 627N = 52kg Upwards!
FN is the force the scale exerts on the person,
and is equal and opposite to the force she
exerts on the scale.
Example 5.9 Weight Loss, 2
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What does the scale read when the
elevator descends at a constant speed
of 2.0m/s?
From Newton’s 2nd law: ∑F = 0
FN – mg = 0  FN = mg = 640N = 65kg
The scale reads her true mass!
NOTE: In the first case the scale reads an
“apparent mass” but her mass does
not change as a result of the
acceleration: it stays at 65 kg
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5.7 Some Applications of
Newton’s Law
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OBJECTS IN EQUILIBRIUM
If the acceleration of an object that can be
modeled as a particle is zero, the object is
said to be in EQUILIBRIUM!!
Mathematically, the net force acting on the
object is zero
F  0
 F  0 and  F
x
y
0
Example 5.10 Equilibrium
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A lamp is suspended from a
chain of negligible mass
The forces acting on the lamp
are
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Force of gravity (Fg)
Tension in the chain (T)
Equilibrium gives
F
y
 0  T  Fg  0
T  Fg
Example 5.10 Equilibrium, final
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The forces acting on the chain
are T’ and T”
T” is the force exerted by the
ceiling
T’ is the force exerted by the
lamp
T’ is the reaction force to T
Only T is in the free body
diagram of the lamp, since T’
and T” do not act on the lamp
Example 5.11 Traffic Light
at Rest
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Example 5.4 (Text Book)
This is an equilibrium problem
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No movement, so a = 0
Upper cables are not strong as
the lower cable. They will break
if the tension exceeds 100N.
Will the light remain or will
one of the cables break?
mg =122N
Example 5.11 Traffic Light
at Rest, 2
Apply Equilibrium Conditions:
ΣFy = 0  T3 – Fg = 0 
T3 = Fg = 122N
 Find Components:
T1x = – T1cos37 T1y = T1sin37
T2x = T2cos53 T2y = T2sin53
T3x = 0
T3y = – 122N
 Apply Newton’s 2nd Law:
ΣFx = 0
ΣFy = 0
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T2y
T1y
-T1x
T2x
Example 5.11 Traffic Light at
Rest, final
ΣFx = – T1cos37 + T2cos53 = 0
(1)
ΣFy = T1sin37 + T2sin53 – 122N = 0 (2)
 Solving for T1 or T2:
From Eqn (1) solve for T2
T2 = (cos37/cos53)T1 =1.33T1
 Substituting this value into Eqn (2)
T1sin37 + (1.33T1)sin53 =122N 
T1 = 74.4 N and T2 = 97.4N
Both values are less than 100N, so
the cables will not break!!!
T2y
T1y
-T1x
T2x
Objects Experiencing a Net
Force, Examples
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If an object that can be modeled as a
particle experiences an acceleration
(a ≠ 0), there must be a nonzero net
force (F ≠ 0) acting on it.
Draw a free-body diagram
Apply Newton’s Second Law in
component form
Example 5.12 Net Force
FR 
F12  F22 
F

1
  tan  2
 F
 1
1002  1002  141N

  tan 1(1)  45
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Forces on The Crate
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Forces acting on the
crate:
A tension, the magnitude
of force T
 The gravitational force, Fg
 The normal force, n,
exerted by the floor
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Forces on The Crate, 2
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Apply Newton’s Second Law in
component form:
F
x
F
y
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 T  max
 n  Fg  0  n  Fg
ONLY in this case n = Fg
Solve for the unknown(s)
If T is constant, then a is constant and the
kinematic equations can be used to more
fully describe the motion of the crate
Example 5.13 Normal Force
The normal force, FN is
NOT always equal to
the weight!!
m = 10.0 kg  Fg = 98.0N
Find: ax ≠ 0? FN?
if ay = 0
m=10kg
Example 5.13 Normal
Force, final
FPy = FPsin(30) = 20.0N
FPx = FPcos(30) = 34.6N
ΣFx = FPx = m ax 
ax = 34.6N/10.0kg 
ax = 3.46m/s2
ΣFy = FN + FPy – mg = m ay 
FN + FPy – mg = 0 
FN = mg – FPy= 98.0N – 20.0N
FPy
FPx
FN = 78.0N
Example 5.14 Conceptual
Example: The Hockey Puck
Moving at constant velocity, with NO friction.
Which free-body diagram is correct?
(b)
Inclined Planes
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Choose the coordinate
system with x along the
incline and y perpendicular
to the incline
Forces acting on the
object:
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The normal force, n, acts
perpendicular to the plane
The gravitational force, Fg , acts
straight down
Example 5.15 The Runaway
Car (Example 5.6 Text Book)
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Replace the force of gravity
with its components:
Fgx = mgsin Fgy = mgcos
With: ay = 0 & ax ≠ 0
(A). Find ax
Using Newton’s 2nd Law:
y-Direction
ΣFy = n – mgcos = may = 0
 n = mgcos
Example 5.15 The Runaway
Car, final
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x-direction
ΣFx = mgsin = max  ax = gsin
Independent of m!!
(B) How long does it take the front of the car to reach the
bottom?
2
1
x f  xi  vxi t  2 ax t 
d  12 axt 2  t 
2d
t 
ax
2d
g sin 
(C). What is the car’s speed at the bottom?
vxf 2  vxi 2  2ax d  vxf 2  2( g sin  )d 
vxf  2 gd sin 