Download TOPIC 5: DYNAMIC FORCES SUPPLEMENTAL INDEPENDENT

TOPIC 5: DYNAMIC FORCES
SUPPLEMENTAL INDEPENDENT PRACTICE
The goal of this document is to provide extra support for students that
might need help and to provide enrichment to students that want to go
deeper into the topic. This is a great source of practice problems as you
prepare for the trimester final too.
Table of Contents:
KEY PROBLEMS with FULL SOLUTIONS
EXTRA PROBLEMS with ANSWERS
VOCABULARY TERMS
Dynamic Forces – Web Supplement
Pages 2 - 8
Pages 9-15
Pages 16-20
Page 1 of 20
Topic 5
Key Problem -5.1 Horizontal Linear Problems
A toy train consists of a 7 kg engine and two 4 kg cars. The tension between
the last two cars is 50 N. What is the acceleration of the train? What is the
other tension value?
EXPLANATION
SOLUTION
a)
F
FT
Fy  0
U D
F  Fg
Fg
40N  40N
Fx  ma
FT 1  ma
50N  (4kg)a
a  12.5m / s 2
b)
F
FT
FT
Fg
Fx  ma
W  L  ma
FT2  FT 1  ma
FT2  50N  (4kg)(12.5m / s 2 )
FT2  100N
Dynamic Forces – Web Supplement
a) Start with a free body diagram
for the last car. We can tell from
context that the cars are in
equilibrium in the y direction. That
means the weight is equal to the
normal force. There is only one
force in the x direction, and so it
will produce an acceleration. Note
that the FBD was drawn for the
last car only, so I only use his mass
in my math.
b) The first car has a slightly
different free body diagram. It is
being pulled forward by the engine
BUT it is also being pulled
backward by the second car. The
vertical forces cancel out again,
but the horizontal ones don’t. The
forward tension is winning. The
total force is always the “winner
minus the loser.” Since both cars
are connected, the have the same
acceleration.
Both cars are the same mass, so it
makes sense that the first tension
is twice as big as the second one.
Page 2 of 20
Topic 5
Key Problem - 5.2 Elevators
A 60 kg girl is standing on a bathroom scale in an elevator. At a particular
moment she sees that the scale reads 800 N. What is the magnitude and
direction of the elevator’s acceleration at that moment?
SOLUTION
EXPLANATION
a)
a) Start with a free body diagram.
The girl has a weight and the floor
is pushing up on her. A 60 kg girl
at rest on a scale should only
experience a normal force of 600
N, so she is clearly NOT in
equilibrium.
F
Fg
b)
Fy  ma
W  L  ma
F  Fg  ma
800N  600N  (60kg )a
200N  60a
a  3.3m / s 2
Dynamic Forces – Web Supplement
b) Write out the second law and
use your diagram to tell which
force is the winner and which is
the loser. Substitute in all of the
values given and solve for the
acceleration. The math tells us
that the magnitude of the
acceleration is 3.3 ms/2 and we can
tell that the direction is “up”
because the “up” force was
winning.
Page 3 of 20
Topic 5
Key Problem -[after 5.2] Chunky Dynamics
A 30 kg box is being pulled up a 20° incline with a force of 70 N. The µ value
between the box and the ramp is 0.2. What is the acceleration of the box?
SOLUTION
a)
EXPLANATION
F
FF
FT
Fg
a) Once again, we start with the
free body diagram. Sensing a
theme yet? There should be four
forces in your picture: Weight,
normal force, tension and friction.
b) Since this is a ramp problem,
we do our math in the ramp
F  ma
direction. Presumably, the box will
accelerate uphill, since it is being
W  L  ma
pulled uphill. BE CAREFUL! The
FT  FF  F  ma
tension is winning, but there are
300N  mg cos   mg sin   (30kg)a
300  (.2)(30)(10)(cos 20)  (30)(10) sin(20)  30a TWO losers: Friction AND the
300  56  102  30a
parallel force! (You always have a
142  30a
parallel force if the normal force
2
a

4.7m
/
s
and the weight don’t perfectly

cancel out.) Finishing the math,
we see that the acceleration of
the box is 4.7 m/s2 uphill.
b)
R
Dynamic Forces – Web Supplement
Page 4 of 20
Topic 5
Key Problem - 5.3 Uniform Circular Motion
A 200 kg go-car is traveling at 12 m/s and is taking laps around a 30 m radius
circular track. What is the period of the motion? What is the centripetal
acceleration of the cart? What is the centripetal force necessary to keep the
car on this circle?
EXPLANATION
SOLUTION
a)
² s C

² t
T
12m / s  (2)( )(30m) / T
T  15.7 s
v 
a) The period is the amount of
time it takes for one lap. You can
do this with the first Galilean
definition of speed or the new
version that replaces the generic
“distance” with the circumference
of the motion and the generic
“time interval” with the period.
b)
2
v
r
(12m / s )2
ac 
30m
ac  4.8m / s 2
ac 
c)
Fc  mac
Fc  (200kg)(4.8m / s 2 )
b) The centripetal acceleration can
be found with this new definition.
It is pretty easy to use.
c) The centripetal force definition
is just the second law rewritten
with the “c” subscripts. It works
the same way as the basic version
of the second law.
Fc  960N
Dynamic Forces – Web Supplement
Page 5 of 20
Topic 5
FT
Key Problem - 5.3 Uniform Circular Motion
Tarzan (75 kg) is swinging on a 4 m long vine. At the bottom of the arc he is
traveling forward at 12 m/s. What is the tension in the vine at that point?
SOLUTION
a)
EXPLANATION
FT
Fg
a) We will begin with the Free
Body Diagram. We know he has a
weight and that there is tension in
the vine, but should they be equal?
If not, which is winning? He is
traveling in an arc so, he needs a
centripetal force. The tension is
going the “right way” so it must be
winning.
b)
Fy  Fc
W  L  Fc
mv 2
FT  Fg 
r
(75kg )(12m / s )2
FT  750N 
4m
FT  750  2700
FT  3450N
Dynamic Forces – Web Supplement
b) The sum of the unbalanced
forces IS our centripetal force.
We can substitute values given in
the problem into our expression of
the second law and arrive at our
answer. Note that if Tarzan were
just hanging off the vine in an
equilibrium state, the tension
would only be 750 N. The vine is
much more likely to break when he
is swinging.
Page 6 of 20
Topic 5
Key Problem - 5.4 Universal Gravitation
What is the force of gravity between two 4000 kg dump trucks that are
parked 15 meters apart?
SOLUTION
EXPLANATION
a)
a) To find the gravitational
attraction between two masses, we
use the definition of Universal
Gravitation.
Fg 
Gm1m2
r2
b) All of the values that we need
were given in the problem. There
are two parts to this that students
find tricky:
b)
Fg 
Fg 
Gm1m2
r2
(6.6x 10 11)(4000kg )(4000kg)
(15m)2
Fg  4.7x 10 6 N
1) Make sure you use parentheses
in your math when you type on your
calculator. Your calculator reads
2/3*4 and 2/(3*4) VERY
differently.
2) Make sure you remember to
square the distance between the
objects!!!
Dynamic Forces – Web Supplement
Page 7 of 20
Topic 5
Key Problem - 5.4 Universal Gravitation
A weather satellite is supposed to orbit the earth 4 times a day. What is
the proper radius for that orbit? How high up in the air is that?
EXPLANATION
a) Orbits are situations in which an
object moves in a circle because of
gravity. This means we can solve
orbit problems by setting the
universal gravitation definition
equal to the centripetal force
definition.
SOLUTION
a)
Fg  Fc
Gm1m2
r2
mv 2

r
b) After doing some algebra with
the initial definitions, we can
arrive at an expression that solves
for the radius of the orbit. Note
that to get the answer you need
the CUBE root of the expression
under the radical sign.
b)
r 
3
r 
3
(G)(m)(T )2
4 2
(6.6x 10 11)(6x 10 24 )(4 * 3600)2
4
2
r  2x 10 21
3
r  1.25x 10 7 m
c) The answer in part b) is from
the center of the object to the
center of the earth. To get the
height of the orbit above the
ground, you need to subtract away
the radius of the earth.
c)
1.25x 10 7 m  (6x 10 6 m)  6.5x 10 6 m
Dynamic Forces – Web Supplement
Page 8 of 20
Topic 5
EXTRA PROBLEMS with ANSWERS for TOPIC 5: DYNAMIC FORCES
Section 5.1 Linear Dynamics
1) A 40 kg crate is pushed across the floor with a force of 500 N, but the
acceleration of the cart is only 4 m/s2. What is the force of friction
between the crate and the floor? What is µ?
2) A soccer player has cleats on that give her a µ value of 1.35 with the turf.
What is the fastest horizontal acceleration she can have?
3) A 10 kg curling stone is released with a speed of 6 m/s and it comes to
rest on its own after traveling 78 meters. What is the force of friction
between the stone and the ice? What is µ?
4) A toy train engine has a mass of 12 kg and is pulling two cars. The first
car is 5 kg and the second car is 10 kg. The engine’s wheels are pulling the
whole train forward with a force of 200 N. What is the acceleration of the
train? What is the tension in each coupling?
ANSWERS:
1) F= 340 N. µ=.85
2) 13.5 m/ss
3) µ = 0.023 F = 2.3 N
4) a = 7.4 m/ss, T1 = 111 N T2= 74 N
Dynamic Forces – Web Supplement
Page 9 of 20
Topic 5
Section 5.2 Vertical Dynamics
1) A 4 kg toy rocket leaves the ground due to an upward thrust of 200 N.
What is the net force on the rocket? What is the acceleration of the
rocket?
2) A 70 kg kid is standing on a bathroom scale in an elevator. At a
particular moment, the scale reads 450 N. What is the magnitude and
direction of the elevator’s acceleration at that moment?
3) A kid is standing on a bathroom scale in an elevator. The scale reads 600
N and the kid knows that the elevator is accelerating up at 3 m/s2. What is
the mass of the kid?
4) An Atwood machine is created by hanging a 30 kg ball of fungus and a 24
kg sack of bananas over a pulley. After the system is released from rest,
what will the tension in the connecting string be? What will the magnitude
of the acceleration of the objects be?
ANSWERS:
1) 160 N, 40 m/ss
2) 3.6 m/ss, down.
3) 46 kg
4) FT= 266 N, a = 1.1 m/ss
Dynamic Forces – Web Supplement
Page 10 of 20
Topic 5
[After Section 5.2 Chunky Dynamics]
1) A force of 50 N is used to push a 20 kg block up a 10° frictionless ramp.
What is the magnitude and direction of the net force on the block? What is
the acceleration of the block? Draw a FBD…
2) A force of 150 N is used to push a 20 kg block up a 10° ramp. The µ value
between the block and the ramp is 0.5. What is the magnitude and
direction of the net force on the block? What is the acceleration of the
block? Draw a FBD…
3) A 4 kg block on a table is connected to a hanging
10 kg block. The µ value between the 4kg block and
the table is 0.2. When released, what will the
acceleration of the blocks be? Draw FBD’s!
4) A 40 kg block is accelerating down a 30° incline at only 2 m/s2. What is µ
between the ramp and the block? Draw a FBD!
ANSWERS:
1) F = 15.3 N, a = 0.76 m/ss
2) F = 76 N, a = 3.8 m/ss
3 ) FT= 34 N a= 6.6m/s2
4) µ = 0.69
Dynamic Forces – Web Supplement
Page 11 of 20
Topic 5
Section 5.3 Circular Motion
1) A 2 kg ball on a 2.3 m long string is swung in a circle and makes 10
revolutions every 13 seconds. What is the period of the ball? What is the
speed of the ball in m/s?
2) What is the centripetal acceleration of a 900 kg car traveling around a 15
meter radius turn at 10 m/s? What is the centripetal force required for
this motion?
3) A 75 kg passenger on a merry-go-round is sitting on a bathroom scale.
After one complete trip around, he observed that the greatest reading of
the scale was 900 N. The radius of the ride is 12 m. How fast is the ride
moving. Draw a FBD!
4) A 35 kg child on a swing is traveling at 3 m/s at the bottom of the swing’s
motion. What is the tension in EACH of the 2 m long chains of the swing at
that point? Draw a FBD!
ANSWERS:
1) T=1.3 s, v = 11 m/s
2) ac= 6.6 m/ss , Fc = 5940 N
3) v = 4.9 m/s
4) each tension is 253 N.
Dynamic Forces – Web Supplement
Page 12 of 20
Topic 5
Section 5.4 Universal Gravitation
1) Calculate the gravitational force between the sun and the earth. Get the
values for the masses and the distance from your reference table.
2) A 5000 kg block is at the origin. Four meters to it’s left there is am
8000 kg block and three meters to its right there is a 12,000 kg block.
What is the magnitude and direction of the net force on the block at the
origin?
3) A satellite is put into an orbit around the earth such that it makes one
trip around the planet every 14 days. What is the radius of that orbit?
4) An 80 kg astronaut goes to a new planet and discovers that she weighs
1200N. The radius of this planet is 4 x 106 meters. What is the mass of
this new planet?
ANSWERS:
1) 3.5 x 1022 N
2) 2.75 x 10-4 N to the right
3) 2.4 x 108 meters
4) m = 3,6 x 1024 kg
Dynamic Forces – Web Supplement
Page 13 of 20
Topic 5
Vocabulary Supplement for Topic 5
Acceleration An object is accelerating if its speed or velocity is changing
during some time interval. Accelerations are measured in “m/s2” which is
pronounced “meters per second each second.” The word “acceleration” can
be used as either a scalar or a vector term.
Acceleration Due to Gravity This is a different value on each planet. On
Earth the value is about 10 m/s2.
Agent-Object Notation A method of labeling forces in free body diagrams
to highlight the interactions based on the Third Law.
Centrifugal force Oh, no! No, no, no, no, no…..
Centripetal acceleration In order to move in a circle, an object must have an
acceleration that is directed toward the center of the circle. This
acceleration will change the direction but not the magnitude of the object’s
velocity.
Centripetal force Any force such as weight, friction, normal force, or
tension that is responsible for creating a centripetal acceleration for an
object.
Coefficient of Friction A unitless value that indicates how much the
surfaces of a set of materials will interact when in contact. It is based on
chemistry.
Coefficient of Kinetic Friction This represents the amount of friction that
exists between two surfaces that are being slid passed one another
Coefficient if Rolling Friction This represents the amount of friction that
exists between a round object and the surface it is rolling over. Rolling
friction coefficients are the smallest coefficients of friction.
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Page 14 of 20
Topic 5
Coefficient of Static Friction This represents the amount of friction that
exists between two objects that are attempting to slide passed one another.
Static coefficients of friction are the highest coefficients of friction.
Components Any vector can be broken up into 1 to 3 orthogonal pieces using
trigonometry. Each piece is called a component.
Concurrent A synonym for simultaneous.
Coordinate System This is a mathematical construct where we assign values
to directions in space. In physics we use a right-handed coordinate system.
Forward and Up are “positive directions.” Backwards and Down are “negative
directions.”
Cosine This is the ratio of the adjacent side of a triangle to the
hypotenuse of the triangle. It can tell you the percentage of the
hypotenuse that is on the adjacent side of the triangle.
Delta This is how you pronounce the name of the Greek letter “D.” The
delta symbol, Δ, is used in math and physics to represent the word “change.”
Diameter The diameter of a circle is the width of the circle taken between
any two points along the circumference that are 180° apart.
Direction
In order to be fully defined, a vector must include a direction.
The direction can be stated cardinally, but is more commonly expressed as a
right-handed angle.
Dynamic Forces These are situations in which the net force acting on an
object is not zero, and the second law is used in analysis.
Ellipse An ellipse is the proper name for an oval. It is a smooth curve that is
made up of the locus of all points in a plane whose distances tow two specific
points (called foci) add up to a constant. A circle is a special case of an
ellipse.
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Page 15 of 20
Topic 5
Elliptical Orbit All of the planets and comets orbit their respective stars in
elliptical orbits. Planets usually have relatively low eccentricities, while
comets often have large eccentricities.
Equilibrium An object is in equilibrium if the sum of the forces acting on it
adds up to zero. This means that the object is not accelerating. It does
NOT mean the object is not moving.
Equilibriant The vector that has the same magnitude but opposite direction
of the resultant of a set of vectors.
First Law Newton’s first law states that if the sum of the forces acting on
an object is zero that the object is in equilibrium; it is not accelerating.
Free Body Diagram A simple sketch that represents the forces acting on an
object as a set of labeled, scaled vectors.
Free Fall An object is in “free fall” if the only influence on it is gravity.
Force A general term that represents any flavor of push or pull on an object.
Friction A force that opposes the motion of an object.
Geosynchronous Orbit An orbital distance around a planet at which the
satellite’s period matches the rotational period of the planet so that the
satellite “does not move” relative to a fixed point on the surface of the
planet.
Gravity Gravity is the pull that makes massive objects come together.
Hanging Mass Method A way to determine the coefficient of static friction
between two surfaces using a string, a pulley and a pale of sand.
Horizontal Component A piece of a vector that is parallel to the ground.
Inclined Plane This is a flat surface that is raised at one end to some angle
relative to the horizontal direction.
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Page 16 of 20
Topic 5
Inertia An archaic synonym for mass. It represents how much an object
resists the influence of forces.
Inverse Square Law The radiant effects of point like sources decrease as
you get farther away from them. In these specific situations, effect
Kepler’s First Law This law describes how all orbital motion is elliptical in
shape.
Kepler’s Second Law This law describes how a planet moves faster when it is
closer to the sun and slower when it is farther from the sun.
Kepler’s Third Law This law says that every object in orbit around a given
parent body has the same ratio of T2 to r3.
Local Gravity This is the acceleration due to gravity at a given point in space.
It is DIFFERENT in DIFFERENT places.
Magnitude This is the “size” of a measurement, always consisting of a
number AND some units. Scalar quantities only have magnitudes. The
“magnitude” of a vector is what you are left with if you strip its direction
off.
Mass This is the modern version of Inertia. Mass is a fundamental property
of objects. In Physics mass is measured in kilograms.
Net Force The total vector force acting on an object.
Newtons The standard unit for Force is kgm/s2, but as a convenient
shorthand, this string of units is referred to as a “Newton,” N.
Newton’s Laws This set of three equations guide us as we examine situations
in which forces are acting on objects.
Normal This term is a synonym for the word perpendicular.
Normal Force This represents that push that a surface gives to an object at
the point of contact. This force is always perpendicular to the surface.
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Page 17 of 20
Topic 5
Orbit An orbit is circular motion which is due to gravity acting as the
centripetal force.
Parallel Force The vector sum of an object’s weight and the normal force
from the surface it is on.
Period The specific time interval that a repetitive motion takes to complete
one cycle.
Perpendicular Two vectors are perpendicular if there is a 90° in between
them.
Radial acceleration This is a synonym for centripetal acceleration. These
accelerations only change the direction of the velocity vectors, NOT the
magnitude.
Radius Half of the width of a circle.
Ramp Another name for an inclined plane.
Ramp Method A way to determine the coefficient of static friction between
two surfaces by measuring the angle at which they slip passed each other
when inclined.
Resultant
The vector sum of two or more vectors is called the resultant.
Scalar Addition
This is simple arithmetic addition where all the quantities
must have the same units.
Scalar Quantity
be fully defined.
Any measured quantity that only requires a magnitude to
Second Law Newton’s second law tells us that if there is a net force acting
on an object, that object will accelerate in the direction of that net force.
Sigma The greek letter S,  is used in math and science to represent the
idea of “summation.”
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Page 18 of 20
Topic 5
Simultaneous When events take place at the exact same time.
Sine This is the ratio of the opposite side of a triangle to the hypotenuse of
the triangle. It can tell you the percentage of the hypotenuse that is on
the opposite side of the triangle.
Static Forces These are situations in which the net force on an object is
zero and the first law is used in analysis.
Tangent A line is tangent to a curve if it only touches the curve in one place.
Tangential acceleration If an object that is moving along a curve has a
tangential acceleration, it is speeding up or slowing down.
Tangential velocity When an object moves in a circle, its velocity is always
tangent to the circle.
Tension Forces that pull on an object through a medium such as a string,
rope, cable or chain.
Third Law All forces occur in simultaneous pairs which are equal in size and
opposite in direction
Third Law Pair A set of two forces that are related to each other through
agent/object notation. “A on B” and “B on A” form a third law pair.
Uniform Circular Motion Circular motion that has radial acceleration but no
tangential acceleration.
Universal Gravitation This is the idea that all masses feel an attraction to all
other masses, with a strength that depends on the square of their
seperation.
Addition that takes direction in to account. Many times
Vector Addition
this operation looks just like the Pythagorean Theorem.
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Topic 5
Vector Quantity Any measured quantity that requires both magnitude and
direction to be fully defined.
Vertical Component The piece of a given vector that is perpendicular to the
ground.
Weight The force with which we are gravitationally attracted to the planet.
X Direction In our standard coordinate system, we associate the X
direction with horizontal motion.
Y Direction In our standard coordinate system, we associate the Y direction
with vertical motion.
Dynamic Forces – Web Supplement
Page 20 of 20
Topic 5