Chapter(4

Dynamics:*Newton�s*Laws*of*Motion

Example: A"person"jumps"from"the"roof"of"a"house"3.6"m"high."When"she"strikes"the"ground"below,"she"bends"her"knees"so"that"her"torso"decelerates"over"an"approximate"distance"of"0.70"m."If"the"mass"of"her"torso"(excluding"legs)"is"44"kg,"a)"find"her"velocity"just"before"her"feet"strike"the"ground,"and"b)"find"the"average"force"exerted"on"her"torso"by"her"legs"during"deceleration.

The$Normal$Force

3.6"m

}"0.70"m

a)"""""""""""""""b)

m"="44"kg

The$Normal$Force

3.6$m

}$0.70$m

a)$$$$$$$$$$$$$$$b)

m$=$44$kg

a)#Find#vy

vy ?

Assume +y is upv0 y = 0, y = −3.6 m, ay = −g

vy2 = v0 y

2 − 2gy

vy = − −2gy = − −2(9.80)(−3.6) = −8.4 m/s

x

y

The$Normal$Force

3.6$m

}$0.70$m

a)$$$$$$$$$$$$$$$b)

m$=$44$kg

b)#Find#Fave

Fave ?

Fy∑ = FN −W =may ⇒ FN =m(g+ ay )

Find ay assuming constant accelerationv0 y = −8.4 m/s from a), y = −0.70 m, vy = 0

vy2 = v0 y

2 + 2ayy ⇒ ay =−v0 y

2

2y=−(−8.4)2

2(−0.70)= 50 m/s2

∴Fave = FN =m(g+ ay ) = 44(9.8+ 50) = 2600 N = 580 lb

FN

W

aym

+y

mg

x

y

Types&of&Forces:&An&Overview

Examples)of)Nonfundamental Forces ,,All)of)these)are)derived)from)the)electroweak)force:

normal)or)support)forces

friction

tension)in)a)rope

The$Tension$Force

Cables'and'ropes'transmit'forces'through'tension.

A'force'T is'being'applied'tothe'right'end'of'a'rope.

For'a'massless'rope,'all of'the'force'is'transmitted'to'the'box'from'the'left'end'of'the'rope.

The'box'exerts'an'equal'andopposite'force'to'the'left'endof'the'rope'via'Newton�s'3rd law.

For a massless rope, F∑ =m

a = 0 ⇒ tension force the same on both ends

Example: Two$masses$are$connected$together$and$being$pulled$to$the$right.$

m2m1

!a

!T1!

T2?

If$the$force$applied$to$the$right$side$of$m1 isfind$the$force$applied$to$the$right$side$of$m2,$i.e.$find$$$$$in$terms$of$m1,$m2 and$$$

!T1

!T2

The$Tension$Force

!T1

The$Tension$Force

A"massless"rope"will"transmittension"undiminished"from"oneend"to"the"other.

If"the"rope"passes"around"amassless,"frictionless"pulley,"thetension"will"be"transmitted"tothe"other"end"of"the"ropeundiminished.

Example. A"supertanker"of"mass"m"="1.50"x"108 kg"is"being"towed"bytwo"tugboats."The"tensions"in"the"towing"cables"apply"the"forces"T1and"T2 at"equal"angles"of"30.0o with"respect"to"the"tanker�s"axis."Inaddition,"the"tanker�s"engines"produce"a"forward"drive"force"D,"whosemagnitude"is"D ="75.0"x"103 N."Moreover,"the"water"applies"an"opposingforce"R,"whose"magnitude"is"R ="40.0"x"103 N."The"tanker"movesforward"with"an"acceleration"that"points"along"the"tanker�s"axis"andhas"a"magnitude"of"2.00"x"10J3 m/s2."Find"the"magnitudes"of"the"tensions"T1 and"T2.

Application*of*Newton�s*Laws*of*Motion

Application*of*Newton�s*Laws*of*Motion

m ="1.50"x"108 kg"""""R ="40.0"x"103 N""""""D ="75.0"x"103 N

ax ="2.00"x"1003 m/s2 ay ="0

Application*of*Newton�s*Laws*of*Motion

Force x"component y component

1T

2T

D

R

0.30cos1T+

0.30cos2T+

0

0

D+

R−

0.30sin1T+

0.30sin2T−

Fy =∑ +T1 sin30.0 −T2 sin30.0 =may = 0 ⇒ T1 = T2 = T

Fx =∑ +T1 cos30.0 +T2 cos30.0+D− R =max

∴T = max + R−D2cos30.0!

=1.50×108( ) 2.00×10−3( )+ 40.0×103 − 75.0×103

2 0.866( ) =1.53×105 N =17.2 Tons (US)

The$Tension$Force

Example: A"window"washer"standing"in"a"bucket"pulls"herself"up"with"a"rope"over"a"pulley"as"shown."The"bucket+window"washer"have"acombined"mass"of"65"kg."Starting"from"rest"at"the"bottom,""how"much"constant"force"must"she"exert"on"the"rope"to"reach"a"window"160"m"above"her"in"oneChalf"of"a"minute?

160"m

+y

ay

T

W

m="65"kg

!

Fy = 2T −W =may∑

T

x

y

Fy = 2T −W =may∑ ⇒ T =m ay + g( )

2

The$Tension$Force

Find%ay using%constant%acceleration%equation:

y = v0 yt + 12 ayt

2

y =160 m, v0 y = 0, t = 30 s

y = 0+ 12 ayt

2 ⇒ ay =2yt2 =

2(160)302 = 0.36 m/s2

∴T =m(ay + g)

2=

65(0.36+ 9.80)2

= 330 N

≈ 74 lbs

mg