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5 Additional Applications of Newton’s Laws
• Friction
• Drag Forces
• Motion Along a Curved Path
• The Center of Mass
• Hk: 31, 43, 53, 57, 67, 81, 91, 101.
Friction
• Surface Force opposing relative motion
• Component of Contact Force (other component is the Normal Force)
• Characterized by coefficients (mu)
• Static (varies 0 to max)
• Kinetic (~ constant)
• Rolling (~ one tenth of kinetic)
Surface Dependence of Friction
Normal Force Dependence of Friction
• Contact area ~ to Normal Force
• Frictional Force ~ Normal Force
• Summary:
• Two factors affect frictional force;
• Surface composition
• Normal Force
• /
5
Equations for Friction
Nss Ff max,Friction Static
Nkk Ff Friction Kinetic
Nrr Ff Friction Rolling
6
Example Friction• Ex. 10kg block. FN = weight = mg = 98N.
Static coef. = 0.50; Kinetic coef. = 0.30.
N
Nf s49
)98)(50.0(max,
N
Nfk29
)98)(30.0(
Block at rest. Draw a Force Diagram for the block.
Three boxes are pushed by force F with v > 0 along a horizontal surface with k = 0.291.
F=26N
3kg5kg
2kgfk
NF
F
w
xx aF )10()98)(291.0(26
0mgFF Ny NmgFN 98
ssmax //25.010
)98)(291.0(26
Derive the Angle of Repose relation:
Maximum angle block remains at rest:
0sinmax mgfF sp
0cos mgFF Nn
sinmax mgFf Nss
cosmgFN
tancos
sin
sincos
s
s mgmg
Atwood with Friction. m1=1kg m2=2kg. Kinetic friction = 0.5.
CWCW ammfgmF )( 212
ssmmm
fgmaCW //9.4
21
8.9)5.0(6.19
21
2
Drag Forces
speedhigh at 2n and speed lowat 1n
:speedVary with
n
d bvF
throughmovesit fluid of type theand
object theof shape on the depends
when speed lat termina is
object fallingA :Speed Terminal Ex.
/1
b
b
mgv
mgbvn
T
n
Motion Along a Curved Path
• Force required turn and to change speed
• Coordinates usually used are F/B (tangential) and L/R (radial)
• Sum forces tangential = mass x tangential acceleration
• Sum forces L/R (radial centripetal) = mass x centripetal acceleration
• /
What is the fastest speed the car can go without sliding? Assume the car has m = 1200kg and s = 0.92.
r
vmmgfF ssc
2max
r
vgs
2
smgrv s /3.20)7.45)(8.9)(92.0(
A block loops the loop. Which force diagram is correct for when it passed point D?
Center of Mass Definition
21
2211
mm
xmxmxcm
21
2211
mm
vmvmvcm
21
2211
mm
amamacm
Center of Mass Acceleration
221121 amamFF
21
221121, )(
mm
amammmF extnet
cmextnet MaF ,
Center of Mass when Net External Force is Zero
• Zero Net Force implies center of mass acceleration is also zero. So if CM originally at rest, it remains at rest. If CM moving, its velocity remains same.
• Example: two people standing on ice push off one another
• /
Ex. Center of Mass. A 100kg person walks 6 feet forward in a 50kg canoe. How far did he move relative to the shore?
21
2211
21
2211
mm
xmxm
mm
xmxm
ftxx 621
22112211 xmxmxmxm zero beboth and Let 21 xx
02211 xmxm
ftxx 621 0)6( 2221 xmxm
06)( 1221 mxmm
ftmm
mx 4
50100
)100(66
21
12
ftftftx 2641
8Ch05ISM#13 toCompare
Summary
• Friction depends on Surface Composition and Normal Force
• Drag Force vary with speed
• Force required to move along curved path even at constant speed
• Center of Mass stays same when only internal forces operate
Can you stop in time?
Buggy rolls.You slide.
xnk
Mmx
aMmF
F
)(1
xnk aMmF )(1
Given m = 75kg, M = 20kg, D = 3.5m, vo =1.1m/s. What frictional coefficient is needed?
xavv xo 222
Insert values, determine ax.
2
22
/1728.0
)5.3(2)1.1(0
sma
a
x
x
xk aMmmg )(
mg
aMm xk
))((
02233.0
)75(
)1728.0)(2075(
k
k g
Diagramming Refresher:
Accelerating with F2WD.
Stopping with 4W Disc-Brakes
A 3kg box at rest on level surface with s = 0.55. What is the largest F acting 60° below horizontal for which the box remains at rest?
xy
060cos max sx fFF
04.2960sin FFF Ny
NF
F60
w
fs
NsFF 60cos
4.2960sin FFN
)4.2960sin(60cos FF s
17.16)4763.05.0( F
NF 683 (60° is close to maximum angle)
vpc
vpg
Relative vs. Absolute Velocity
Which force diagram applies to the object at Point B?
Which force diagram applies to the object at Point C?
Assume mass = 1.2kg and radius = 45cm.If speed at Point D is 3.6m/s, what is the size of the normal force acting at Point D?
r
vmmgFF Nc
2
45.0
6.32.1)8.9)(2.1(
2
NF
NFN 8.22
-cen
+cen
Q. Assume mass = 1.2kg and radius = 45cm.If speed at Point B is 5.1m/s, what is the size of the normal force acting at Point B?
r
vmmgFF Nc
2
45.0
1.52.1)8.9)(2.1(
2
NF
NFN 81
+cen
-cen
Net
Given: T = 50N, = 30°, r = 1mFind: mg and v.
NTTFc 255.030sin 030cos mgTFy
NTmg 3.4330cos5030cos
18.9
3.4325
22 v
r
vmN
smv /378.23.43
)8.9)(25(
The speed is now 6.5m/s and r = 1.0m. Angle, tension, mass?
r
vmTFc
2
sin mrL
Lr
03.1sin/
sin
0cos mgTFy
r
vmT
2
sin mgT cos
9.76
8.9
)5.6(tan
22
gr
v
Net
3.42cos// gmT
For example, if m = 1.0kg,
then T = 42.3N.
Q. The speed of a mass on a string of length L is 6.5m/s. The radius r = 2.0m. Find angle, tension, mass, and L.
1.65
6.19
)5.6(tan
22
gr
v
Net
9.23cos// gmT
For example, if m = 1.0kg,
then T = 23.9N.
mrL
Lr
20.2sin/
sin
Practice Q: What is F such that 0.5kg block stays at rest if all surfaces are frictionless?
Banked Turn:
Banked Turn:
Modified Atwood Machine with friction.
gmgmfF kCW 2121 sin
21
211 sincos
mm
gmgmgm
mass
Fa kCWCW
Let m1 = 2kg, m2 = 3kg, = 30°, sliding friction coeff. 0.44
)43.2(3)3(22 TgTgmFCW
NT 1.223.74.29
Objects are in CW motion.
cosmgFf kNkk
ssmggg
aCW //43.232
330sin)2(30cos)2)(44.0(
Q. Recalculate last problem with m1 = 6kg m2 = 1kg. (All else remaining the same)
)0.6(1)1(22 TgTgmFCW
NgT 8.15)0.6()1(
ssmggg
aCW //0.616
130sin)6(30cos)6)(44.0(
Practice Q: Find the variable relationships.
Figures
Q. Assume the car has m = 1200kg and s = 0.92.
How large is the frictional force if v = 15m/s?
r
vmfF sc
2
r
vmf s
2
Nf s 59087.45
151200
2
5-3Drag Forces
Drag Forces
• Can be approximated as,
• Fdrag = bvn
• where b and n are constants
Example: Air drag, n = 2.
If b = 25N/(m/s)2, at what speed would the object be resisted by 10N?
2bvFd smv /632.025/10 22510 v
At what speed would the same object be resisted by 30N?
22530 v smv /095.125/30
Drag force grows quickly with v:
Terminal Velocity:
Reached when drag force equals weight force
