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Forces
Homework:
Physics: Handout #1 – 9, 12, 14, 15,
but all are highly recommended.
Forces:
push
pull
A force is a _____________
or a ____________
on an object by
another object.
Forces can be applied either through direct contact, as shown in the
picture below (a), (b), and (c).
Forces shown on the right are
called “action at a distance”,
where no physical contact is
required between the objects to
cause a force.
inertia
A force is tied to the motion of an object through ________________
.
resist
Inertia is the ability of an object to ________________
a change in its
motion.
mass
Inertia is measured as the__________
of an object. The standard unit
is the kilogram, kg.
The more massive an object is, the more it is able to resist a
change in its motion. It is easier to push a 10 kg object into
motion than it is a 3000 kg car.
dynamics
All of this begins the topic of _______________,
or why objects move.
Newton’s Laws of Motion:
The relationship between motion and force was stated clearly by Isaac
Newton in his 3 laws. Each law relates a different aspect of motion.
Newton’s First Law:
If the total, or net, force on an object is zero, then an object will
not accelerate.
First variation: If an object is at rest, it will continue to remain at
rest until acted upon by some external agent.
Example: A book placed on a desk will remain on the
desk until someone removes it. There are forces acting
on the book: gravity pulls down on it and the table
pushed upwards on it. These forces balance out, and the
object does not move.
Newton’s First Law: continued
Second variation: If an object is moving, it will continue to
move with a constant velocity. Since velocity is a vector, this
means the speed (how fast) and the direction remain
unchanged.
Example: Take a look at a hockey game. When the puck
(small black projectile) is hit, the puck travels at nearly
the same speed and in a straight line across the ice. All
real objects are subject to friction for anything sliding on a
surface, and an icy surface is as close to frictionless as
we can get.
Newton’s Second Law:
If the net force on an object is not zero, then an object will accelerate.
The acceleration of the object is directly proportional to the total force
acting upon it, and the acceleration is inversely proportional to the mass
of the object.
As an equation:



Fnet  F  ma
Here, m is the mass of the object, measured in kg.
a is the acceleration of the object, measured in m/s2. This is a
vector quantity
F refers to the amount of force. Fnet or F is the sum of the
forces acting on an object. Force is a vector, so it will have
magnitude and direction. The net force and the acceleration
point in the same direction.
Units: Force is measured in units called newtons, and this unit is
represented by the letter N.
Newtons can be written in term of fundamental units through the
statement of Newton’s Second Law:
kg m
 m
1N  1kg 1 2   1 2
s
 s 
Example #1: What is the acceleration of a 2.00 kg mass if a force of
4.00 N acts upon it?
F 4.00 N
a 
 2.00 m s 2
m 2.00kg
Example #2: An object with a mass of 3.00 kg moves from rest to a
speed of 20.0 m/s in a time of 5.00 s. a. What is the acceleration of the
object?
Start with:
v  vo  at
and solve for a:
v  vo 20.0 ms  0
a

 4.00 m s 2
t
5.00s
b. What is the net force acting on the object?
Fnet  ma  3.00kg 4.00 m s 2 
 12.0N
Both force and acceleration are vectors. The net force points in the same
direction as the acceleration of the object.
Example #2: c. Let the object’s acceleration point to the right. If there
is a drag force of 30.0 N opposing the motion, what other applied force
must there be on the object to give the desired acceleration?
A picture helps!

Fdrag

Fapplied



Fnet  Fapplied  Fdrag
For one dimension, the vector signs are dropped.
It is understood all vectors point along the same line.
Just use a (+) sign for the vectors pointing right and a
(–) sign for the vectors pointing left.
The variables are assumed to be only the magnitude
of the vectors.
Fnet  12.0 N   Fapplied  Fdrag
Fapplied  Fnet  Fdrag  12.0 N  30.0 N
Fapplied  42.0 N
Newton’s Third Law:
If one object (object A) exerts a force on another object (object B), then
object B exerts the same force back on object A. The force is equal in
magnitude, but opposite in direction.
Alternate interpretation: For every action there is an equal but
opposite reaction.
Example: If you drive your fist against a wooden wall as hard
as you can, you will possibly leave a mark on the wall with
your fist. The mark is made by the force you exerted against
the wall. The wall also exerts an equal force on your fist. You
perceive this as the severe pain felt in your knuckles.
object A
object B
Drawing:


FAon B  FB on A
FB on A
FA on B
Example #3: A student pushes with a 30.0 lb force towards the South
on a heavy box. There is a friction (drag) force between the floor and
the box that prevents the movement of the box. What is the magnitude
and direction of the friction force?
Since the box does not move, the net force must be zero.

Fdrag

Fapplied



Fnet  0  Fapplied  Fdrag


Fdrag  Fapplied  30.0 lb north
Example #4: A baseball is thrown at a speed of 40.0 m/s towards the
right. The ball is hit by a baseball bat, and then travels at 50.0 m/s
towards the left. The ball is in contact with the bat for 2.00 ms and the
ball has a mass of 200 grams. What is the net force acting on the ball?
Initial information: Let right be positive and left be negative.
vo = +40.0 m/s, v = – 50.0 m/s, t = 2.00 x 10-3 s, m = 0.200 kg
v  vo  at
v  vo  50.0 ms   40.0 ms 
m
a



45
,
000
s2
3
t
2.00 10 s
Fnet  ma  0.200kg  45,000 m s 2   9000N
Example #5
A horse is hitched to a wagon. Which statement is correct?
A. The force that the horse exerts on the wagon is greater than the
force that the wagon exerts on the horse.
B. The force that the horse exerts on the wagon is less than the
force that the wagon exerts on the horse.
C. The force that the horse exerts on the wagon is just as strong as
the force that the wagon exerts on the horse.
D. The answer depends on the velocity of horse and wagon.
E. The answer depends on the acceleration of horse and wagon.
Example #5
A horse is hitched to a wagon. Which statement is correct?
A. The force that the horse exerts on the wagon is greater than the
force that the wagon exerts on the horse .
B. The force that the horse exerts on the wagon is less than the
force that the wagon exerts on the horse.
C. The force that the horse exerts on the wagon is just as strong as
the force that the wagon exerts on the horse.
D. The answer depends on the velocity of horse and wagon.
E. The answer depends on the acceleration of horse and wagon.
Example #6
You are standing at rest and begin to walk forward. What
force pushes you forward?
A. the force of your feet on the ground
B. the force of your acceleration
C. the force of your velocity
D. the force of your momentum
E. the force of the ground on your feet
Example #6
You are standing at rest and begin to walk forward. What
force pushes you forward?
A. the force of your feet on the ground
B. the force of your acceleration
C. the force of your velocity
D. the force of your momentum
E. the force of the ground on your feet
Table Cloth Pull: An example of the Law of Inertia.
Tension, Gravity, and Normal
Force
Homework:
Physics Day 1: #1 – 7, 10
Physics Day 2: #8, 9, 11 – 14,
15 a,b,c.
Definitions: Forces are any kind of push or pull on a system or on an
object. There are some special types of forces that we will define today.
The first is the weight of an object.
gravity
The weight of an object is the force of ___________
acting on the
object by Earth.
mass
The amount of force is directly proportional to the ____________
of the
object.
The equation for weight is as follows:
w  mg
w = weight of the object. This is a force measured in N (newtons).
m = mass of the object, in kg (kilograms).
g = acceleration of gravity, 9.80 m/s2.
The direction of the force is downwards, towards the center of Earth.
Example #1: What is the weight of a 55.0 kg person?
w  mg  55.0kg 9.80 m s 2 
 539 N
Example #2: In the British system, mass is measured in slugs and force
is measured in pounds. Weight is still measured as mg, with one slug
mass weighing 32.2 pounds. Convert the mass of the person above
into slugs.
Conversion Factor: 1.0000 kg mass weighs 2.2046 lbs
w 121lb
m 
g 32.2 ft s2
 3.77 slugs
normal force on an object is defined as the force
Definition: The __________
exerted on an object by a surface. The term “normal” is another term
used in mathematics for perpendicular. The normal force exerted by a
surface is always perpendicular to that surface.
n
The normal force is represented by the letter _________
. Do not
confuse this with newtons of force!
Problem Solving: Always draw a picture, and label all the forces acting
on the object. This is referred to as drawing a “Free Body Diagram”.
Only draw the forces acting on the object. By Newton’s third law,
the forces on an object result also in forces exerted by the object on
the surroundings. We are only interested in the forces on the object
by the environment.
Example #3: A 1.4 kg book is placed on a flat surface at rest. What is
the normal force acting on the book?
Start with a labeled picture:
Next apply Newton’s Second Law:


Fnet  ma
m = 1.4 kg
For one dimension, define up as
positive and down as negative. The
forces may be written as follows:
and
normal = n
weight = mg
The variables show the magnitude, or size of
the vector. The ± sign gives the direction.
The net force becomes:

 
Fnet  n  w   n  mg
Since the object is at rest, the acceleration of the object is zero.


Fnet  ma  0
Combine the above equations and solve for n.
Fnet  0  n  mg
n  mg  1.4kg 9.80 m s 2 
 13.7 N
Example #4: {Classic Physics Problem!!!} A person stands in an
elevator on a scale. (Don’t ask why…..) When the elevator is at rest,
the scale reads 588 N. What will the scale read if the elevator
accelerates upwards at 2.20 m/s2?
Solution: The scale is designed to measure a force. The
scale does not measure the weight of the person, rather the
force exerted upwards on the person to support them. In
other words, the scale measures the normal force acting on
the person.
When the person is at rest, the normal force equals the
weight of the person. When the person is accelerated in the
elevator, the net force on the person is no longer zero. For
an upward acceleration, the normal force must be larger
than the weight of the person. Let’s find out why.
Before anything, determine the mass of the person:
At rest, the person weighs 588 N. This equals mg, so the mass is:
w 588 N
m 
 60.0kg
g 9.80 m s 2
up = positive
Apply Newton’s 2nd Law:
Fnet  ma  n  mg
normal = n
person, m
Draw in a coordinate axis and
choose a positive direction.
elevator
Start by drawing a complete
picture and then apply Newton’s
Second Law.
n  ma  mg
n  ma  g 
weight = mg
down = negative
Substitute numbers: Solve for the new normal force for the accelerated
elevator.
n  ma  g   60.0kg 2.20 m s 2  9.80 m s 2 
n  720.N
The person appears to weigh more when the elevator accelerates
upwards.
Example #5: What is the motion of the elevator if the person’s weight
appears to be 360 N?
Just follow the same reasoning as above and show again:
Fnet  ma  n  mg
Solve this expression for a.
n
360.N
a g 
 9.80 m s2
m
60.0kg
 3.80 m s 2
Since up was chosen as the positive direction, a negative acceleration
means that the elevator is accelerating downwards.
Example #6: What would the scale read if the elevator was in freefall?
Just follow the same reasoning as above and show again:
n  ma  g 
Solve this expression for n. Note that a = – g.
n  ma  g   m g   g   0
For freefall, the scale would read zero. The person would feel
weightless.
pulling
tension force is a ____________
Definition: A __________
force, usually
applied through a string or rope.
Ropes are to be treated as ideal, meaning that the mass of the ropes
can be ignored and the ropes do not stretch or sag when used.
The force of tension is then the same throughout the rope.
Example #7: Two masses rest on a frictionless horizontal surface. The
mass on the left is 60.0 kg and the mass on the right is 40.0 kg. The
masses are tied together with a light cord and a 25.0 N force is applied
towards the right on the 40.0 kg mass. Find the acceleration of the
system and the tension force in the cord.
m1 = 60.0 kg
m2 = 40.0 kg
cord
applied force
Fa = 25.0N
Solution: First treat the two masses as a single object.
They move the same: same velocity and acceleration.
mtotal = m1 +m2
Apply Newton’s 2nd Law:
a
 Fapplied
mtotal
applied force
Fa = 25.0N
Fnet  mtotala   Fapplied
 Fapplied
25.0 N


m1  m2 100kg
 0.250 m s 2
Now isolate one mass to solve for the force of tension in the cord. If the
mass on the left (m1) is chosen, the solution is as follows:
m1
T = tension force
Apply Newton’s 2nd Law and solve for the tension force.
Fnet  m1a  T
T  m1a  60.0kg 0.250 m s 2 
 15.0N
An alternate solution would be to choose the mass on the right (m2).
T = tension force
m2
applied force
Fa = 25.0N
Apply Newton’s 2nd Law and solve for the tension force. Let the positive
direction be towards the right.
Fnet  m2 a   Fapplied  T
T   Fapplied  m2 a  25.0 N  40.0kg 0.250 m s 2 
T  25.0N 10.0N
 15.0N
Example #8: A mass (m1 = 10.0 kg) hangs from the ceiling of an
elevator from a light cord. A second mass (m2 = 20.0 kg) is suspended
from the first mass by another cord. If the elevator accelerates upwards
at a rate of 4.70 m/s2, determine the tension in each cord.
up = positive
elevator
cord with
tension T1
m1
cord with
tension T2
m2
down = negative
Solution: First treat the two masses as a single object and solve for the
tension in the top cord. Start with a simple picture.
Apply Newton’s 2nd Law and solve
for the tension force. Let the
positive direction be upwards.
T1
Fnet  mtot a  T1  mtot g
T1  mtot a  mtot g
T1  30.0kg 4.70 m s 2  9.80 m s 2 
T1  435N
mtot
mtotg
Now isolate one mass to solve for the force of tension in the lower cord.
If the mass on the bottom (m2) is chosen, the solution is as follows:
Apply Newton’s 2nd Law and solve
for the tension force. Let the
positive direction be upwards.
T2
Fnet  m2 a  T2  m2 g
T2  m2a  m2 g
m2
T2  20.0kg 4.70 m s 2  9.80 m s 2 
T2  290.N
m2 g
If the mass on the top (m1) is chosen, the solution is as follows:
Apply Newton’s 2nd Law and solve for the tension force. Let the positive
direction be upwards.
Fnet  m1a  T1  T2  m1 g
T1
T2  T1  m1a  m1 g
T2  435 N   10.0kg 4.70
T2  290.N
m
s2
 9.80
m
s2

m1
T2
m1 g
Q4.1
An elevator is being lifted at a constant
speed by a steel cable attached to an electric
Cable
motor. There is no air resistance, nor is
there any friction between the elevator and
the walls of the elevator shaft.
Motor
v
Elevator
The upward force exerted on the elevator
by the cable is
A. greater than the downward force of gravity.
B. equal to the force of gravity.
C. less than the force of gravity.
D. any of the above, depending on the speed of the elevator.
A4.1
An elevator is being lifted at a constant
speed by a steel cable attached to an electric
Cable
motor. There is no air resistance, nor is
there any friction between the elevator and
the walls of the elevator shaft.
Motor
v
Elevator
The upward force exerted on the elevator
by the cable is
A. greater than the downward force of gravity.
B. equal to the force of gravity.
C. less than the force of gravity.
D. any of the above, depending on the speed of the elevator.
Q4.9
A woman pulls on a 6.00kg crate, which in turn is
connected to a 4.00-kg
crate by a light rope. The
light rope remains taut.
Compared to the 6.00–kg crate, the lighter 4.00-kg crate
A. is subjected to the same net force and has the same acceleration.
B. is subjected to a smaller net force and has the same acceleration.
C. is subjected to the same net force and has a smaller acceleration.
D. is subjected to a smaller net force and has a smaller acceleration.
E. none of the above
A4.9
A woman pulls on a 6.00kg crate, which in turn is
connected to a 4.00-kg
crate by a light rope. The
light rope remains taut.
Compared to the 6.00–kg crate, the lighter 4.00-kg crate
A. is subjected to the same net force and has the same acceleration.
B. is subjected to a smaller net force and has the same acceleration.
C. is subjected to the same net force and has a smaller acceleration.
D. is subjected to a smaller net force and has a smaller acceleration.
E. none of the above
2 – Dimensional Forces
Homework:
Handout #3,
all problems.
ConcepTest 4.17 Three Blocks
Three blocks of mass 3m,
1) T1 > T2 > T3
2m, and m are connected
2) T1 < T2 < T3
by strings and pulled with
constant acceleration a.
What is the relationship
between the tension in
3) T1 = T2 = T3
4) all tensions are zero
5) tensions are random
each of the strings?
a
3m
T3
2m
T2
m
T1
ConcepTest 4.17 Three Blocks
T1 pulls the whole set of
1) T1 > T2 > T3
blocks along, so it must
2) T1 < T2 < T3
be the largest. T2 pulls
3) T1 = T2 = T3
the last two masses, but
4) all tensions are zero
T3 only pulls the last
5) tensions are random
mass.
a
3m
T3
2m
T2
m
T1
vector
magnitude
Forces are _________
quantities: they have both a _____________
direction
and a ____________.
For Newton’s 2nd Law, both the net force and the acceleration are
vectors. It is easier to resolve the vectors into components to solve 2 –
dimensional problems. On any given problem, draw in a coordinate set
and label which directions are positive .
Example #1: A 40.0 kg block rests on a smooth surface. A force of 400 N
is applied to the block at an angle of 36.9o above the horizontal. a. Label all
forces acting on the object, and resolve these forces into horizontal and
vertical components.
Fa
n = normal
q = 36.9o
m
mg = weight
Fa
n = normal
q = 36.9o
m
mg = weight
The normal force has only a y – component: nx = 0, ny = +n
The weight has only a y – component: wx = 0, wy = –mg
The x – component of the applied force, Fa, is: Fa,x = +Facosq
The y – component of the applied force, Fa, is: Fa,y = +Fasinq
+y
+x
Ex. #1 b. Determine the size of the normal force acting on the block.
Apply Newton’s 2nd Law to the y – components and solve for the normal
force.
Fnet y  may  0
The y – component of the acceleration is zero. The y – component of
the applied force is not larger than the weight of the object, so the object
will not be lifted from the surface. It will only slide parallel to the surface.
0  Fnet y  n  mg  Fa sin q
n  mg  Fa sin q
n  40.0kg9.80
m
  400N sin 36.9 
o
s2
n  152N
Ex. #1 c. What is the acceleration of the block? How fast will it be
moving after traveling 10.0 meters?
Apply Newton’s 2nd Law to the x – components and solve for the
acceleration.
Fnet x  max   Fa cosq
Fa
400 N
ax   cos q  
cos 36.9o  8.00 m s 2
m
40.0kg
Reach back to chapter 2 for a motion equation:
vx  vox  2ax x
2
2
v x  vox  2a x x  0 2  28.00 m s 2 10.0m 
2
vx  12.6 ms
ConcepTest 4.10 Normal Force
Below you see two cases: a
1) case 1
physics student pulling or
pushing a sled with a force F
which is applied at an angle q. In
which case is the normal force
greater?
2) case 2
3) it’s the same for both
4) depends on the magnitude
of the force F
5) depends on the ice surface
Case 1
In Case 1, the force F is pushing
down (in addition to mg), so the
normal force needs to be larger.
In Case 2, the force F is pulling
up, against gravity, so the
normal force is lessened.
Case 2
Example #2: An object has a two forces acting upon it. One force is
454 N, pointing due north, and the other force is 844 N, pointing due
west. What is the net force on the object and the acceleration of the
object? Give the answer as magnitude and direction. What one force
would counter this net force?
a. solve with Pythagorean theorem.
N, +y
2
2
total
1
2
F
 F F
Ftotal
F1
q
Ftotal  958 N
1 454 N
q  tan
 28.3o NW
844 N
E, +x
F2
Fopposite  958N @ 28.3o SE
Example #3: Statics. A lady with a weight of 522 N balances at the
center of a long rope, as shown. If the rope deflects 10.0o from the
horizontal, determine the tension in the rope.
+y
+x
T2
T1
mg
find the components of each vector.
 
 T sin 10.0 
T1x  T1 cos 10.0 o
T1 y
o
1
 
 T sin 10.0 
T2 x  T2 cos 10.0o
T2 y
o
2
This is a static system, meaning there is no acceleration. Solve this
with Newton’s second law. Start with the x – component.
Fnet x  max  0  T1x  T2 x



0  T1 cos 10.0o  T2 cos 10.0o
therefore
T1  T2

Next use the y – component.
Fnet y  may  0  T1y  T2 y  mg




0  T1 sin 10.0o  T2 sin 10.0o  mg


2T1 sin 10.0o  mg
mg
522 N
T1 

 1503N
o
o
2 sin 10.0
2 sin 10.0




Example #4: A mass of 125 kg is supported by two ropes. One rope
extends horizontally, towards the left from the object to a support wall.
The second rope extends towards the right of the object, at an angle of
53.1o above the horizontal. Find the tension in each rope.
draw each force vector.
find the components of
each vector.
tension T2
T1x  T1
q = 53.1o
tension T1
T1 y  0
 
 T sin 53.1 
T2 x  T2 cos 53.1o
T2 y
+y
mg
o
2
+x
This is a static system, meaning there is no acceleration. Solve this
with Newton’s second law. Start with the y – component. (Random
choice. You could start with the x – component.)
Fnet y  may  0  T1y  T2 y  mg


0  0  T2 sin 53.1o  mg
125kg 9.80 m s 2 
mg
T2 

o
sin 53.1
sin 53.1o




 1530N
Next use the x – component.
Fnet x  max  0  T1x  T2 x

0  T1  T2 cos 53.1o

T1  T2 cos 53.1o
therefore


T1  920N
Example #5: A 615 kg mass is suspended as shown below. Determine
the tension in each rope.
q1 = 20.0o
q2 = 60.0o
draw each force vector.
tension T2
tension T1
find the components of
each vector.
 
 T sin 20.0 
 T cos60.0 
 T sin 60.0 
q1 = 20.0o
q2 = 60.0o
T1x  T1 cos 20.0 o
T1 y
T2 x
T2 y
o
1
o
2
o
2
+y
mg
+x
This is a static system, meaning there is no acceleration. Solve this
with Newton’s second law. Start with the x – component. (Random
choice. You could start with the y – component.)
Fnet x  max  0  T1x  T2 x



0  T1 cos 20.0o  T2 cos 60.0o





cos 60.0o
T2   0.532089T2 
T1 
o
cos 20.0
This is one equation with two unknowns. Use the y – component to get
another equation with the same two unknowns.
Next use the y – component.
Fnet y  may  0  T1y  T2 y  mg




0  T1 sin 20.0o  T2 sin 60.0o  mg




 T1 sin 20.0  T2 sin 60.0  mg
o
o
Substitute the equation from the x – direction.




 0.532089T2 sin 20.0o  T2 sin 60.0o  mg





 T2 0.532089sin 20.0o  sin 60.0o  mg
Solve for T2.
mg
T2 
0.532089sin 20.0o  sin 60.0o





615kg 9.80 m s 
T2 
0.532089sin 20.0o   sin 60.0o 
2
T2  5751N
Substitute the equation from the x – direction.
T1  0.532089T2 
 3060 N
2 – Dimensional Forces
Day #2
Homework:
Handout #4,
problems #1 – 3, 12
Example #6: Two masses are connected by a string over a pulley, as
shown. The bottom mass (m1) is 25.0 kg and the hanging mass (m2) is
10.0 kg. What is the normal force on the bottom mass?
m2
m1
Example #6: Two masses are connected by a string over a pulley, as
shown. The bottom mass (m1) is 25.0 kg and the hanging mass (m2) is
10.0 kg. What is the normal force on the bottom mass?
Draw in the forces and a coordinate line.
+y
This system is at rest. The suspended
mass is too small to make the bottom
mass rise upwards on its own.
T
m2
Solve this problem by isolating
each mass and apply Newton’s 2nd
law to each mass.
T
n
m2 g
m1
m1 g
–y
Start with the hanging mass, m2.
T
Fnet  m2 a  T  m2 g
Set the acceleration equal to zero and solve for T.
m2
m2 g
0  T  m2 g
T  m2 g  10.0kg 9.80 m s 2   98.0 N
Next look at the bottom mass, m1:
Fnet  m1a  T  n  m1 g
T
Set the acceleration equal to zero and solve for n.
n
0  T  n  m1 g
m1
n  m1 g  T  25.0kg 9.80 m s 2   98.0 N
n  147N
m1 g
Example #7: Given the two masses shown below, find the acceleration
of the system and the tension in the string. Ignore friction.
m1
T
Mass m1 moves and accelerates towards the
right.
First determine the motion of
each mass:
Let these directions define the
positive direction for each
mass. Now label the forces.
T
m2
Mass m2 falls and
accelerates downwards.
m2 g
Apply Newton’s 2nd law
to each mass separately.
Start with mass m1. There are two unknowns: acceleration and tension
m1
T
Fnet  m1a  T
Next look at mass m2. This mass also has the same two unknowns:
acceleration and tension. This will give us a second equation to solve
the two unknowns.
T
m2
m2 g
Fnet  m2 a  m2 g  T
By choosing the positive directions to match
the directions of the accelerations, the two
objects will have the same acceleration, both
in magnitude and direction.
Solve the simultaneous equations:

Notice T’s will
subtract out
m1a  T
m2a  m2 g  T
m1a  m2a  T  m2 g  T
m1  m2 a  m2 g
m2 g
a
m1  m2 
Solve for the tension by substituting into one of the initial equations:
m1a  T
m2 g
T  m1a  m1
m1  m2
m1m2 g
T
m1  m2
Example #8: Determine the acceleration of the system and the tension
in the string. Assume the mass on the right is the heavier mass.
Ignore friction.
m  m1
M  m2

m1a  T  m1g
m2a  m2 g  T
m1a  m2a  T  m1 g  m2 g  T
m1  m2 a  m2  m1 g

m2  m1 
a
g
m1  m2 
Solve for the tension by substituting into one of the initial equations:
m1a  T  m1g
T  m1a  m1g
m2  m1
 m1
g  m1 g
m1  m2
 m2  m1 
 m2  m1 m2  m1 
 m1 g 
 1  m1 g 


 m1  m2 
 m1  m2 m1  m2 
2m1m2 g
T
m1  m2
Force of Friction
Homework:
Handout #4
problems #5 and complete chapter
reading.
Test Friday
drag
Definition: Friction is a form of a ____________
force. This is a force
that opposes the motion of an object. In general, this force may be very
complex, but the form we will study will be simple.
static
kinetic
The friction force comes in two types: ___________
and ___________
.
friction
Static
____________
____________
refers to the force that will keep an
object at rest. Static means that there is no motion between the
surfaces in contact. The static friction force is a variable force: The
force will take on whatever value it needs to prevent an object from
moving, up to a maximum amount. The maximum amount of force is
proportional to the normal force between the two surfaces.
Kinetic
friction
____________
____________
refers to the force that resists an object’s
motion. Kinetic friction only operates when the object is moving. For
our purposes, this force is constant, regardless of the speed of the
object.
Friction always tries to
oppose the motion of two
contacting surfaces relative
to one another.
Static friction balances any
applied force, up to a
maximum limit.
limit of static friction
forces balance
When the applied force
exceeds the maximum static
friction, the object breaks free
and moves with constant
kinetic friction.
The size of the maximum static friction force is given in terms of the
normal force the surface exerts on the object:
Fs  Fs ,max  n
Fs is the static friction force between two surfaces.
Fs,max is the maximum static friction force between two surfaces.
n is the normal force between two surfaces.
The maximum static friction force can be made equal to the normal
force through a proportionality constant. This constant is called the
coefficient of static friction, and it is represented by the symbol ms.
Fs ,max  m s n
The coefficient of static friction rates the “roughness” of a
surface. The higher the value of the coefficient, the stronger the
static friction force can be. The coefficient does not have any
units! This value is unique to a given surface, and it is usually
given in some way.
Example #1: A 400 lb block rests on an icy horizontal surface. The
coefficient of static friction between the block and the ice is 0.100.
a. What will be the static friction force if a horizontal force of 20.0 lb is
applied to the block? b. What if the applied force is 40.0 lb? c. 50.0 lb?
SOLUTION: Start by finding the maximum static friction force
available.
The coefficient has a value of ms = 0.100, so the maximum static friction
force is:
Fs ,max  m s n
For a block resting on a flat surface, the normal force equals the weight
of the object.
n  mg
Combining the above, get:
Fs ,max  m s n  m s mg
Fs ,max  m s mg  0.100400lb 
 40.0lb
When the applied force is 20.0 lb, the static friction force will
match it at 20.0 lb in the opposite direction.
When the applied force is 40.0 lb, that is right at the limit that
static friction can hold the object still, or move it at constant speed.
When the applied force is 50.0 lb, the static friction will fail and the
object will accelerate.
Definition: Kinetic friction refers to the force that opposes a moving
object’s motion. The kinetic friction force is proportional to the normal
force between the object and the surface. The kinetic friction force is
constant in magnitude. This force is independent of the speed of the
object. It is also independent of the surface area of the contact area.
Fk  n
Fk is the kinetic friction force between two surfaces.
n is the normal force between two surfaces.
The kinetic friction force can be made equal to the normal force through
a proportionality constant. This constant is called the coefficient of
kinetic friction, and it is represented by the symbol mk.
Fk  mk n
Example #2: A 40.0 kg box is made to slide on a horizontal surface that
has a coefficient of kinetic friction of 0.510. How much force must be
applied to the object to make it accelerate at 3.00 m/s2 horizontally?
n
Fk
Fapplied
mg
a
Since the block never leaves the surface, the vertical forces balance.
n  mg
Use the definition of kinetic friction:
Fk  mk n  mk mg
Now use Newton’s 2nd law in the horizontal direction:
Fnet  ma   Fapplied  Fk
ma   Fapplied  m k mg
Solve for the applied force:
Fapplied  ma  m k mg
Fapplied  ma  m k g 
Fapplied  320 N
1. (Review) A force of 90.0 N pulls horizontally towards the right on
an object with a mass of 20.0 kg along a surface with a coefficient of
kinetic friction of 0.300. (a) What is the acceleration of this block?
n
Fk
Fapplied
mg
a
Since the block never leaves the surface, the vertical forces balance.
n  mg
Use the definition of kinetic friction:
Fk  mk n  mk mg
Now use Newton’s 2nd law in the horizontal direction:
Fnet  ma   Fapplied  Fk
ma   Fapplied  m k mg
Solve for the acceleration:
a
Fapplied
m
 mk g
 1.56 m s 2
(b) Now use kinematics to get the final velocity: v 2
v  vo  2ax
2
 vo  2ax
v  5.59 ms
2
2. A force of 90.0 N pulls the same object towards the right at an angle
of 16.7 degrees above the horizontal. If the mass is still 20.0 kg and the
coefficient of kinetic friction is still 0.300, (a) determine the normal
force and kinetic friction force acting on the object.
n
Fo
q
Fo cos q
Fk
mg
Fo sin q
There is no upward acceleration:
0  may  Fy  n  Fo sin q  mg
n  mg  Fo sin q
The normal force is smaller than the weight of the mass since the
applied force is also lifting upwards. A smaller normal force
corresponds to a smaller friction force.
Fk  mk n  mk mg  Fo sin q 
Fk  mk mg  mk Fo sin q
(b) Determine the horizontal acceleration.
max  Fx   Fo cos q  Fk
max   Fo cos q  mk mg  mk Fo sin q 
A couple of algebraic steps later…….
Fo
a x   cos q  m k sin q   m k g
m
a x  1.76 m s 2
Force of Friction
Day #2
3. A 1.20 kg book is pushed against a vertical wall by a horizontal
applied force. The coefficient of static friction between the book and
the wall is 0.425. What is the amount of applied force needed to hold
the book up without sliding?
The forces balance in each direction:
Fo  n
F fr  mg
Use the definition of friction to combine the
two equations:
mg  F fr  m s n  m s Fo
 Fo 
mg
ms
1.20 kg 9.80 m s 

0.425
Fo  27.7 N
2
4. Two masses are tied together, with one suspended over a pulley as
shown. Let m1 = 20.0 kg and m2 = 5.00 kg. Let the coefficient of
kinetic friction between m1 and the surface be 0.150. (a) Determine the
acceleration of the system, assuming m2 is initially moving downwards.
n
T
Fk
Mass m1 moves and accelerates
towards the right.
T
m1 g
m2 g
Mass m2 moves and
accelerates downwards.
Let these define the
positive direction for the
motion of each of the
two masses.
Friction
Positive
Tension
m2g
Positive
Isolate the first mass. The vertical forces balance.
n  m1 g
Fk  mk n  mk m1 g
Solve the acceleration in the horizontal direction:
m1a  T  Fk  T  mk m1 g
Isolate the second mass and solve the acceleration :
m2a  m2 g  T
Solve the simultaneous equations.
m1a  T  mk m1 g

m2a  m2 g  T
m1a  m2 a  T  mk m1 g  m2 g  T
m1  m2 a  m2  mk m1 g

m2  m k m1 
a
g
m1  m2 

5.00 kg  0.150 20.0 kg 
9.80 m s 
a
20.0 kg  5.00 kg 
2
a  0.784 m s 2
(b) What is the tension in the string?
Pick one equation and solve for T:
m2a  m2 g  T
T  m2 g  m2a  m2 g  a
T  5.00 kg 9.80 m s 2  0.784 m s 2 
T  45.08 N
(c) What would be the acceleration if the coefficient of kinetic friction
increased to 0.300? What does this mean physically?

m2  m k m1 
a
g
m1  m2 

5.00 kg  0.300 20.0 kg 
9.80 m s 
a
20.0 kg  5.00 kg 
2
a  0.392 m s 2
The acceleration is negative. If the first mass were initially moving
towards the right, then it would slow to a stop. If the system starts
at rest, it would remain at rest {static friction}.
Complete Homework Set 5
Problems.
Inclined Planes
Homework: Set 6
{take two days to finish…}
Introduction: An inclined plane (or incline or ramp) is a flat surface set
to some angle relative to the horizontal.
q = angle of the incline to
the horizontal.
Motion of the object will be
parallel to the surface of the
incline.
F  0
F||  ma
The following slides include examples of inclined planes.
Johnstown, Pennsylvania
Reconstruction of Galileo’s device for determining the law of odd
numbers.
Problem Solving Strategy: Since the motion is parallel to the ramp,
take coordinates that are parallel and perpendicular to the ramp:
 direction
|| direction
This choice puts the acceleration only along one axis. This will
simplify the problems.
When the cart is released on the ramp, it accelerates down the ramp.
Note the water level when at rest or accelerating down the ramp.
http://www.physics.umd.edu/lecdem/services/demos/demosc4/c412.htm
Ex. #1: Determine the components of gravity for a mass on an inclined
surface. The weight force pulls straight downwards. Determine the
components parallel and perpendicular to the surface of the incline.
90  q q
 mg cos q
mg
ǁ= mg sin θ
Ex. #2: A mass m slides down a ramp without friction. (a) What is the
normal force acting on the mass and (b) what is the acceleration of the
mass?
The perpendicular
components balance:
n  mg cos q
The parallel components
produce the acceleration:
ma  mg sin q
a  g sin q
Ex. #3: A mass m slides down a ramp with friction. (a) What is the
normal force acting on the mass and (b) what is the acceleration of the
mass?
The perpendicular
components still balance.
n  mg cos q
By definition, the friction
force is:
Fk  mk n
Fk  mk mg cos q
The parallel components produce the
acceleration. Take to the bottom of the
ramp as the positive direction.
ma  mg sin q  Fk
Substitute values and solve for the acceleration:
ma  mg sin q  mk mg cosq
a  g sin q  mk cos q 
Note: The acceleration will be zero when:
sin q  mk cosq  mk  tan q
Ex. #4: A mass m is pulled to the top of a frictionless ramp by an
applied force f. (a) If the mass is pulled to the top at a constant
acceleration of a = 4.90 m/s2, what is the required force?
Let q = 36.9° and m = 80.0 kg.
Since there is no drag, the perpendicular
components still balance are not needed.
Apply Newton’s 2nd law to the parallel
components:
ma   f  mg sin q
Solve for f:
ma   f  mg sin q
f  ma  mg sin q
f  80.0 kg 4.90 m s 2  9.80 m s 2 sin 36.9
f  863 N
Ex. #5: Repeat the previous problem, but this time include friction on
the surface of the ramp. The coefficient of kinetic friction is 0.400.
There is an additional
drag force pointing to
the bottom of the ramp.
Fk  mk n
n  mg cos q
By definition, the
friction force is:
Fk  mk n
Fk  mk mg cos q
Now compute the net force parallel to the incline:
ma   f  mg sin q  Fk
f  ma  mg sin q  Fk
f  ma  mg sin q  mk mg cos q
f  ma  g sin q  mk cosq 
plug in values:
f  1110 N
Ex. #6: A mass m is placed on a frictionless incline of angle q to the
horizontal. What horizontal force, F, is needed to hold the mass at rest?
Values are given in the diagram.
Balance the vertical
forces:
n cos q  mg
n
q
Balance the horizontal
forces:
n sin q  F
mg
Divide the two equations to eliminate the normal force:
n sin q
F

n cos q mg
 tan q
F  mg tan q
F  100 kg 9.80 m s 2  tan 30.0
F  566 N
Ex. #7: Solve for the acceleration of this system. Mass m1 is 30.0 kg,
mass m2 is 25.0 kg, and the coefficient of kinetic friction mk is 0.200.
Assume m2 is initially moving downwards.
T
Tackle this beast with
the same technique as
used on the Atwood’s
machine.
The perpendicular
components for
mass m1 balance:
T
n
Fk
n  m1 g cosq
Fk  mk m1 g cos q
m1 g
m2 g
Taking to the top of the ramp as the positive direction, solve for the
parallel direction of incline:
m1a  T  m1 g sin q  Fk
m1a  T  m1 g sin q  mk m1 g cos q
Now find Fnet on mass m2.
m2a  T  m2 g
Add the two equations together:
m1a  T  m1 g sin q  mk m1 g cos q

m2a  m2 g  T
m1a  m2 a  m2 g  m1 g sin q  mk m1 g cos q
m1  m2 a  g m2  m1 sin q  mk m1 cosq 
m

ag
2
 m1  sin q  mk cos q  
 m1  m2 
Substituting numbers gives:
a  0.856 m s 2
Ex. #8: {Bonus type question} Solve for the acceleration of this
system. Assume the ramp is frictionless.
m1 
 m2
q  30.0
  60.0
Solution Hints:

m1a  T  m1 g sin q
m2a  T  m2 g sin 
m1  m2 a  m2 sin   m1 sin q g

m2 sin   m1 sin q 
a
g
m1  m2 
 0.455 m s 2