-
MOTION AND FORCE:DYNAMICS
We’ve been dealing with the fact that objects move. Velocity,
acceleration, projectile motion, etc. WHY do they move? Forces act
upon them, that’s why! The connection between Force and motion
isDYNAMICS.
FORCEA push or pull on an object. An object falls due to the
force of gravity. Forces do not always
give rise to motion.! Spring Scale--a device used to measure
force! Weight--a measure of the FORCE on an object, carries a unit
of Newtons (N).! Forces--are vectors that carry a magnitude and
direction.
NEWTON’S FIRST LAW OF MOTION [Law of Inertia]
An object in motion remains in motion along a straight line path
until a force acts upon that object. Anobject at rest remains at
rest until a force acts upon that object.
! Inertia--tendency of a body to maintain its state of rest or
uniform motion.! Friction--a force that acts “against” motion!
Inertial Reference Frame--Newton’s first law holds true. Rest a cup
on a
dashboard in your car before leaving your garage--back out of
the garage--splash! That is NOT an inertial frame of reference!
! Mass--a property; measure of the inertia of a body (kg) !
Weight--a force; Fg = mg measured in Newtons. A word that has a
casual
meaning on Earth since g is relatively constant @ 9.8 m/s2. Mass
and weighthave often been used interchangeably during your short
lifetime!
NEWTON’S SECOND LAW OF MOTION [F = ma]
The acceleration of an object is directly proportional to the
net force acting on it and is inverselyproportional to its mass.
The direction of the acceleration is in the direction of the net
force acting onthe object.
a = EF m
In English:When a net force acts upon an object, it’s rest or
uniform motion is changed. That means it experiencesan
acceleration, technically defined as a change in velocity. !!!! E F
= ma! E F = net force! Force--an action capable of accelerating an
object; measured in Newtons! 1 N = kg Cm/s2! ONE Dimensional L F =
ma! TWO Dimensional L Fx = max & Fy = may & Fz = maz
-
Rene’ McCormick, AP Strategies, Inc.Forces 2
Example 4.1Estimate the net force needed to accelerate a 1000-kg
car at 1/2 g.
Example 4.2What net force is required to bring a 1500-kg car to
rest from a speed of 100 km/h within a distance of55 m?
NEWTON’S THIRD LAW [action-reaction]
Whenever on object exerts a force on a second object, the second
object exerts an equal and oppositeforce on the first object. Often
stated, “To every action there is an equal and opposite
reaction.”
The second law quantifies how forces affect motion. But, where
do forces come from? A force isexerted ON an object and BY another
object. Hit a nail with a hammer--the force is applied ON the
nailBY the hammer BUT, the nail exerts a force back on the hammer.
How do we know? 1) the speed ofthe hammer is rapidly reduced to
zero! 2) this force must be strong to so rapidly reduce the speed
of thehammer 3) the nail rarely sinks all the way into the wood on
the first try!
Example 4.3What makes a car go forward?
-
Rene’ McCormick, AP Strategies, Inc.Forces 3
Example 4.4Michelangelo’s assistant has been assigned the task
of moving a block of marble using a sled. He saysto his boss, “When
I exert a forward force on the sled, the sled exerts an equal and
opposite forcebackward, so how can I ever start it moving? No
matter how hard I pull, the backward reaction forcealways equals my
forward force, so the net force must be zero. I’ll never be able to
move this load.” Isthis a case of a little knowledge being
dangerous? Explain.
WEIGHT--THE FORCE OF GRAVITY; AND THE NORMAL FORCE
! Weight--The force of gravity originates at the center of the
Earth.Fg = Fw = mg always directed downward, toward the center of
the Earth.
! g--9.8 m/s2
The moon’s gravity is 1/6 that of the Earth since the moon’s
mass is 1/6 that of theEarth’s. Your weight on the moon is 1/6 that
of your weight here on Earth--I findthis a pleasant thought!
So why don’t we go crashing through the floor on to the poor
souls below? Theforce of the floor exerts force “right back at ‘ya”
due to it’s elasticity. This force isthe normal force and is NOT a
case of action-reaction!
! Normal force-- “normal” means z (usually thought of as z to
the motion!). FN! Since E F = 0 for an object at rest and g is
acting downward this force often “balances” the force
of weight for a stationary object.
-
Rene’ McCormick, AP Strategies, Inc.Forces 4
Example 4.5A friend has given you a special gift, a box of mass
10.0 kg with a mystery surprise inside. It’s a reward for your fine
showing on the physics final. The box is resting on a smooth
(frictionless) horizontal surface of a table.a) Determine the
weight of the box and the normal force acting on it.
b) Now your friend pushes down on the box with a force of 40.0
N. Again determine the normal force acting on the box.
c) If your friend pulls upward on the box with a force of 40.0
N, what now is the normal force on the box?
Notice that the normal force is elastic in origin! THAT’S WHY IS
NOT AN ACTION-REACTIONPAIR WITH THE WEIGHT!
Example 4.6What happens when a person pulls upward on the box in
Example 4.5 c) with aforce equal to, or greater than, the box’s
weight, say Fp = 100.0 N rather than the40.0 N shown in figure 4-16
c)?
-
Rene’ McCormick, AP Strategies, Inc.Forces 5
SOLVING PROBLEMS WITH NEWTON’S LAWS:VECTOR FORCES AND FREE-BODY
DIAGRAMS
Net force = vector sum = E F and is % to acceleration ofan
object
Example 4.7Calculate the sum of the two forces acting on the
boat shown in Fig. 4-19a.
Free-Body diagram:! Show all forces acting on EACH object
involved! Point masses for now..... until we start rotating
stuff!Example 4.8A hockey puck is sliding at constant velocity
across a flathorizontal ice surface that is assumed to be
frictionless. Which of the sketches is the correct free-body
diagram forthis puck? What would your answer be if the puck
wasslowed down?
-
Rene’ McCormick, AP Strategies, Inc.Forces 6
Example 4.9Suppose a friend asks to examine the 10.0 kg box you
were given earlier, hoping to guess what’s inside. You respond,
“Sure, pull the box over to you.” She then pulls the box over by an
attachedribbon/string/cord/rope along the smooth surface of the
table. The magnitude of the force exerted by theperson is 40.0 N
and is exerted at a 30.0° angle with the table top. Calculatea) the
acceleration of the box.
b) the magnitude of the upward force, FN exerted by the table on
the box. Assume that friction can beneglected.
! Tension--FT--when a flexible cord/rope/etc. pulls on an object
it is said to be under “tension”. These can only pull, NOT push
since they are flexible pieces of matter!
Example 4.10Two boxes are connected by a lightweight cord and
are resting on a table. The boxes have masses of12.0 kg and 10.0
kg. A horizontal force Fp of 40.0 N is applied by a person to the
10.0 kg box. Find a) the acceleration of each box
-
Rene’ McCormick, AP Strategies, Inc.Forces 7
b) the tension in the cord.
Example 4.11Two masses suspended over a pulley by a cable is
sometimes referred to as anAtwood’s machine. Consider the real-life
application of an elevator (m1) andits counterweight (m2). To
minimize the work done by the motor to raise andlower the elevator
safely, m1 and m2 are similar in mass. We leave the motorout of the
system for this calculation, and assume the cable’s mass is
negligibleand the pulley is frictionless and massless, which
assures that the tension, FT, inthe cord has the same magnitude on
both sides of the pulley. Let the mass ofthe counterweight be m2 =
1,000 kg. Assume the mass of the empty elevator is850 kg, and its
mass when carrying 4 passengers is m1 = 1150 kg. For the lattercase
(m1= 1150 kg), calculate a) the acceleration of the elevator
and
b) the tension in the cable.
-
Rene’ McCormick, AP Strategies, Inc.Forces 8
Example 4.12Muscleman is trying to lift a piano (slowly) up to a
second-story apartment. He is using a rope loopedover two pulleys
as shown. How much of the piano’s 2000 N weight does he have to
pull on the rope?
Example 4.13Finding her car stuck in the mud, a bright graduate
student of a good physics course ties a strong rope tothe back
bumper of the car, and the other end to a tree. She pushes at the
midpoint of the rope with hermaximum effort, which she estimates to
be a force Fp . 300 N. The car just begins to budge with therope at
an angle 2 which she estimates to be 5°. With what force is the
rope pulling on the car? Neglectthe mass of the rope.
-
Rene’ McCormick, AP Strategies, Inc.Forces 9
APPLICATIONS INVOLVING FRICTION; INCLINES
Friction--exists between 2 solid surfaces because even the
smoothest lookingsurface is quite rough on a microscopic scale.
When we try to slide an objectacross another surface, these
microscopic bumps impede motion. Additionally,the atoms snuggle up
next to one another and have interactions of an attractivenature
and can further impede motion.
kinetic friction--sliding friction; the friction that
persistseven once the object is in motion. This force actsopposite
to the body’s velocity and is determined by thenature of the two
surfaces. The force of kinetic friction isapproximately
proportional to the normal force. [the normal force is the
forcethat either object exerts on the other perpendicular to their
common surface ofcontact]
Ff % FN insert a proportionality constant and presto, an equals
sign appears along with a constant!
Ff = :K FN
:K--the coefficient of kinetic friction; its value depends on
the nature of the 2 surfaces; it’s anapproximation since polishing/
sanding, etc. alters the surfaces.
This is not a vector equation since the two vectors act z to one
another. It is also an experimentalrelationship NOT a fundamental
law. The force of friction depends very little on surface area.
Itdoesn’t matter whether you slide your book flat along the table
or on its spine along the table--it’sfrictional force is
essentially unaffected.
-
Rene’ McCormick, AP Strategies, Inc.Forces 10
Static friction--force applied to get an object moving; it is
always present between 2stationary objects and increases when the
force applied increases until the forceapplied overpowers the
static frictional force and the kinetic frictional force takesover.
At the point it begins to move you have applied the maximum force
of staticfriction
FMAX # :S FN you’ve probably noticed it’s easier to keep an
object movingthan get it to move!
Example 4.14Our 10.0 kg mystery box rests on a horizontal floor.
The coefficient of static friction is :s = 0.40 and thecoefficient
of kinetic friction is :K = 0.30. Determine the force of friction,
Ff, acting on the box if ahorizontal external applied force FA is
exerted on it of magnitue [answer in 2 sig. figs for each part!]a)
0 N
b) 10 N
c) 20 N
d) 38 N
e) 40 N
-
Rene’ McCormick, AP Strategies, Inc.Forces 11
Notice that BOTH the normal force and thefrictional forces are
exerted by one surface ONanother.
FN is perpendicular to the contact surface
Ff is parallel to the contact surface and in theopposite
direction of the x velocity.
Example 4.15Your little sister wants a ride on her sled. If you
are on flat ground, will ou exert less force if you pushher or pull
her? Assume the same angle 2 in each case.
Example 4.16Two boxes are connected by a cord running over a
pulley. The coefficient of kinetic friction betweenbox I and the
table is 0.20. We ignore the mass of the cord and pulley and any
fricion in the pulley,which means we can assume that a force
applied to one end of the cord will have the same magnitude atthe
other end. We wish to find the acceleration, a, of the system,
which will have the same magnitudefor both boxes assuming the cord
doesn’t stretch. As box II moves down, box I moves to the
right.
-
Rene’ McCormick, AP Strategies, Inc.Forces 12
OBJECTS MOVING ON AN INCLINE
Easier to choose the xy coordinate system so that the x-axis is
parallel to theincline surface and the y-axis is therefore,
perpendicular to the incline. Thishelps because then a has only one
component and if friction is present, two of theforces will have
only one component: Ff along the plane, opposite ot the
object’svelocity, and FN which is NOT vertical but is perpendicular
to the plane. Apicture will help!
F FF F mg
N W
applied W
== =
cossin sinθ
θ θ
Example 4.17A skier has just begun descending a 30.0° slope.
Assuming the coefficient of kinetic friction is 1.10, a) draw the
free body diagram
b) calculate her acceleration
c) calculate the speed she will reach after 4.0 seconds.
Example 4.18Suppose the snow is slushy and the skier moves down
the 30.0° slope at constant speed. Determine thecoefficient of
friction, :K?