Download PHYS 1114: Physics I

PHYS 1114: Physics I
Lecture 8:
Concepts of Force & Net Force
Professor Kenny L. Tapp
Balancing$Act
The$Concepts$of$Force$and$Net$Force
A force is something that is capable of changing an
object’s state of motion, that is, changing its velocity.
Any$par(cular$force$may$not$actually$change$an$
object’s$state$of$mo(on,$as$there$may$be$other$
forces$that$prevent$it$from$doing$so.$
However,$if$the$net$force—the$vector$sum$of$all$
forces$ac(ng$on$the$object—is$not$zero,$the$
velocity$will$indeed$change.
The$Concepts$of$Force$and$Net$Force
• Forces'are'vectors!
• Balanced'forces
– “cancel$out”
– No$accelera(on
– DOES$NOT$MEAN$no$mo(on!
• Unbalanced'forces
– “something$leK$over”
– Cause$acclera(ons
Forces are Vectors so Directions are Important
Total$Force$
Force$#2
Force$#1
This$figure$illustrates$what$
happens$in$the$presence$of$zero$
and$nonzero$net$force.
Force$#1
Total$Force$=$0
Forces Add
Force$#2
Forces Cancel
The$Concepts$of$Force$and$Net$Force
We'dis5nguish'two'types'of'forces:
1. 'A'contact$force,'such'as'a'push'or'pull,'
fric5on,'tension'from'a'rope'or'string,'
and'so'on.
2. 'A'force$that$acts$at$a$distance,'such'as'
gravity,'the'magne5c'force,'or'the'
electric'force.
Mass or Inertia
– Inertia is the tendency of an object to remain at rest or
in motion with constant speed along a straight line.
– Mass (m) is the quantitative measure of inertia. Mass is
the property of an object that measures how hard it is to
change its motion.
– Units: [M] = kg
Force Balance Example
An airplane is flying from Buffalo airport to O'Hare.
Many forces act on the plane, including weight
(gravity), drag (air resistance), the thrust of the
engine, and the lift of the wings. At some point
during its trip the velocity of the plane is measured to
be constant (which means its altitude is also
constant). At this time, the total (or net) force on the
plane:
li:
drag
1. is pointing upward
2. is pointing downward
3. is pointing forward
4. is pointing backward
5. is zero
correct
Quick Question 1:
Quick Question 2:
Find the x and y components of this vector:
Find the sum of these two forces:
F = 15 N at 40 degrees N of E
A = 20 N at 30 degrees W of N
B = 15 N at 20 degrees N of W
thrust
weight
Quick Question 3:
CHALLENGE
A 5.0 N horizontal force (to the right) pulls a 20 kg box on a
horizontal surface. A 3.0 N friction force retards the motion.
What is the net force of the object?
Create a Photo-Essay using 2
different examples of Forces on
Campus.
Develop a sketch of those
photos that show all of the
forces.
Kerri McAffrey & Emily Grubis
Read Chapter 4 of Cutnell & Johnson.
Forces
Forces:$$GRAVITY
In order to work with forces, we have to identify the common
forces we find, both as to magnitude and direction:
• gravity (near earth’s surface, this is called weight, W)
magnitude = m*g; direc?on = down
• contact$force: magnitude = balances$up$to$point$
of$collapse; direc?on = perpendicular$to$the$
surface that supplies the contact.
• The force that holds the planet in orbit around the Sun, the Moon in
orbit around the Earth
• Always attractive, tries to pull objects together
• Acts between pairs of objects
Newton measured gravity and developed his law of Universal
Gravitation
Every object in the Universe is attracted to, and attracts, every
other object in the Universe by a force that we call gravity.
Quick Question 4:
Forces:$$GRAVITY
• What is the force of gravity exerted by the earth on a typical physics student?
Find the magnitude of the weight of a 3.50 kg object on the
surface of the Earth.
– Typical student mass m"="55kg
– g"="9.8"m/s2.
– Fg"="mg"="(55"kg)x(9.8"m/s2")
– Fg"="539"N
Fg
The$force$that$gravity$exerts$on$any$object$is$
called$its$Weight
W"="539"N
Quick Question 5:
Quick Question 6:
Find the mass of this object on the Earth’s surface:
a) W = 800 N
How much does an 80 kg astronaut weigh on Earth?
...on the surface of the moon (where g=(1/6)gearth)?
Forces:$$GRAVITY
Forces:$$GRAVITY
Force$on$mass:
m
m2
F2,1
F1,2
mass$on$surface
of$Earth
m1
r12
Re
}
Me
g
F1,2$=$force$on$m1$due$to$m2$=
=$F2,1$=$force$on$m2$due$to$m1
Fg$≡$W$=$mg
Direc(on:$$along$line$connec(ng$the$masses;$aVrac(ve
G$=$universal$gravita(on$constant$=$6.673$x$10[11$N$m2/kg2
In the news...
Quick Question 7:
(March, 2011)
A 5.0-kg rock and a 3.0x10-4-kg pebble are held near the surface
of the earth. (a) Determine the magnitude of the gravitational
force exerted on each by the earth.
!
International Space Station
(June, 2010)
http://spaceflight.nasa.gov/realdata/tracking/index.html
In the news... Quick Question 8:
The mass of the International Space Station is 344,378 kg.
Determine the weight of the station (a) when it was resting on
the Earth and (b) as it is in orbit 360 km above the Earth’s
surface.
Forces:$$NORMAL
FN
book$at$rest$on$table:
What$are$forces$on$book?
W
Weight$is$downward
System$is$“in$equilibrium”$(accelera(on$=$0$⇒$net$force$=$0)
Therefore,$weight$balanced$by$another$force
FN$=$“normal$force”$=$force$exerted$by$surface$on$object
FN$is$always$perpendicular$to$surface$and$outward
For$this$example$$$$FN$=$W
Forces:$$NORMAL
Forces
• fric?on, Ff: magnitude: balances up to a point, and
then reaches a constant$value that depends on the two
surfaces and how hard the two surfaces are being pressed
together ( Ff$≤$µFc ), $direc?on: parallel$to$
surface.
Forces:$$KINETIC$FRICTION
A force, fk, between two surfaces that
Forces:$$KINETIC$FRICTION
FN
opposes motion.
FN
direc(on$of$mo(on
$fk$=$µkFN$
fk
direc(on$of$mo(on
F
µk$=$coefficient of kinetic friction
(a property of the two surfaces)
fk
W
F
W
Kerri McAffrey & Emily Grubis
Forces:$$STATIC$FRICTION
FN
Forces:$$STATIC$FRICTION
FN
A force, fs, between two surfaces that
prevents motion.
fs
F
W
fs$≤$fsmax=$µsFN
$$$$$$$
force$just$before$breakaway$$
fs
F
W
µs$=$coefficient of static friction
(a property of the two surfaces)
Isaac Henry & Matt English
Quick Question 9:
Forces:$$FRICTION$&$ICE
The coefficients of static and kinetic friction between a 3.0 kg
box and a desk are 0.40 and 0.30, respectively. What is the
net force on the box and the acceleration of the box when
each of the following horizontal forces is applied to the box:
fk
fk
(a) 5.0 N
(b) 10 N
(c) 15 N
F
Quick Question 10:
Professor Tapp on Ice...If
the kinetic coefficient
of friction between ice
skates and ice is 0.01,
what is the average
forward force
necessary for a 500 N
skater to maintain a
constant forward
velocity?
fk$=$µkFN
FNET$=$Fskater$$[$fk$
Quick Question 11 HINT:
Quick Question 11:
Clark Griswold coats his saucer sled with a non-nutritive cereal
varnish and smokes down a slope with an angle of inclination of
31 degrees.
a) Neglecting
friction, how many
forces are acting on
the sledder?
c) If Clark has a speed of 4 m/s at the top of the
b) What is the
slope, what is his speed when he reaches the
acceleration of Clark?
bottom of the 32 m slope?
Quick Question 12:
µk$=$0.2
v0$=8$m/s
Find*stopping*distance
Normal$force$is$balanced$by$gravity$because$there$is$no$ver(cal$mo(on,$
i.$e.,$N$=$Mg,$if$M$is$the$mass$of$the$object
Kine(c$fric(onal$force$that$decelerates$the$block$is,$f$=$µk$N$=$µk$Mg
Therefore,$decelera(on$(direc(on$opposite$of$v0),$a$=$[f/M$=$[µk$g
Given$decelera(on,$use$kinema(cs$equa(on$to$obtain$the$answer.
v 2 = v02 + 2aΔx ⇒ Δx =
Forces
Forces:$$TENSION
• tension, T: magnitude: pulls$same$at$one$end$
as$another unless rope is being accelerated;
direc?on:$parallel$to$rope.
€
Phuong Nguyen,
Nancy Nguyen,
Charles Ung,
Star Kamal
v 2 − v02 v 2 − v02
0 − 82
=
=
m = 16.3 m
2a
−2µk g −2 ⋅ 0.2 ⋅ 9.8
T
• Tension: force exerted by a rope.
• Magnitude: same everywhere in
rope (Not changed by pulleys).
• Direction: same as direction of
rope.
ΣFy = may
T - W = may
T = W + may
y
T
ay = 0
W
Forces:$$TENSION$&$PULLEYS
Forces:$$TENSION$&$PULLEYS
Net Force = F + T + (-W)
Net Force = T + T + (- W)
We know that W = mg, so...
0 = 2T - W
F + T = mg
We know that W = mg, so...
We know that F = T, so...
0 = 2T - mg
2F = mg
mg = 2T
F = (mg)/2
(mg)/2 = T
Quick Question 14:
Determine the force exerted by the hand to suspend the 45
kg mass as shown in the picture.
Remember the magnitude of
the tension is the same
everywhere along the rope!
Quick Question 15:
Determine the force on the ceiling.
∑F = 0 = T + T + (- W)
∑F = 0 = Fc -T - T - T
2T = W
Fc = 3T
T = (mg)/2
Fc = 3 x 220 N
T = (45 x 9.8)/2
Fc = 660N
T = 220 N
Quick Question 16:
Brad was able to make his dad, Tim
Taylor, fly around the backyard with
the help of a pulley system.
Assuming Tim (mass = 75 kg) is
lifted at a constant rate using a
single frictionless pulley, with how
much force is Brad pulling on the
cable?
Quick Question 17:
The mass of an average Holstein cow that produces milk in the US
is 580 kg and the mass of an average mature bull is 800 kg.
Determine the tension on the rope for each cow.
Drawing a diagram may help.
Quick Question 18:
If your professor’s mass is 80 kg, determine the tension on the
rope at the mid and final points of the descent.
Think about Fnet!
Drawing a diagram may help.
Forces
• Spring$Force, Fs: magnitude:$kx,$the force depends on the
materials and geometry; direc?on: the spring force is opposite the
Spr
direction of the pull or push.
in
gC
Hooke’s Law :
Fs$=$Vkx
Restor
ing Fo
rce
ons
tan
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s
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he sp rium
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f
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Disp d from i sition
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Quick Question Extra:
# End of Lecture
Provide a detailed solution to the following problem...
Quick Question Extra:
Jack Sparrow (mass = 75 kg) escapes an East India Trading
Company ship. With one hand, he is gripping the rope that is
tied to the end of a canon ( mass = 330 kg). Ignore the
friction between the rope and the wooden mast over which it
slides, and find (a) the acceleration with which Jack is pulled
upward and (b) the tension in the rope while Jack escapes.