Transcript Newton Law Application Lecture

Newton 2nd Law problems
- Atwood Machines
-Incline Planes
-Tension Problems
-Other Object Connected problems
Tension
Ropes- Assumptions
• Ignore any frictional effects of the rope
• Ignore the mass of the rope
• The magnitude of the force exerted along the rope is called the
tension
• The tension is the same at all points in the rope
Section 4.5
Tension Forces
• A taut rope has a force exerted on it.
• If the rope is lightweight and flexible the force
is uniform over the entire length.
• This force is called tension and points along
the rope.
forces on the block
FN
forces on the rope
FT
Frope
-FT
Ffr
m
m
Fg
Frope = -FT by the law of reaction
Tension or Normal Force
• Tension and normal forces are different.
– A pull on an object - tension
– A push from a surface - normal force
• Either one or both may be present.
Normal force
Tension force
FN
FT
Ffr
m
Fg
Coupled Motion
v2
v1 = v2
FT2
Ffr2
m2
FT2
Ffr1
FT1
m1
• Objects linked by tension move together
– Same velocity and acceleration
• Tension may not be the same on two ropes
– FT2 = Ffr2
– FT1 = Ffr1 + FT2 = Ffr1 + Ffr2
Pulley
• A pulley uses tension to
transfer a force to another
direction.
FT
forces on the rope
m1
FT
m2
Ffr
Frope
forces on block 1
m1
forces on block 2
m1
m2
Frope
m2
Fg
Tension is equally distributed in a rope and can be bidirectional
Draw FBD block
30N
Tension does not always equal weight:
Draw FBD hand
Derive a general formula for the acceleration of the system and tension in
the ropes
m1
m2
Derive a general formula for the acceleration of the system and tension in
the ropes
Draw FBDs and set your sign
conventions: Up is positive down is
negative
m1
m1
m2
m2
Derive a general formula for the acceleration of the system and tension in
the ropes
Draw FBDs and set your sign
conventions: If up is positive down is
negative on the right, the opposite is
happening on the other side of the
pulley.
-T
+T
m1
m1
m2
m2
+m1g
-m2g
Then, do ΣF= ma to solve for a if possible.
Derive a general formula for the acceleration of the system and tension in
the ropes
-T
+T
m1
m1
m2
m2
+m1g
Σ F1 = - T + m 1 g = m 1 a
-m2g
Σ F2 = T – m2 g = m 2 a
To find acceleration I must eliminate T:
so solve one equation for T and then
substitute it into the other equation (This is called “solving simultaneous equations” )
• Σ F1 = - T + m1g = m1 a
Σ F2 = T – m2g = m2 a
• Σ Fnet = m1 a+ m2 a = - T + m1g + T – m2g
• -T and T Cancel !!!!!
• Σ Fnet =m1 a+ m2 a = m1g – m2g
• Now get all the a variables on one side and
solve for a. Solve!!!!!
m1 g - m1 a – m2 g = m 2 a
m1 g – m 2 g = m 2 a + m1 a
Factor out g on left and a on right
g(m1– m2) = a (m2 + m1)
Divide this away
(m1– m2)
(m2 + m1)
g =a
Now you can find the tension formula by eliminating a. Just plug the new
a formula either one of the original force equations.
Σ F1 = - T + m 1 g = m 1 a
-T
m1
m1
m2
+m1g
Σ F2 = T – m2 g = m 1 a
Pulley Acceleration

Ffr
• The normal force on m1 equals
the force of gravity.
Consider two masses linked
• The force of gravity is the only
by a pulley
force on m2.
• m2 is pulled by gravity
• Both masses must accelerate
• m1 is pulled by tension
together.
• frictionless surface
Fnet  ma
Frope
m2 g  (m1  m2 )a
m1
m2
m2
a
g
m1  m2
Atwood’s Machine Problem 1
• In an Atwood machine both
masses are pulled by gravity, but
the force is unequal.
Fnet  ma
m1 g - m2 g  (m1  m2 )a
m1 - m2
a
g
m1  m2

The heavy weight will move downward at
• (3.2 - 2.2 kg)(9.8 m/s2)/(3.2 + 2.2 kg) = 1.8 m/s2.

Using y = (1/2)at2, it will take t2 = 2(1.80 m)/(1.8 m/s2)
• t = 1.4 s.
Atwood’s Machine (simple)
Forces of Friction
• When an object is in motion on a surface or
through a viscous medium, there will be a
resistance to the motion
– This is due to the interactions between the object
and its environment
• This is resistance is called friction
Section 4.6
More About Friction
• Friction is proportional to the normal force
• The force of static friction is generally greater than
the force of kinetic friction
• The coefficient of friction (µ) depends on the
surfaces in contact
• The direction of the frictional force is opposite the
direction of motion
• The coefficients of friction are nearly independent of
the area of contact
Section 4.5
Static Friction, ƒs
• Static friction acts to
keep the object from
moving
• If F increases, so does ƒs
• If F decreases, so does ƒs
• ƒs  µs n
– Use = sign for impending
motion only
Section 4.5
Kinetic Friction, ƒk
• The force of kinetic
friction acts when the
object is in motion
• ƒk = µk n
– Variations of the
coefficient with speed
will be ignored
Section 4.5
Friction, final
• Adjust the force and see
where you are on the
graph
• Note especially where ƒ
= Fs
Section 4.5
Some Coefficients of Friction
Section 4.5
Other Types of Friction
• Friction between the
moving car’s wheels
and the road is static
friction
– Unless the car is skidding
• Also have the air
resistance,
Section 4.5
Inclined Planes
Axes for Inclined Planes
• X axis is parallel to the inclined plane
• Y axis is perpendicular to the inclined plane
• Friction force (f or Ff) is on the x axis opposite
the motion
• Weight or Force of Gravity (Fg) is straight down
and must be split into x and y components.
• Normal force (FN) is upward along the y axis
• Applied force (Fa) may be in either direction on
the x axis or at angle (which would require it to
be split into x and y components)
Forces Acting on the Object
y
Note: The applied force and
the force of friction can be
in either direction as long as
the friction force is opposite
to the motion.
F
Fa
N
f
Fgy
Fgx

Fgx
Fg

x
Remember: Weight or force of gravity = F g= mg
(some may use different symbols such as W or FW for weight)
Useful Equations for Inclined Plane
Problems
•
•
•
•
•
•
•
Formulas
Fnet,x = Fx = max
Fnet,y = Fy = may
Fg = mg
Fgx = mg sin θ
Fgy = mg cos θ
Ff= FN
* Assumes that Ɵ is as shown on
previous diagram.
•
•
•
•
•
•
•
•
Explanation of Symbols
Fnet = net force
m = mass
Fg = weight (force of gravity)
Ff or f = friction force
 = coeff. of friction
FN = normal force
Fa = applied force
Problems With No Acceleration
When the object on the plane does not have
any acceleration, then the forces acting on the
object in both x and y must be equal to zero.
Fx = Fa - f - Fgx = max = 0
y
Fa
Fy= FN - Fgy= may = 0
F
N
f
Fgy
Fgx
Fgx
Fg

Note: There might not be
an applied force or it may
be down the incline (so f
would be up the incline).
x
Problems With Acceleration
When the object on the plane has acceleration, there is
a net force along the incline (x) but the net force
perpendicular to incline (y) is still zero.
Fx = Fa - f - Fgx = max
y
Fa
Fy= FN - Fgy= may =0
F
N
f
Fgy
Fgx
Fgx
Fg

Note: There might not
be an applied force or
it may be down the
incline (so f would be
up the incline).
x
Solving an Inclined Plane Problem # 1
A box slides down an plane 8 meters long inclined at 300 with a coefficient of
friction of 0.25. (a) what is the acceleration of the box? (b) what is its velocity
at the bottom of the incline?
Center of Mass
Upward force of plane
Friction
force
Force parallel
Normal force

FParallel = W sin 
FNormal = W cos 
FParallel = w sin 300
Ffriction =  Fnormal
Ffriction = 0.25 x w cos 300
Fnet = P – Ffriction
Fnet = 0.5 w – 0.17 w
weight
•
•
•
•
(a) Fnet = ma, w = mg , m = w /g , Fnet = (w/g) a
0.5 w – 0.17 w = (w/9.8) a , a = 2.77 m/s2
(b) Vo = 0 (the box starts at rest), a = (Vi2 – Vo2) / 2 S
Vi =( 2 S a + Vo)1/2 , Vi = ( 2 x 8 x 2.77 + 02)1/2 , Vi = 6.7 m/s
Inline Plane Problems # 2
Connected Objects
• Apply Newton’s Laws
separately to each object
• The magnitude of the
acceleration of both objects
will be the same
• The tension is the same in
each diagram
• Solve the simultaneous
equations
Section 4.5
Modified Atwood’s Machine Problem 2
A 15 kg cart is attached to a hanging 25 kg
mass. Friction is negligible. What is the
acceleration of the 15 kg cart?
What is the acceleration of the system? What is the tension developed in the
connector?
Tension Problems
What is Tension?
• Tension is defined as a force transmitted along
a rope, chain, or wire.
• Tension will remain constant throughout the
length of the rope.
• Tension is treated as a force in force diagrams
and calculations
• Tension is measured in force units. (Newtons,
dynes, or pounds)
Weight or mass?
What’s the difference?
• Mass is the amount of
matter that an object is
made up of, measured in
mass units (g, kg, slugs)
• The mass of an object
remains constant at any
location in the universe
unless matter is added or
removed
• Weight is the force of
gravity acting on an
object, measured in force
units (N, dynes, or
pounds)
• Weight depends on the
acceleration due to gravity
for the location of the
object.
• Weight (w) = mg
Problems in Equilibrium
• Equilibrium means that the object has a net
force and acceleration of zero.
• Since Fnet,x or Fx= 0, the x-components of
tension should cancel each other.
(Fright = Fleft)
• Since Fnet, y or Fy= 0, the y-components of
tension should cancel the objects weight.
(Fup = Fdown )
Force Diagram
(Free-Body Diagram)
1
T1y=T1sin1
T1x=T1cos1
2
T1
1
T2
T2y=T2sin2
2
T2x=T2cos2
Fg = mg
Applying Newton’s 2nd Law
Since the sign is in equilibrium,
the net force must be equal to
zero.
θ1
θ2
T1
T2
Fy = T1 sin θ1+T2 sin θ2 – mg =
0
Fx =T1 cos θ1 –T2 cos θ2 = 0
Fg = mg
Example 4.4
Given that Mlight = 25 kg, find all three tensions
T3 = 245.3 N, T1 = 147.4 N, T2 = 195.7 N
Other objects connected
Objects connected
Example: Three objects, m1, m2, and m3 are tied together by a
rope and pulled along a level surface by an applied force Fa.
All three objects have the same coefficient of friction, μ. Find
the tensions in the ropes connecting the masses.
m1
m2
m3
Some real life situations that this model applies to include:
train cars connected by couplings, find the tension in the
couplings; pulling sleds tied together; pulling a trailer
(only 2 objects)…
Objects connected
Drawing a force diagram
FN1
FN2
m1
f1=μFN1
T1
T1
FN3
m2
f2=μFN2
Fg1=m1g
T2
T2
m3
f3=μFN3
Fg2=m2g
Fg3=m3g
Fa
Objects connected
Write the force equations for each object.
FN1
FN2
m1
T1
f1=μFN1
T1
F
F
FN3
m2
T2
f2=μFN2
Fg1=m1g
For mass m1:
y1
 FN 1 - Fg1  0
x1
 T1 - f1  m1a
T2
m3
Fa
f3=μFN3
Fg2=m2g
For mass m2:
F
F
a
y2
 FN 2 - Fg 2  0
x2
 T2 - T1 - f 2  m2 a
Fg3=m3g
F
F
For mass m3:
y3
 FN 3 - Fg 3  0
x3
 Fa - T2 - f 3  m3a
Objects connected
Case 1: Constant velocity
Objects connected
constant velocity, summary
So we started with a force diagram, then wrote equations for each mass. We did not
need to evaluate the equations for the last mass because we had solved for both
tensions already, so we were done.
Note: We could have started at mass 3 and worked back to
mass 1. You could solve this problem for applied force to
keep the object moving at constant velocity in this method.
Applied force did not need to be given.
Objects connected
Case 2: Accelerated motion
Objects connected
Write the force equations for each object.
FN1
FN2
m1
T1
f1=μFN1
T1
F
F
FN3
m2
T2
f2=μFN2
Fg1=m1g
For mass m1:
y1
 FN 1 - Fg1  0
x1
 T1 - f1  m1a
T2
m3
Fa
f3=μFN3
Fg2=m2g
For mass m2:
F
F
a
y2
 FN 2 - Fg 2  0
x2
 T2 - T1 - f 2  m2 a
Fg3=m3g
F
F
For mass m3:
y3
 FN 3 - Fg 3  0
x3
 Fa - T2 - f 3  m3a
Objects connected
with constant velocity, solving T2
Let m1 = 10 kg, m2 = 20 kg, m3 = 15 kg, Fa = 110, μ = 0.25
F
y3
 FN 3 - Fg 3  0
FN 3  Fg 3  ( 15kg)( 9.81 m s 2 )  147.2 N
F
x3
 Fa - T2 - f 3  m3 a
110 - T2 - (0.25)(147.2 N )  0
T2  73.2 N
Tension between masses m2 and m3.
Objects connected
constant velocity, solving mass T1
Let m1 = 10 kg, m2 = 20 kg, m3 = 15 kg, Fa = 110, μ = 0.25
T 2 = 73.2 N
F
y2
 FN 2 - Fg 2  0
FN 2  Fg 2  (20kg)(9.81 m s 2 )  196.2 N
F
x2
 T2 - T1 - f 2  m2 a
73.2 N - T1 - (0.25)(196.2)  0
T1  24.2 N Tension in the string between
masses m1 and m2.
Objects connected
accelerated motion, solving mass 1
On the last example we started with mass 3 and worked toward mass 1. This
time I am beginning at mass 1 and moving toward mass 3 for no other reason
than to show that you can work these problems either way.
Let m1 = 10 kg,
m2 = 20 kg,
m3 = 15 kg,
Fa = 150N,
μ = 0.25
a = 0.88 m/s2
F
y1
 FN 1 - Fg1  0
FN 1  Fg1  (10kg)(9.81 m s 2 )  98.1N
F
x1
 T1 - f1  m1a
T1 - (0.25)(98.1N )  (10kg)(0.88 m s 2 )
T1 - 24.5 N  8.8 N
T1  33.3N
Objects connected
accelerated motion, solving mass 2
Let m1 = 10 kg,
m2 = 20 kg,
m3 = 15 kg,
Fa = 150N,
μ = 0.25
a = 0.88 m/s2
T1 = 33.3 N
F
y2
 FN 2 - Fg 2  0
FN 2  Fg 2  (20kg)(9.81 m s 2 )  196.2 N
F
x2
 T2 - T1 - f 2  m2 a
T2 - 33.3N - (0.25)(196.2 N )  (20)(0.88)
T2 - 33.3N - 49.1N  17.6 N
T2  100 N
Objects connected
accelerated motion, must find acceleration first
To find acceleration to use for this problem you can treat the
whole system together as one object being accelerated under the
influence of the applied force.
f
FN
m1 + m2 + m3
F
y
Fa
FN  Fg  (10  20  15)(9.81)  441.5 N
F
x
Fg = (m1 + m2 + m3)*g
 FN - Fg  0
 Fa - f  mt a
150 N - (0.25)( 441.5 N )  (10  20  15) * a
150 N - 110.4 N  45 * a
a  0.88 m s 2
Now we have acceleration of each mass to use as we solve for tensions.