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Unformatted text preview: Chapter 5 Using Newton’s Laws: Friction, Circular Motion,
Drag Forces Forces of Friction . When an object is in motion
on a surface, there will be a
resistance to the motion,
called the force of friction  Fm, (not a constant),
n1a:»:imun1=_ui5 F,, . F,,f,,= pk F,, (constant) no
motion “—F"'—'sliding—I mi P—n gradLari ':. 61"er Quad ( ‘0“) Some Coefficients of Friction TRBLE H Coefﬁcients ef brittle“i Coefﬁcient ef {Inefﬁcient of
Surfaces Slutle Friction. ﬂ. Kinetic I‘rletiuu. pk
Weed en weecl {L4 i1:
lee eu iee [Ll ilﬂﬁ
Metal en metal {lubricated} {1.15 M)?
Steel en steel {unlubrieateu} EL? be
Rubber e11 erg.r eenerete Hi {18
Rubber en wet eunerete Ill? [LS
Rubber urn ether eulid surteees 1—4 1
Tenant eu Tbﬂeu in air em em
Teﬂun un steel in air 0.04 ilﬂt
Lubricated ball bearings {eel {:1th
Syncvial joints [in human limbs} ilﬂl [Li]! *Vetues are epprmtimele and intended only as e guide. Friction in Newton’ 5 Laws
Problems  Friction is a force, so it simply is
included in the Fir: (we may use f for
simplicity) in Newton’s Laws  The direction of the force of friction
is against motion  The determination of magnitude of
the force of friction needs the normal force, F" (or n). a Two blacks eennected by a rape bf negligible mass are being
dragged by a horizontal farce F. Suppose that F = 100.0 N, m1 = 10.0 kg, mr2 = 20.0 kg, and the caefficient bf kinetic
frictien between each black and the surface is 0.200.  (a) Draw a freebnd'yr diagram fur each black.
— (b) Determine the tensien T and the magnitude ef the aceeleratinn of the system. _. m;
{4111 IN”
Edﬂmk: WW Aaif'mihn 9‘15”“1 :0 .EL
“I '" 01.3 6; FL» F
“: mu. ““3 1 ft. T E
in FF VF.“
15 1
J;
CD T “Wits—.min. ‘— «Q03: ﬂﬁlJ
mini
WM
(9 FIT— lama"; W‘s“
iea— Sq 1.
a: (3.. Askew): E T ‘3‘: \Jl'ﬁ's
[HIPm:
L)
T: (’Q.ﬁ+I‘JHt
:L qu.ﬁ+in)m
: 1.1m; Q Box A, of mass 10.0 kg, rests on a surface inclined at 37° to the
horizontal. It is connected by a lightweight cord, which passes over a massless and frictionless puliey, to a second box B, which hangs
freelyr as shown. If the coefﬁcient of kinetic friction is 0.30, and m3
= 10.0 kg, determine the acceleration of the system. Flawmates FT— MHJSing  ﬁm“5wﬁ=ﬂ“h Masma me «wastes: we)... A 3.0kg hiock sits on top of a 5.0kg block which is on a horizontal
surface. The 5.0kg block is pulled to the right with a force F=lﬂﬂN
as shown in the figure . The coefﬁcient of static friction between all
surfaces is 0.60 and the kinetic coefﬁcient is 0.50. What is the
acceleration of each block? a as: IJ~= ““5 “Pid “I T E .'. Jim
‘..'lI Uniform Circular Motion—Kinematics Uniform circular motion: motion in a circle of constant radius at
constant speed Instantaneous yelocity is always tangent to the circle. Looking at the change in velocity
in the limit that the time interyal becomes inﬁnitesimally small, we
see that I‘ll. .1!" _'_—_'_'—u 1": RB . r‘ This acceleration is ca _=_ __:. centripetal, or radial,
:: _‘ 1 _ k“; “93V. and it points toward th center of
’5‘" "" ...————L" ~ _...— the circle.
N: i“
0x EV HG if ArrW4 The rater of an ultracentrifuge rotates at 60,000 rpm (reunlutiene per minute
cr 5280 reefs). A particle at the tap cf a test tube is 5.00 cm from the ratatien axis. Calculate its centripetal acceleratien, in “g’s.” Dynamics of Uniform Circular Motion For an object to be in uniform circular motion, there must be a net
force acting on it. We already know the acceleration, so
can in'Imetliatei'l»F write the force: EFR = milk = I?!" We can see that the force must
be inward by thinking about a ball on a string. Stringe only:r pull;
they never push. There is no centrifugal force painting outward; what happens is that the
natural tendency cf the object to mauve in a straight line must he
evercerne. If the centripetal farce vanishes, the abject ﬂies off at a tangent to the
circle. .i" DUESNT
HAPPEN O Highwayr curves are marked with a suggested speed. If this speed is based
an what wculd be safe in wet weather, estimate the radius of curvature fer a curve marked 45 milesfh {er 2i} mfs}. The caefﬁcient of static fricticn of
rubber en wet cencrete is {3.3, the ceeFficient cf kinetic frictien ef rubber an
wet cancrete is 0.5. r GA:
(F;Mﬁ>
L
“11““ Asmﬁ: P1 ”F
P“ 1
:2 mi“; .L .. L21:
Fm .1145 ﬂﬂ‘lt)
”31:44.55 ’H‘Elm
i.
4M1; ﬁg”
3. “”59”“3Jtsu'. a
3 )(L ngmu +b41Fi:i55"=i"'i.53:t
L ‘ "
“its“
Hwrr‘a‘os)‘ I A reller caaster car has a mass ef E,
see kg when fully,i leaded with
passengers.  a? If the vehicle has a speed af
.0 mfs at aint A, what is the
fence exerte b the track an the
car at this pain ? ... {b} What is the maximum speed_
the vehicle can have at B and 5t!“
remain an the track? at: V1 2.
i it 1': Hams...)
n: U1
"(37)
“5 T" 5"" 4*2'18512)
up
: qgﬂaaIN
mu: J. ‘9 in: {CmL MAI {pr‘4
I} an" A
I
P31} J ”I r
_ u‘ ——
11L .______ __
f V F3
V: iﬁf—w An amusement park ride censists cf a retating
circular piatferm 8.00 m in diameter from
which 10.0kg seats are suspended at the end
of 2.50m massless chains. When the system
retates, the chains make an angle 6' = 30.ﬂ°
with the vertical. (a) iWhat is the speed of each
seat? (In) Draw a freehady diagram of a 40.0—
kg child riding in a seat and ﬁnd the tension in
the chain. hauls) ‘— KM*h£}a 'CiﬁﬂsTﬁlqﬁﬂsjﬂ' t3) W1 [m5 .mgsihhmg)ﬁggnﬁ A model airplane of mass 0.?50 kg flies in a horizontal circle at the
end of a 60.ﬂm control wire with a speed of 35.0 me. Compute the
tension in the wire assuming that it makes a constant angle of 20.0“
with the horizon al. The forces exerted on the air ne are the pull of
the control wire, e gravitational force, and odynamic lift, which
acts at 20.0“ inwa from the 1vertical as s wn in Figure. Motion with Resistive Forces I Motion can be through a medium
— Either a liquid or a gas
I The medium exerts a resistive force, FD, which depends on the medium and is
opposite to the direction of motion. I FD nearly always increa s with increasing
Speed ﬂ FD = @v”
. n=1; Good approximation for slew motions or small
objects  11:2; Good approximation for large objects FD Proportional To v, I Analyzing the
motion results in “T = 31'” = "3 I b Ham. 1' is; called time Eﬂﬂﬁ'i’ﬂﬂf FD Proportional To v o=ﬂ Initially, v = o and dvfdt = g H ° 1
As 1‘ increases, R increases
and a decreases 0 * I;
The acceleration approaches
0 when FD 9 mg
At this point, v approaches 0 ‘ i
the terminal speed of the
object 0 ‘ ii 1! FD=4n o ‘ ' A small, spherical bead ef mass 4.00 g is released from rest at I:
[J in a battle ef liquid. The terminal speed is ebserved te be vT =
2.00 emfs. Find — (a) the value of the eenstant b, — (b) the time at whieh the bead reaehes 0.632 vT
— (e) the value nf the resistive fnree when the bead reaches terminal speed. 2F=mg— 2m 1 E
ﬂ=g_[DpA FD Proportional To v2  For objects moving at
high speeds through air,
the resistive force is
approximately equal to
the square of the spe FD = to ﬂow?
.0 is a dimensi
empirical
Iled t
' 'ent p is t sity of air
A i the eros seotional
area of the o ‘ect u is the speed of t
object antity that
drag £39sz = mo rhearind
Him —mhm Haastrod Drag commits FD Prop ional To 1:52, Termi s Speed 1 The terminal speed ' occur 1when the FD
acceleratinn goes to zern  Solving the equatipn gi‘u’ES 'I'tI'II'IiIInE HIMHI fur l’nrinum {H'Ilil'l‘IJ«i MilEing 'I'hl'uugl'l Air ﬂhject. Hm {kg} Malian1m {m3} mum;
5ij diver 1'5 me an
Baseball [radius 3.? cm] 0.115 4.? K 1133 43
Gnifhall {radius 2.1 cm} (1.1145 1.4 3:: III—5 44
Haitmne {radius 0.5:} cm} 4.3 an: 111'" 19 x 10*5 I4 Raindrnp {radius ﬂit} cm} 3.4 F: “1‘5 1.3 H. 11115 9J1 ...
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 Spring '11
 GUERRA
 Physics, Resistance, Circular Motion, Force, Friction

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