Final Tufree Mechanics 2.2558

8/17/2019 Final Tufree Mechanics 2.2558

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STATICS & DYNAMICS

!"#$%& FINAL


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!"!"#$ MECHANICS FINAL EXAM

1 MOTION OF PARTICLE VELOCITY AND ACCELERATION

Velocity

Acceleration

with Constant Acceleration

RECTANGULAR COORDINATE(x-y)

v =ds

dt =

!s

a =dv

dt = !v

ads =vdv

v = v 0 +at

s = s 0 +v

0 t +

1

2at

2

v 2= v

2

0 + 2a (s ! s 0 )

!r = x

+ y

!v =

!"r = v

x

+v y j

!a =

!""r = a

x

+ ay j

v x =

v y =

ax =

ay =

v = !

v =

a =

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Motion of Projectile!"#$%"&'()*+,-.// projectile

Horizontal :

Vertical :

ax = ay =

v x = v

x ( )

0

x = x 0 + v

x ( )0 t

v y = v y ( )0!gt

v 2

y = v

y ( )2

0! 2g (y ! y 0)

y = y 0 + v y ( )0 t ! 1

2gt 2


y

x

a g

(v0) y

(v0) x

v0

v x

v yv

r

y0

y

x0

x



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1. 012&343,()5+67+8&"9!6)&:6;, 869&(") )


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2. O"P,B+6"&'Q6R0SCD:6-GA A T6UI 869&V1K+6(W,0&3>&K+6O"P,-1+6GA A M/.,


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NORMAL AND TANGENTIAL COORDINATE (n-t)

.", tangential (t) ]+

.", normal (n) ]+

,

Note: 1. (^+_>`()*+,-(a,(W,>'6

2. (^+_>`()*+,-4b6J


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3. [final 2554] I:+"()*+,-8&" A c6 B>&1'&64b6 /,'%,&/ de[ 16 m J


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POLAR COORDINATE (r-#) Velocity

4A3-

Acceleration

4A3-

Note: 1. "&'()*+,-(a,


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4. [final 2556] +mn&) P ()*+,-HIA&1(W,f&6(o,UIp,9+3>&1UI 4A3- R=4+0.25t 2 m .:% ! =2 " t rad (^+ t ]+(


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5. s,HA'ET/(gt6u"++".//1&CDvA++"J


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RELATIVE MOTION (Translation Axes)

4A3-

4A3-

6. (z+4{f0+6|++"8&"L&C,(&1kef&6T6UI }& v A = 20 m/s .:% vB = 15 m/s 86~,


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7. C,Y&,€.•696 (z+'/ A :&A>'%(


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2 FORCE AND ACCELERATION FRICTION

EQUATION OF MOTION

F = µ

N

P

mg

Pm

! = m

P




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8. [final 2556] 01969O" 60 kg Ž+"Z+6 9 kg v,+‡C,t-[)":6;,869&.'6 R .:% L *

R = 846 N, L = 110.4 N ; R,L$0


Problem 3/4

9 kg

60 kg

g––4



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9. "Z+6!69O" 50 kg 86~,


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MOTION OF CONNECTED PARTICLES

CURVILINEAR MOTION

!F n = ma

n

!F t =ma

t

an =

at =

!F r =ma

r

!F " =ma

"

a r =

a ! =


an

O¿

P

at

a

O

r

ar

au

a

u



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10. .'6 P = 40 N "'%‘-"Z+6 B T6UI -(”1B,"Z+6 A .:% B 9A–6M/- 869&)


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9. [final 2/2554] 'E1+(>+‹H—˜B+6™/F&1š,›,/,E,,K$%œ&, Hš&$8'&8' 4A3‰9,ACDš,›,[de[)


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10. .L6 ABO 9V,C,.,


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3 WORK AND ENERGY WORK

Definition of Work

Work of a Constant Force

Work of Friction

Note :

dU = F !ds

U 1!2

= PLcos!

P

dr

x

y

α

L

1 2

U 1!2 = PL cos! !

µ k NL


Fc

F c cos u s2 s1

su



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WORK AND ENERGY EQUATION

Kinetic Energy

Potential Energy

Large Change in Altitude

No Force acting on the Particle

POWER

( )

EFFICIENCY

T 1+V

1+U

1!2 =T

2 +V

2

T =

1

2mv 2

V =V e +V g

V e =

1

2kx

2

V g =mgh

T 1 +V 1 =T 2 +V 2

P =dU

dt = F !v 1 W = 1 J/s

e =P output

P input


V e 0

Unstretchedposition, s 0

s

s

k

k

k

V e ks21

2

V e ks21

2

V g = 0

V g = mgh

mg

h

Earth

m e

r

m

R

V g = –mgR2—–—r

mgR2—–—r2

V g = !mgR

2

r



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11. I:+".9


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12. I:+"1&1.",.,


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13. [final 2/2554] µ"¶11


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4 IMPULSE AND MOMENTUM

4A3- (º3"Š&

Note :

!F = ma = m dv

dt

!F dt t 1

t 2

" = m dv

v 1

v 2

"

mv 1+ !F dt

t 1

t 2

" =mv 2

!F dt t 1

t 2

"

mv x 1+ !F dt

t 1

t 2

! = mv x 2

mv y 1 + !F dt t 1

t 2

! = mv y 2

!F dt t 1

t 2

"


ΣF dtt1

t2

G1 =mv1

G2 =mv2

+ =



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14. µ"(/0/+:1


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CONSERVATION OF MOMENTUM

Impact

!"#$%&'()*$+, (Coefficient of Impact)

)*$+,-.. 2 /0

!mv 1 = !mv

2

m Av A +m B v B =m A !v A +m B !v B

e =

!v B " !v

A

v A "v

B




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16. 'EHK,Q6Á,b& A 1


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17. 869&š1I'%ÁfÅ"&'2, e }&V19!62,(a,)Æ696K+6Ç+,"&'2, ! T6UI ‰9,A ! = 40° *

e = 0.434


Problem 3/253

v v′

θ θ

––2



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18. f'6":11


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Final Tufree Mechanics 2.2558

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