Introduction to Newton’s Laws
• Introducing Forces
• A force is a push or pull on an object. Forces are
what cause an object to accelerate, or to change its
velocity by speeding up, slowing down, or changing
direction.
• It is important that we learn to identify all the
forces acting on an object, and to draw these forces
as vectors.
• Draw the forces on the book
The table pushes up on the book.
FTable
A book rests on a table.
FG
Gravity pulls down on the book.
• There are two main methods for drawing forces
Force Diagram
Free Body diagram
FTable
FTable
FG
FG
• Draw a force diagram and a free body diagram for a
monkey hanging motionless by one arm from a vine
attached to a tree.
Force Diagram
FTension
FG
Free Body diagram
FTension
FG
• Draw a force diagram and a free body diagram for a
monkey hanging motionless by two arm from two
vines hanging from a tree.
Force Diagram
Free Body diagram
FTension1 FTension2
FTension1
FG
FTension2
FG
• Newton’s First Law (Law of Inertial)
– A body in motion stays in motion at constant
velocity and a body at rest stays at rest unless
acted upon by a net external force.
– It is often said that the Law of Inertia violates
“common sense”. Why do you think some people
say that?
• Newton’s 1st Law
– If there is zero net force on a body, it cannot
accelerate, and therefore must move at constant
velocity. This means
• the body cannot turn.
• the body cannot speed up.
• The body cannot slow down.
• An example of zero net force
FTable
A book rests on a table.
FG
Even though there are forces on the book, they are
balanced. Therefore, there is no net force on the book.
∑F = 0
• Mass and Inertia
– Chemists like to define mass as the amount of
“stuff” or “matter” a substance has.
– Physicists define mass as inertia, which is the
ability of a body to resist acceleration by a net
force.
– What is the relationship between mass and
inertia?
Sample Problem: A heavy block hangs from a string
attached to a rod. An identical string hangs down from the
bottom of the block. Which string breaks
a) when the lower string is pulled with a slowly increasing
1st string
force?
FTension
2nd string
Fweight
Fpull
Top will break slow pull
all forces act
FTension
Fpull
b) when the lower string is pulled with a quick jerk?
A quick jerk will break the 2nd string because of the inertia
of the block harder to get the block to move quickly
• Newton’s Second Law
– A body accelerates when acted upon by a net
external force. The acceleration is proportional
to the net force and is in the direction which the
net force acts.
– ∑F = ma
• where ∑F is the net force measured in
Newtons (N)
• m is mass (kg)
• a is acceleration (m/s2)
• Units of force
– Newton (SI system)
• 1 N = 1 kg m /s2
– 1 N is the force required to accelerate a 1 kg mass
at a rate of 1 m/s2
– Pound (British system)
• 1 lb = 1 slug ft /s2
• Working 2nd Law Problems
1. Identify the system being accelerated.
2. Define a coordinate system.
3. Identify forces by drawing a force or free body
diagram.
4. Explicitly write ∑F=ma
5. Replace SF with the actual forces in your free
body diagram.
6. Substitute numeric values, where appropriate,
and solve for unknowns.
• In a grocery store, you push a 14.5-kg cart with a force of
12.0 N. If the cart starts at rest, how far does it move in
3.00 seconds?
m = 14.5 kg
F = ma
F = 12.0 N
t = 3.00s
a = F / m = 12.0N / 14.5 kg = 0.828 m / s2
x = x0 + v0t + 1/2at2
x = 1/2at2 = ½(0.828 m/s2)∙(3.00s)2 = 3.72 m
• A catcher stops a 45 m/s pitch in his glove, bringing
it to rest in 0.15 m. If the force exerted by the
catcher is 803 N, what is the mass of the ball?
m = ? kg F = 803 N v0 = 45 m/s
v = 0 m/s
∆x = 0.15 m
v2 = v02 + 2a∆x
a = -v02/2∆x = -(45 m/s)2/(2∙0.15m) = -6750 m/s2
Fmitt
F = ma
Fball
m = F /a = -803 N / -6750 m/s2 = 0.12 kg
• A catcher stops a 48 m/s pitch in his glove, bringing it to
rest in 0.15 m. If the force exerted by the catcher is 803 N,
what is the mass of the ball?
v0= 48 m/s v= 0 m/s x= 0.15 m F= 803 N
a= ? m/s2
v2 = v02 + 2a∆x
a = (v2 - v02) / (2∆x)
a = ((0m/s)2 – (48m/s)2) / (2∙0.15m) = 7680 m/s2
F= ma
m= F / a = 803 N / 7680 m/s2 = 0.105 kg
• A 747 jetliner lands and begins to slow to a stop as it moves
along the runway. Its mass is 3.50 x 105 kg, its speed is 27.0
m/s, and the net braking force is 4.30 x 105 N a= ? m/s2
5N
F
=
4.30x10
a) What is time it takes the jet to stop?
v0= 27 m/s v= 0 m/s x= ? m
t= ? s m= 3.50 x105 Kg
a= F / m = 4.30x105 N / 3.50x105Kg = -1.23 m/s2
v = v0 + at
t = (v - v0) / a = (0 m/s – 27m/s) / -1.23 m/s2 = 21.98s
b) How far has it traveled in this time?
v0= 27 m/s
v= 0 m/s a= -1.23 m/s2
v2 = v02 + 2a∆x
∆x = (v2 - v02) / (2a)
∆x = ((0 m/s)2 – (27m/s)2) / (2 ∙ -1.23 m/s2) = -296.31 m
• A 747 jetliner lands and begins to slow to a stop as
it moves along the runway. Its mass is 3.50 x 105 kg,
its speed is 27.0 m/s, and the net braking force is
4.30 x 105 N
c) What is its speed 7.50 s later?
v0= 27 m/s v= ? m/s t= 7.50 s
a = -1.23 m/s2
v = v0 + at
v = 27 m/s + (-1.23 m/s2 ∙ 7.50s = 17.78 m/s
• Newton’s Third Law
– For every action there exists an equal and
opposite reaction.
– If A exerts a force F on B, then B exerts a force of
-F on A.
• You rest an empty glass on a table.
a) Identify the forces acting on the glass with a free body
diagram.
FTable
FG
b) Are these forces equal and opposite?
Yes, because the glass is in equilibrium
c) Are these forces an action-reaction pair? Why or why
not?
Yes, because it is not acceleration the forces are equal.
The table is pushing up with the same force that the
glass is pushing down on the table with.
• Requirements for Newton’s Laws
– The 1st and 2nd laws require that ONE system be
analyzed and that ALL the forces on the system
be accounted for.
– The 3rd law requires that TWO systems be
analyzed and that the forces of interaction
between the two be accounted for.
• A force of magnitude 7.50 N pushes three boxes as
shown. Find the acceleration of the system.
Copyright James Walker, “Physics”, 1st edition
F = Ma
M = m1 + m2 + m3
a = F / (m1 + m2 + m3)
a = 7.5N / (1.3 kg + 3.2kg + 4.90 kg)= 0.798 m/s2
• force of magnitude 7.50 N pushes three boxes as
shown. Find the force that box 2 exerts on box 3.
Copyright James Walker, “Physics”, 1st edition
a = 0.798 m/s2
Box 2
exerts
on 3
∑F = F - F23
∑F =F2,3 = (m1 + m2)a
F = (1.3 + 3.20) •0.798
= 3.59 N
∑F = F – m3a
F = 7.50 - 4.90•0.798
= 3.59 N
• force of magnitude 7.50 N pushes three boxes as
shown. Find the force that box 2 exerts on box 3.
Copyright James Walker, “Physics”, 1st edition
a = 0.798 m/s2
Box 1
exerts
on 2
∑F =F1,2 = m1a = 1.3 ∙ 0.798 = 1.04 N
∑F = F - F23 – F12
∑F = F – m3a – m2a
F = 7.50 - 4.90•0.798 – 3.20•0.798
= 1.036 N
• Newton’s 2nd Law in 2-D
– The situation is more complicated when forces
act in more than one dimension.
– You must still identify all forces and draw your
force diagram.
– You then resolve your problem into an x-problem
and a y-problem
(remember projectile motion????).
• Draw a force diagram and a free body diagram for
the man pushing the chair across the floor.
FFloor
Ffriction
Fperson
Fchair
Copyright James Walker, “Physics”, 1st edition
• An object acted on by three forces moves with constant velocity. One
force acting on the object is in the positive x direction and has a
magnitude of 6.5 N; a second force has a magnitude of 4.4 N and
points in the negative y direction. Find the direction and magnitude
of the third force acting on the object.
Fx = 6.5 N
Fx = 6.5N
Fy = 4.4 N R2 = Fx2 + Fy2
F= ? N
Fy = 4.4 N
c = √(Fx2 + Fy2 ) = √((6.5N)2 + (4.4N)2 ) = 7.84 N
θ = tan-1 (Fy / Fx) = tan-1 (4.4 N / 6.5 N) = 34.1°
Now we use the inverse vector
F = 7.84 N
θ = 34.1°
Fx = 6.5N
Fy = 4.4 N
Mass, Weight, and Apparent Weight
• Mass versus Weight
– Mass is inertia, or resistance to acceleration.
– Weight and mass are not equivalent!
– Weight is gravitational force, which is equal to mg
near the earth’s surface.
• A man weighs 150 pounds on the surface of the
earth at sea level. Calculate his
a) mass in kg.
1 kg = 2.2 lbs
150 lbs
1 kg
2.2 lbs
= 68.18 kg
b) weight in Newtons.
F = ma
W = ma
F = 68.18 kg ∙ -9.8 m/s2 = 668.18 N
• Apparent weight
– Apparent weight is a force that acts in opposition
to gravitational force in order to prevent a body
from going into freefall.
– When you stand on the floor, the floor pushes up
on your feet with a force equal to your apparent
weight.
• Elevator rides
– When you are in an elevator, your actual weight
(mg) never changes.
– You feel lighter or heavier during the ride
because your apparent weight increases when
you are accelerating up, decreases when you are
accelerating down, and is equal to your weight
when you are not accelerating at all.
v = 0 m/s
a = 0 m/s2
V ≥ 0 m/s
a > 0 m/s2
V > 0 m/s
a = 0 m/s2
V ≥ 0 m/s
a < 0 m/s2
Heavy feeling
Normal feeling
Normal feeling
Lighter feeling
Ground floor Just starting up Between Arriving
floors
at top
floor
v = 0 m/s
a = 0 m/s2
V ≤ 0 m/s
a < 0 m/s2
V < 0 m/s
a = 0 m/s2
V ≤ 0 m/s
a > 0 m/s2
Lighter feeling
Normal feeling
Normal feeling
Heavy feeling
Top floor
Beginning
descent
Between Arriving
floors
at
ground
floor
• Sample problem: An 85-kg person is standing on a bathroom scale in
an elevator. What is the person’s apparent weight
Wapp
a) when the elevator accelerates upward at 2.0 m/s2?
∑Fy: Wapp - W = ma
Wapp = ma + W
Wapp = ma + mg
Wapp = m(a + g)= 85 kg(2.0 m/s2 + 9.8 m/s2) = 1003 N
W
b) when the elevator is moving at constant velocity between floors?
Wapp = m(a + g)= 85 kg(0 m/s2 + 9.8 m/s2) = 833 N
c) when the elevator begins to slow at the top floor at 2.0 m/s2?
Wapp = m(a + g)= 85 kg(-2.0 m/s2 + 9.8 m/s2)= 663 N
• 5-kg salmon is hanging from a fish scale in an elevator.
What is the salmon’s apparent weight when the elevator is
a) at rest
Wapp = m(a + g)= 5 kg(0 m/s2 + 9.8 m/s2) = 49 N
b) moving downward and slowing at 3.2 m/s2?
Wapp = m(a + g)= 5 kg(-3.2 m/s2 + 9.8 m/s2) = 33 N
Normal Force
• Normal force
– The normal force is a force that keeps one object
from penetrating into another object.
– The normal force is always perpendicular a
surface.
– The normal exactly cancels out the components
of all applied forces that are perpendicular to a
surface.
• derive an expression for the normal force of a box
FN
• on a flat table.
y
x
W
∑Fy: FN - W = 0
FN = W
ma = mg
• Derive an expression for the normal force of a box
sitting on a ramp.
FN
W
Wy= mg cos θ θ
Wx= mg sin θ
W = mg
∑Fy: FN - WY = 0
FN = mg cos θ
• derive an expression for the normal force an eraser
being pushed up against a whiteboard by a force F.
f
∑Fx: FN - F = 0
FN = F
FN
F
W
• Derive the normal force for the box in the picture
below. The box is sitting on the floor, but is being
pulled by the force shown. Ignore friction.
FN
F
F
40°
Fx
∑Fx : Fx = 0
W
∑Fy : FN + Fy – W = 0
∑Fy : FN + Fy – mg = 0
FN = mg – Fy
FN = 6.0 kg ∙ 9.8 m/s2 – 12.86 N = 43.48 N
Fx = F cos θ
Fx = 20 N cos 40°
Fx = 15.32 N
Fx = F cos θ
Fx = 20 N sin 40°
Fx = 12.86 N
Fy
• A 2.5 kg book rests on a surface inclined at 28° above the
FN
horizontal
A. Find the normal force.
f
θ = 28°
ΣFy: FN – mg sin θ = 0
ΣFy: FN = mg sin θ
Fgx = mg
θ
Fgy= mg cos θ
Fgx= mg sin θ
Fgx = mg
FN = 2.5 kg ∙ 9.8 m/s2 sin 28° = 11.50 N
B. If the angle of the incline is reduced, do you expect the
normal force to increase, decrease, or stay the same?
Increase as the angle approaches 0° the FN will approach the
weight
• A skier skis down a slope with an acceleration of
3.50 m/s2. If friction can be ignored, what is the
angle of the slope with respect to the horizontal?
θ = ?°
θ
Fgy= mg cos θ
Fgx= mg sin θ
Fg = mg
ΣFx: - Fgx = ma
ΣFx: – mg sin θ = ma
– g sin θ = a
θ = sin-1 (a/ – g)
θ = sin-1(-3.5m/s2 /– 9.8 m/s2) = 20.92°
• How long will it take a 1.0 kg block initially at rest to slide
down a frictionless 20.0 m long ramp that is at a 15° angle
L = 20 m m=1.0 kg
FN
with the horizontal?
Fg
θ = 15°
θ
Fgy= mg cos θ
Fgx= mg sin θ
ΣFx: - Fgx = ma
ΣFx: – mg sin θ = ma
Fg = mg
– g sin θ = a
a = – 9.8 m /s2 sin (15°) = – 2.54 m /s2
t
unk
v
V0
0 m/ s ?
∆x
20m
a
-2.54 m/s2
t
unk
v
?
V0
0 m/ s
∆x
-20m
a
-2.54 m/s2
v2 = v02 + 2a(∆x)
v2 = 0 + 2a(∆x)
v2= 2a(∆x)
v= √(2a(∆x))
v= ±√(2∙(-2.54 m/s2)∙(-20m))= - 10.08 m/s
v = v0 + at
t =(v - v0) / a
t =(-10.8 m/s - 0) / -2.54 m/s2= - 3.97 s
• Friction
– Friction is the force that opposes a sliding
motion.
– Friction is highly useful. It enables us to walk and
drive a car, among other things.
– Friction is also dissipative. That means it causes
mechanical energy to be converted to heat. We’ll
learn more about that later.
• What causes friction?
FN
friction
f
W
Big view:
Surfaces look
perfectly
smooth.
Fpush
Small view:
Microscopic
irregularities
resist
movement.
• Friction may or may not exist between two surfaces.
If it exists, it opposes the direction object “wants”
to slide. It is parallel to the surface.
• Friction depends on the normal force.
– The friction that exists between two surfaces is
directly proportional to the normal force.
– This has several implications, such as…
• Friction on a sloping surface is less than friction on a
flat surface (since the normal force is less on a slope).
• Increasing weight of an object increases the friction
between the object and the surface it is resting on.
• Weighting down a car over the drive wheels
increases the friction between the drive wheels and
the road (which increases the car’s ability to
accelerate).
• Static Friction
– This type of friction occurs between two surfaces
that are not slipping relative to each other.
– Static friction is tricky. It can range from zero up
to a maximum allowed value for two surfaces.
– fs ≤ μsFN
• fs : static frictional force (N)
• μs: coefficient of static friction
• FN: normal force (N)
Warning: fs ≤ μsFN is an inequality!
– The fact that the static friction equation is an
inequality has important implications.
– Static friction between two surfaces is zero unless
there is a force trying to make the surfaces slide
on one another.
– Static friction can increase as the force trying to
push an object increases until it reaches its
maximum allowed value as defined by μs.
– Once the maximum value of static friction has
been exceeded by an applied force, the surfaces
begin to slide and the friction is no longer static
friction.
• Static friction and applied horizontal force
Force Diagram
FN
Surface
W
There is no static friction since there is no applied
horizontal force trying to slide the book on the
surface.
• Static friction and applied horizontal force
Force Diagram
FN
fs
F
Surface
W
Static friction is equal to the applied horizontal force,
and there is no movement of the book since ∑F = 0.
• Static friction and applied horizontal force
Force Diagram
FN
fs
F
Surface
W
Static friction is at its maximum value! It is still equal
to F, but if F increases any more, the book will slide.
• Static friction and applied horizontal force
Force Diagram
FN
F
fs
Surface
W
Static friction cannot increase any more! The book
accelerates to the right. Friction becomes kinetic
friction, which is usually a smaller force.
• Static friction on a ramp
θ
W = mg
Without friction, the
book will slide down
the ramp. If it stays
in place, there is
sufficient static
friction holding it
there.
At maximum angle before the book slides,
let’s prove that μs = tan θ.
FN
θ
Fgy= mg cos θ
Fgx= mg sin θ
Fg
ΣFx: - Fgx + Fs= 0
– mg sin θ + μsFN = 0
μsFN = mg sin θ
FN = mg sin θ / μs
mg cos θ = mg sin θ / μs
Fg = mg
ΣFy: - Fgy + FN = 0
- mg cos θ + FN = 0
FN = mg cos θ
μs = mg sin θ / mg cos θ
μs = sin θ / cos θ
μs = tan θ
• Kinetic Friction
– This type of friction occurs between surfaces that
are slipping past each other.
– fk = μkFN
• fk : kinetic frictional force (N)
• μk: coefficient of kinetic friction
• FN: normal force (N)
– Kinetic friction (sliding friction) is generally less
than static friction (motionless friction) for most
surfaces.
A 10-kg box rests on a ramp that is laying flat. The
coefficient of static friction is 0.50, and the coefficient of
kinetic friction is 0.30.
a) What is the maximum horizontal force that can be
applied to the box before it begins to slide?
μs = 0.50
m = 20 kg
fs = μsFN = 0.5 ∙ 98 N = 49 N
FN= mg = 10 kg ∙ 9.8 m/s2 = 98 N
b) What force is necessary to keep the box sliding at
constant velocity?
μk = 0.30
m = 20 kg
FN= mg = 10 kg ∙ 9.8 m/s2 = 98 N
fk = μkFN = 0.3 ∙ 98 N = 29.4 N
A 10-kg wooden box rests on a wooden floor. The
coefficient of static friction is 0.50, and the coefficient of
kinetic friction is 0.30. What is the friction force between
the box and floor if
a) no force horizontal force is applied to the box?
The frictional force is 0 N
A 10-kg wooden box rests on a wooden floor. The
coefficient of static friction is 0.50, and the coefficient of
kinetic friction is 0.30. What is the friction force between
the box and floor if
b) 20 N horizontal force is applied to the box?
μs = 0.50
m = 20 kg
fs = μsFN = 0.5 ∙ 98 N = 49 N
The frictional force is 20 N
FN= mg = 10 kg ∙ 9.8 m/s2 = 98 N
A 10-kg wooden box rests on a wooden floor. The
coefficient of static friction is 0.50, and the coefficient of
kinetic friction is 0.30. What is the friction force between
the box and floor if
c) a 60 N horizontal force is applied to the box?
μs = 0.50
m = 20 kg
fs = μsFN = 0.5 ∙ 98 N = 49 N
FN= mg = 10 kg ∙ 9.8 m/s2 = 98 N
When it begins to move the frictional force is 49 N, so the box
is accelerating.
μk = 0.30
m = 20 kg
FN= mg = 10 kg ∙ 9.8 m/s2 = 98 N
fk = μkFN = 0.3 ∙ 98 N = 29.4 N
The horizontal for exceeds the kinetic friction, so it will
continue to accelerate until other forces like air resistance
balance out the horizontal force.
A 10-kg wooden box rests on a wooden ramp. The
coefficient of static friction is 0.50, and the coefficient of
FN
kinetic friction is 0.30. What
is the friction force between
the box and ramp if
a) the ramp is at a 25° angle?
F
N
θ
θ = 25°
Fg
Fgx= mg sin θ
Fg = mg
ΣFy: - Fgy + FN = 0
- mg cos θ + FN = 0
FN = mg cos θ
FN = 10 kg ∙ 9.8 m/s2 cos (25°)= 88.82 N
fs = μsFN = 0.5 ∙ 88.2 N = 44.1 N
Fgy= mg cos θ
A 10-kg wooden box rests on a wooden ramp. The
coefficient of static friction is 0.50, and the coefficient of
kinetic friction is 0.30. What is the friction force between
the box and ramp if
a) the ramp is at a 45° angle?
ΣFy: - Fgy + FN = 0 - mg cos θ + FN = 0
FN = mg cos θ
FN = 10 kg ∙ 9.8 m/s2 cos (45°)= 69.29 N
fs = μsFN = 0.5 ∙ 69.29 N = 34.65 N
b) what is the acceleration of the box when the ramp is at
45°? ΣFx: - Fgx + fs= ma Fgx = – mg sin θ= – mg sin θ= – 69.3 N
a = (– Fgx+ fs) / m
Prove weight in x
greater than friction
force
a = (– 69.3 N + 34.65 N) / 10 kg= - 3.74 m/s2
• Tension
– Tension is a pulling force that arises when a rope,
string, or other long thin material resists being
pulled apart without stretching significantly.
– Tension always pulls away from a body attached
to a rope or string and toward the center of the
rope or string.
• A physical picture of tension
• Imagine tension to be the internal force preventing a
rope or string from being pulled apart. Tension as such
arises from the center of the rope or string. It creates
an equal and opposite force on objects attached to
opposite ends of the rope or string.
• Tension examples
Note that the
pulleys shown
are magic! They
affect the
tension in any
way, and serve
only to bend the
line of action of
the force.
A 1,500 kg crate hangs from a crane cable.
a) What is the tension in the cable when the crate is
motionless? Ignore the mass of the cable.
T
ΣFy: - Fg + T= 0
Fg = mg
- Fg + T= 0
T= Fg
T= mg = 1500 kg ∙ 9.8 m/s2 = 14700 N
b) Suppose the crane accelerates the crate upward at 1.2
m/s2. What is the tension in the cable now?
ΣFy: - Fg + T= ma
- Fg + T= ma
T= ma + Fg= 1500 kg ∙ 1.2 m/s2 + 1500 kg ∙ 9.8 m/s2 = 16500 N
• Springs (Hooke’s Law)
• The magnitude of the force exerted by a spring is
proportional to the amount it is stretched.
• F = -kx
– F: force exerted by the spring (N)
– k: force constant of the spring (N/m or N/cm)
– x: displacement from equilibrium (un-stretched and
uncompressed) position (m or cm)
• The negative sign indicates that the direction of the
force is back toward the equilibrium (or unstretched) position.
• 1.50 kg object hangs motionless from a spring with
a force constant of k = 250 N/m. How far is the
spring stretched from its equilibrium length?
m = 1.50 kg
k = 250 N/m
F = ma = 1.5 kg ∙ 9.8 m/s2 = 14.7 N
F = -kx
x =F / -k = - 14.7 N / (250 N / m)= - 0.059 m
The force applied to the string is negative. In the same direction as the
displacement. The force applied by the spring in in the opposite direction of the
applied force so it is positive, so the applied work is positive, so the work by the
spring is negative
• A 1.80 kg object is connected to a spring of force constant
120 N/m. How far is the spring stretched if it is used to drag
the object across a floor at constant velocity? Assume the
coefficient of kinetic friction is 0.60.
μk = 0.60 m = 1.80 kg
FN= mg = 1.80 kg ∙ 9.8 m/s2 =17.64 N
fk = μkFN = 0.6 ∙ 17.64 N = 10.58 N
k = 120 N/m
x =F / -k = - 10.58 N / (120 N / m) = - 0.088 m
• A 5.0 kg object is connected to a 10.0 kg object by a string. If a
pulling force F of 20 N is applied to the 5.0 kg object,
A) what is the acceleration of the system?
10.0 kg
Box 2
5.0 kg
Box 1
T2 = ? N T 1 = ? N
2
F = 20 N
1
∑Fx2 : T2 = m2a
F – m2a= m1a
a = 20 N / (5.0 Kg + 10.0 Kg)
∑Fx1 : F – T1 = m1a F = m1a + m2a
a = 1.33 m/s2
a = F / (m1 + m2)
T1 = T2
B) what is the tension in the string connecting the objects?
We can use either equation to calculate the tension on the cable
between the 2 boxes. Although the tension on the string comes
from box 2
∑Fx1 : F – T1 = m1a
∑Fx2 : T2 = m2a
T1 = F - m1a= 20 N – 5.0 Kg ∙ 1.33 m/s2 = 13.35 N
T2 = 10 kg ∙ 1.33 m/s2 = 13.3 N
(Assume a frictionless surface.)
• Gravity
• A very common accelerating force is gravity. Here is
gravity in action. The acceleration is g.
• Slowing gravity down
• The pulley lets us use gravity as our accelerating
force… but a lot slower than free fall. Acceleration
here is a lot lower than g.
• Pulleys
• Pulleys bend the line of action of the force of gravity
without affecting tension.
• The pulley is “magic” – no mass, no friction, and no
effect on the tension.
• The pulley simply bends the line of action of the
force.
• Derive an expression for the acceleration due to
gravity of the system below, and for the tension in
T in both equations equal,
the string.
so set each equation equal
ΣFy2: T- m2g = m2a2y
T = m2a2y + m2g
to T and set equal to each
other
m1a1x = m2a2y + m2g
a2y = -a a1x = a
FBD
FN
m1a = m2(-a) + m2g
T
m1g
m1a + m2a = m2g
a(m1 + m2) = m2g
+y
FBD
+x
a = m2g / (m1 + m2)
T
ΣF1x: T= m1a1x
T= m1a1x
m2g
• A 10 kg block rests on a table connected by a string to a 5
kg block Find the minimum coefficient of static friction for
which the blocks remain stationary.
ΣFy2: T- m2g = m2a2y = 0
ΣF1x: T – fs = m1a1x = 0
T = m2g
T= fs
m2g = fs
m2g = μsFN m2g = μsm1g m2 = μsm1
μs =m2 / m1 = (5 kg) / (10 kg) = 0.5
FN
fs
m1
T= μsm1g = 0.5∙10 kg ∙ 9.8 m/s2 = 49 N
T
m1g
T
m2
m2g
• Derive the acceleration, assuming a coefficient of
friction of 0.20 between the table and the 3.0 kg
block. Determine the tension in the string.
ΣF1x: T – Fg1x - fk= m1a1x
θ
T – m1g sin θ - μFN = m1a1x
ΣF1y: FN – Fg1y= 0
Fg1x= mg sin θ
F1g = mg
FN = m1g cos θ
T – m1g sin θ - μ m1g cos θ = m1a1x
FN
FBD
fk
a2y = -a
a1x = a
T – m1g sin θ - μ m1g cos θ = m1a
T
+y
Fg1y= mg cos θ
+x
FBD
ΣFy2: T- m2g = m2a2y
ΣFy2: T- m2g = m2 (-a)
T+ y
m1g
+x
m2g
T = -m2a + m2g
T – m1g sin θ - μ m1g cos θ = m1a
T=
T = m1a + m1g sin θ + μ m1g cos θ
-m2a + m2g = m1a + m1g sin θ + μ m1g cos θ
m2a +m1a + = m2g - m1g sin θ - μ m1g cos θ
a = (m2g - m1g sin θ - μ m1g cos θ) / (m1 + m2)
a = (5.0 kg ∙ 9.8 m/s2 – 3.0 kg ∙ 9.8 m/s2 ∙ sin 35° - 0.2∙ 3.0 kg ∙ 9.8
m/s2 cos 35°) / (5.0 kg + 3.0 kg)
a = 3.42 m/s2
T = m2g - m2a
T = 5.0 kg ∙ 9.8 m/s2 - 5.0 kg ∙ 3.42 m/s2 = 31.9 N
• Uniform Circular Motion
– An object that moves at uniform speed in a circle
of constant radius is said to be in uniform
circular motion.
– Question: Why is uniform circular motion
accelerated motion?
– Answer: Although the speed is constant, the
velocity is not constant since an object in uniform
circular motion is continually changing direction.
• Centrifugal Force
• Question: What is centrifugal force?
• Answer: That’s easy. Centrifugal force is the force
that flings an object in circular motion outward.
Right?
• Wrong! Centrifugal force is a myth!
• There is no outward directed force in circular
motion. To explain why this is the case, let’s review
Newton’s 1st Law.
• Newton’s 1st Law and cars
– When a car accelerates forward suddenly, you as
a passenger feel as if you are flung backward.
• You are in fact NOT flung backward. Your
body’s inertia resists acceleration and wants to
remain at rest as the car accelerates forward.
– When a car brakes suddenly, you as a passenger
feel as if you are flung forward.
• You are NOT flung forward. Your body’s inertia
resists acceleration and wants to remain at
constant velocity as the car decelerates.
• When a car turns
• You feel as if you are flung to the outside. You call this
apparent, but nonexistent, force “centrifugal force”.
• You are NOT flung to the outside. Your inertia resists the
inward acceleration and your body simply wants to keep
moving in straight line motion!
• As with all other types of acceleration, your body feels as
if it is being flung in the opposite direction of the actual
acceleration. The force on your body, and the resulting
acceleration, actually point inward.
• Centripetal Acceleration
– Centripetal (or center-seeking) acceleration
points toward the center of the circle and keeps
an object
– This type of acceleration is at right angles to the
velocity.
– This type of acceleration doesn’t speed up an
object, or slow it down, it just turns the object.
• Centripetal Acceleration
– ac = v2/r
• ac: centripetal
ac
• acceleration in m/s2
• v: tangential speed in
m/s
v
• r: radius in meters
Centripetal acceleration always points toward center
of circle!
Centripetal Force
• A force responsible for centripetal
acceleration is referred to as
F
a centripetal force.
• Centripetal force is simply mass
times centripetal acceleration.
• ∑F = m a
• ∑F = m v2 / r
Always toward
– F: centripetal force in N
center of
circle!
– v: tangential speed in m/s
– r: radius in meters
• Any force can be centripetal
– The name “centripetal” can be applied to any
force in situations when that force is causing an
object to move in a circle.
– You can identify the real force or combination of
forces which are causing the centripetal
acceleration.
– Any kind of force can act as a centripetal force
• Static friction
As a car makes a
turn on a flat road,
what is the real
identity of the
centripetal force?
• Tension
As a weight is tied
to a string and spun
in a circle, what is
the real identity of
the centripetal
force?
• Gravity
As the moon orbits
the Earth, what is
the real identity of
the centripetal
force?
• Normal force with help from static friction
As a racecar
turns on a
banked curve on
a racing track,
what is the real
identity of the
centripetal
force?
• Tension, with some help from gravity
As you swing a
mace in a
vertical circle,
what is the true
identity of the
centripetal
force?
• Gravity, with some help from the normal force
When you are riding
the Tennessee
Tornado at
Dollywood, what is
the real identity of
the centripetal force
when you are on a
vertical loop?
• A 1200-kg car rounds a corner of radius r = 45 m. If
the coefficient of static friction between tires and
the road is 0.93, what is the maximum velocity the
car can have without skidding?
FN
+y
fs
+x
w
μsFN = mv2/r
2/r
v = √μsgr
μ
mg
=
mv
FN = mg
s
v = √0.93 ∙ 9.8 m/s2 · 45m = 20.25 m/s must be less
than 13.1 m/s
ΣFx: fs = Fc
• You whirl a 1.0 kg stone in a horizontal circle about
your head. The rope attached to the stone is 1.5 m
long.
a) What is the tension in the rope? (The rope makes
a 10° angle with the horizontal).
ΣFy: T cos θ – mg = 0
L
T = (1 kg ∙ 9.8 m/s2 )/ cos (80°) = 56.44 N
b) How fast is the stone moving?
ΣFx: T sin θ = Fc T sin θ = mv2/r
T sin θ = mv2/(L sin θ)
θ
r
r = L sinθ
T
FBD
v = √ (LT (sin θ)2 / m)
v = √ (1.5 m ∙ 56.44N (sin 80)2 / 1.0 kg) = 82.11 m/s
mg
• Isaac Newton
• Arguably the greatest scientific genius ever.
• Came up with 3 Laws of Motion to explain the
observations and analyses of Galileo and Johannes
Kepler.
• Discovered that white light was composed of many
colors all mixed together.
• Invented new mathematical techniques such as
calculus and binomial expansion theorem in his
study of physics.
• Published his Laws in 1687 in the book
Mathematical Principles of Natural Philosophy.
• The Universal Law of Gravity
– Newton’s famous apple fell on Newton’s famous
head, and lead to this law.
– It tells us that the force of gravity objects exert on
each other depends on their masses and the
distance they are separated from each other.
• The Force of Gravity
• FG = Gm1m2/r2
– Fg: Force due to gravity (N)
– G: Universal gravitational constant 6.67 x 10-11 N
m2/kg2
– m1 and m2: the two masses (kg)
– r: the distance between the centers of the
masses (m)
• The Universal Law of Gravity ALWAYS works,
• whereas FG = mg only works when you’re
• standing on the surface of the earth.
a) How much force does the earth exert on the
moon? b) How much force does the moon exert on
the earth? mearth = 5.9375x1024 kg mmoon = 7.34 x 1022 kg
rmoon = 1738000 m
rearth = 6357000 m
Distance surface to surface = 21605000 m
FG = Gm1m2/r2
FG = 6.67 x 10-11 N m2/kg2 ∙ 5.9375x1024 kg ∙ 7.34 x 1022 kg
/(1738000 m + 6357000 m + 21605000 m)2
FG =3.3 x 1022 N
• Sally, an astrology buff, claims that the position of the planet Jupiter
influences events in her life. She surmises this is due to its
gravitational pull. Joe scoffs at Sally and says “your Labrador Retriever
exerts more gravitational pull on your body than the planet Jupiter
does”. Is Joe correct? (Assume a 50 kg Lab 1.0 meter away, and Sally
is 70 kg).
msally = 70 kg
mjupiter = 1.782X1027 kg
rjupiter = 67100000 m
Distance surface to surface = 6.31 x 1011 m
FG = Gm1m2/r2
FG = 6.67 x 10-11 N m2/kg2 ∙ 70 kg ∙ 1.782 x 1027 kg /(67100000 m
+ 6.31x1011 m)2
FG = 2.1 x 10-5 N
FG = 6.67 x 10-11 N m2/kg2 ∙ 70 kg ∙ 50 kg /(1 m)2
FG = 2.3 x 10-7 N
• What would be your weight (assume 90 Kg) if you were orbiting the
earth in a satellite at an altitude of 3,000,000 km above the earth’s
surface? (Note that even though you are apparently weightless,
gravity is still exerting a force on your body, and this is your actual
weight.)
mearth = 5.9375x1024 kg
rearth = 6357000 m msally = 90 kg
Distance surface to person = 3000000000 m
FG = Gm1m2/r2
FG = 6.67 x 10-11 N m2/kg2 ∙ 5.9375x1024 kg ∙ 90 kg /(1738000 m +
6357000 m + 3.0 x 109 m)2
FG = 3.94 x 10-3 N
• Acceleration and distance
• Surface gravitational acceleration depends on mass
and radius
Planet
Mercury
Planet Radius (m) Mass (kg)
2.43 x 106
3.2 x 1023
g (m/s2)
3.61
Venus
Mars
Jupiter
Saturn
6.07 x 106
3.38 x 106
6.98 x 107
5.82 x 107
4.88 x 1024
6.42 x 1023
1.9 x 1027
5.68 x 1026
8.83
3.75
26.0
11.2
Uranus
Neptune
Pluto
2.35 x 107
2.27 x 107
1.15 x 106
8.68 x 1025
1.03 x 1026
1.2 x 1022
10.5
13.3
0.61
• What is the acceleration due to gravity at the
surface of the moon?
rmoon = 1738000 m
mmoon = 7.34 x 1022 kg
mobject = 1 kg
FG = Gm1m2/r2
FG = 6.67 x 10-11 N m2/kg2 ∙ 1 kg ∙ 7.34 x 1022 kg /(1738000 m)2
FG =1.62N
FG = ma
a = FG / m = 1.62 N / 1 kg = 1.62 m/s2
• Johannes Kepler (1571-1630)
– Kepler developed some extremely important laws
about planetary motion.
– Kepler based his laws on massive amounts of
data collected by Tycho Brahe.
– Kepler’s laws were used by Newton in the
development of his own laws.
• Kepler’s Laws
1. Planets orbit the sun in elliptical orbits, with the
sun at a focus.
2. Planets orbiting the sun carve out equal area
triangles in equal times.
3. The planet’s year is related to its distance from
the sun in a predictable way.
• Kepler’s Laws
• Satellites
• Orbital speed
•At any given altitude,
there is only one speed
for a stable circular orbit.
•From geometry, we can
calculate what this
orbital speed must be.
•At the earth’s surface, if
an object moves 8000
meters horizontally, the
surface of the earth will
drop by 5 meters
vertically.
•That is how far the object
will fall vertically in one
second (use the 1st
kinematic equation to
show this).
•Therefore, an object
moving at 8000 m/s will
•never reach the earth’s
surface.
• Some orbits are highly elliptical
• Some orbits are nearly circular.
•The orbits we
analyze
mathematically will
be nearly circular.
•The analysis starts as
follows
• Using Newton’s Law of Universal Gravitation, derive
a formula to show how the period of a planet’s
orbit varies with the radius of that orbit. Assume a
nearly circular orbit. (This is a derivation of Kepler’s
3rd Law.)
• What velocity does a satellite in orbit about the
earth at an altitude of 25,000 km have? What is the
period of this satellite?
• A geosynchronous satellite is one which remains
above the same point on the earth. Such a satellite
orbits the earth in 24 hours, thus matching the
earth's rotation. How high must a geosynchronous
satellite be above the surface to maintain a
geosynchronous orbit?