Transcript Inclined Plane Problems

Inclined Plane Problems
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
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 – Fg = 0
Fy =T1 cos θ1 –T2 cos θ2 = 0
Fg = mg
Applying Newton’s Laws to
Systems with Multiple Objects.
Atwood Machine
• An atwood machine is a very common system in
classical physics
• It is made of two masses tied together by a string
that passes over a pulley
• Both objects could be hanging and the system
would only have motion in the vertical direction
OR one object could be sliding along a table top
with the other hanging over the side
• Assumptions made (for now): the pulley is
“massless” and “frictionless”. This allows us to
solve the problem as if the pulley was not there at
all. Tension in the string acts equally on both
objects and they both accelerate at the same rate.
A typical Atwood machine problem
m1
m2
Two masses, m1 and m2, are
tied together with a string
that is passed over a light,
frictionless pulley. Mass m2
is accelerating downward. If
the coefficent of friction
between mass m1 and the
table is μ. Find the tension,
T, in the string and the
acceleration, a, of the
masses.
To Solve…simply follow the steps we
have already established for Force
problems.
• Draw the force diagram
– Include all forces acting on each object
– Label all forces with a symbol, NOT NUMBERS
• Define an axis system on each object
• Write the Newton’s 2nd Law equations for each
object.
• Plug in what you know and solve the systems of
equations for what you don’t know. (Algebra)
Setting up the Atwood problem
First with no numbers!
NOTE: When
writing the
equations for
m2, consider the
direction of
acceleration to
be positive.
y
a
FN
f
T
m1
x
T
Fg1=m1g
a=+
m2
Fg2=m2g
Write the equations for m1:
F
F
x1
 T  f  m1a
y1
 FN  Fg  m1a 0
Write the equations for m2:
F
y2
 Fg 2  T  m2a
There are no forces along the x-axis for m2.
Solving the Atwood Problem
Now that you have written the
equations for this problem, it is
time to plug in any values that you
know and solve the system of
equations for Tension and
acceleration. Keep in mind that
there is more than one algebra
L that can be used, this is
technique
an example of substitution. Any
e
properly done method will work.
Substitute
values given in
the problem.
Let…
m1 = 4 kg
m2 = 2 kg
μ= 0.35
Algebra
For mass m1:
F
y1
 FN  Fg  0
FN  Fg  ( 4kg)(9.81 m s 2 )  39.24 N
F
x1
 T  f  m1a
T  (  * FN )  m1a
T  (0.35 * 39.24 N )  ( 4kg) * a
T  13.73  4a
For mass m2:
 Fy 2  Fg 2  T  m2 a
Now solve
for a and T
( 2kg)(9.81 m s 2 )  T  ( 2kg) * a
19.62 N  T  2a
Solving the Atwood Problem
Algebra (continued)
Now that we have plugged in values given in the problem and simplified the
equations, we must solve this system of equations for a and T. This example will
use substitution. Solve one of the equations for T and plug it into the other.
T  13.73  4a
T  4a  13.73
T  4(0.98)  13.73
T  17.65 N
19.62 N  T  2a
19.62  (4a  13.73)  2a
19.62  4a  13.73  2a
19.62  13.73  2a  4a
5.89  6a
5.89
a
 0.98 m s 2
6
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
Solving the equations
To solve this problem further we will need to plug
in any known values and work the equations from
one end to the other. In this problem, the normal
force, FN, is equal to the weight of the block which
can be shown by solving each of the y-equations
for normal force. We will work this problems
under two conditions: 1) constant velocity, and 2)
accelerated motion.
Let m1 = 10 kg, m2 = 20 kg, m3 = 15 kg, Fa = 110, μ = 0.25
Objects connected
Case 1: Constant velocity
Objects connected
with constant velocity, solving mass 3
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 2
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
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
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.
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