233012125-Dyna-Project

EXAMPLES/NOTES: UNIFORM RECTILINEAR MOTION 1. Car A at a gasoline station stays there for 10 minutes after Car B passes at a constant speed of 40 miles per hour. How long will it take Car A to overtake Car B if it accelerates at 4 m/s !olution"  #a  #a $ #% !a $ !10 & !%

Consider B"

v=

s t 

s Sb Tb= = v 17.88 m / s Sb =17.88 Tb

Consider A" 1

Sa =VoTa + aT  ² ² =0 + 2

1 2

( ) 4

 m s

2

( T a ) =2 Ta ² 2

'or !10"

(

S 10= vt =

17.88 m

!u%stitute"

Sa = S 10 + Sb

s

)(

600 s

) =10728 m

= 10728+ 17.88 T 

2 T  ²

Answer: T = 77.85 seconds (. A stone is thrown up from the ground with a velocity of )00ft/s. How long must one wait %efore dropping a second stone from the top of *00ft tower if the two stones are to pass each other (00ft from the top of the tower+

!olution" Consider 1 1

s =Vot − > ² 2

(

=

400 ft 

300 ft 

s

) ( (t )− 1 2

32.2 ft 

s

2

)(

t  ) 2

t = 17.19 s , 1.45 s Consider ( 1

s =Vot + g t 2 2

=0+

200

(

)

1 32.2 ft  2

s

2

( t  ) 2

t =± 3.52 s

How long must one wait+ t $ 1,.1- seconds  ).( seconds

Answer: !."7 seconds

= 10728+ 17.88 T 

2 T  ²

Answer: T = 77.85 seconds (. A stone is thrown up from the ground with a velocity of )00ft/s. How long must one wait %efore dropping a second stone from the top of *00ft tower if the two stones are to pass each other (00ft from the top of the tower+

!olution" Consider 1 1

s =Vot − > ² 2

(

=

400 ft 

300 ft 

s

) ( (t )− 1 2

32.2 ft 

s

2

)(

t  ) 2

t = 17.19 s , 1.45 s Consider ( 1

s =Vot + g t 2 2

=0+

200

(

)

1 32.2 ft  2

s

2

( t  ) 2

t =± 3.52 s

How long must one wait+ t $ 1,.1- seconds  ).( seconds

Answer: !."7 seconds

#ARIA$LE ACCELERATION 4

1. #he motion of a particle is given %y

t 3

s =2 t  − + 2 t 2  where s  is in 6

t   in   seconds . Compute the values of

v  and

feet   and

a  when t =2 seconds .

t 3

4

s =2 t  − + 2 t 2 6

1

v =8 t 3 − t 2 + 4 t  2

2

a =24 t  −t + 4

@ t =2 seconds

ft  1 v =8 ( 2 )3− ( 2 )2+ 4 ( 2 )   v =70 sec 2 a =24 ( 2 )2−( 2 ) + 4  %

a =98

ft  sec

2

(. A particle moves in a straight line according to the law in

s =t 3− 40 t   where s  is

m  and t   in seconds . a. hen

t = 5 seconds 2 compute

v .

%. 'ind the the averag average e velocity velocity durin during g the ) rd to 4th seconds. c. hen the the particle particle comes comes to stop2 stop2 what what is its accel accelerat eration+ ion+

s =t 3− 40 t 

a. 2

v =3 t  − 40 a =6 t 

@ t =5 seconds

2

v =3 ( 5 ) −40  

v ave =

%.

v =35

m sec

total total distanc distancee −96−(−93 ) −3 = =   4 −3 1 totaltime

rd

@ 3 second 3

s =( 3 ) −40 ( 3 )  

s =−93 m

@ 4th second 3

s =( 4 )

if v =0

c. 0

− 40 ( 4 )   s =−96 m

2

=3 t  − 40

t =3.65 seconds

v ave=−3

m sec

a =6 t =6 ( 3.65 )  %

a =21.9

m sec 2

). #he rectilinear motion of a given particle is given %y

m  and

v  is in

v −t  2 and

2

m sec . hen

t =0 2

s =0 2 and

a −t   relations.

s = v 2− 9 ds  dv  =2 v dt  dt  v =2 va a=

1 2

a=

1 2

=

v− vo t  v −(3 )   t  1

s = v o t + a t  =( 3 ) t + 2

2

1

v = t + 3 2

()

1 1 2 2

1

2

t 

 

2

s =3 t + t  4

s = v 2−9  where s  is in v =3

m sec . 3etermine

s −t 

MOTION CUR#ES 1. A particle starting with an initial velocity of *0 ft /s has a rectilinear motion with the constant deceleration of 10 ft/ s t $ - sec.

Solution:

2

 . 3etermine the velocity and displacement at

For &nd :

(. An auto travelled 100 ft in 40sec. #he auto accelerates uniformly and decelerates uniformly at * ft/ s in fps.

2

2 starting from rest at A and coming to stop at B. 'ind the ma5imum speed

Solution:

A *5  *740  56 $ 0 *5  0 $ *740  56 *5 $ (40  *5 ) = '( 6

# '(

). An Auto starts from rest and reaches a speed of *0 ft/s in 1 sec. #he acceleration increases uniformly from 8ero for the 9rst - sec after which the acceleration reduces uniformly to 8ero in the ne5t * sec. Compute for the displacement in this 1 sec interval.

Solution:

PRO*ECTILE MOTION 1. A golf %all is 9red from the top of a cli: 0 m high with a velocity of 10 m/s directed at 4; to the hori8ontal. 'ind the range of the pro
1

?0 $ 10sin4;t ?

range t 

 x t   $

=o cos> $

2

t1 $ ).-- s 7checked6 t( $ ?(. s

10cos4; $

range t 

'or range"

10cos4; $ 'or t"

range

1

y $ =o sin

ө

t?

2

range t 

(

 gt

10cos4; $

3.99

R&n+e = '8.' ,

 7 -.16 t (

(. @n 9gure -?*.102 a %all thrown down the incline strikes it at a distance s $ (4. ft. @f the %all rises to a ma5imum height h $ *4.4 ft a%ove the point of release. Compute its initial velocity and inclination >.

Hori8ontal motion"

√ 10

=

√ 10

 x t 

=o cos> $

3

1

=

y 254.5

 y =80.48 ft .

x

'or ma5imum height"

254.5

H $ *4.4 ft.

 x =241.44 ft .

2

 y = =ertical otion" 1

y $ =o sin

ө

t?

2

2

V  −( V o sin ө )

−2 ( 32.2 ) 2

(

 gt

−( V o sinө ) 64.4 = −2 ( 32.2)

2

( V o sin ө )  $ 7(67)(.(67*4.46

241.44

?0.4 $ =o sin

V o sin ө =√ 64.4 2

241.44

2

?0.4 $ (41.44 tan

 64.4 sin ө

(16.1) 241.44 64.4

'or

7 V o cos ө 6 ?

7)(.(67 V o 2 cos 2 ө 6

V o sin ө =64 .4

V o=

ө

ө:

t?

2

 gt(

ө

?0.4 $ (41.44 tan

2

64.4 tan

ө

=1.33 ;

tan

ө

=−0.27 ;

ө

2

2

ө

=¿  ).0*; ө

=¿  ?1.11;

therefore2

V o=

 64.4 sin ө

=

  64.4 sin 53.06⁰

#o = 8(.57 -/

). 'ind the take?o: velocity that is
 ?

2

cos

(16.1) 241.44

tan

ө

2

1

 ?0.4 $ =o sin

 ?

2

sin ө

ө

ө

1 2

sing hori8ontal motion formula"

 x t 

=o cos> $

17.32

=o cos)0; $

t 

17.32

t$

V o cos 30⁰

sing vertical motion formula" 1

y $ =o sin

ө

?((.($ =o sin

?((.($ =o sin

t?

30⁰

30⁰

2

 gt(

1

t?

2

 7)(.(6t(

17.32

7 V o cos 30⁰ 6 ?

1 2

17.32

 7)(.(67 V o cos 30⁰ 6(

#o = . -/s

4. How high is the hill+

sing hori8ontal motion formula" =o cos> $

 x t  500

100cos*0; $

t 

500

t$

100 cos 60⁰

t $ 10 sec sing vertical motion formula"

1

y $ =o sin

y$ 100sin

ө

t?

60⁰

0 = 17!.27 -.

2

 gt(

1

7106?

2

 7)(.(67106 (

3INETICS 1. 3etermine  that will give the %ody an acceleration of * ft/sec ( D $ 0.(0.

 Σ Fx =ma  R= ma = x− F 

∗

322 6 3.22

= x − F 

∗

322 6

716

3.22

4

= ∗ − 0.2∗ !  5

 Σ Fy =0 " = ! + y

7(6

3

= ! + ∗ 

322

5

!u%stituing ( in 1 we get"

P = 7''.7 4s.

(. 3etermine the acceleration of the system and tension in the chord. D $ 0.)0

Consider (00 E Block 'B3"

 Σ Fv =ma  R= ma = " −T  200

g

∗a= 200−T 

Consider 100 E Block 'B3"

 Σ Fx =ma

 R= ma=T − F  100

g

∗a=T − 0.3∗100

& = 5.5" ,/sec ' T = 8"."7 N ). 'ind the acceleration of the system and tension in the %lock.

Consider )00 E Block 'B3"

 Σ Fv =ma

 R= ma=T −"  300

g

Consider 100 and (00 E Block 'B3"

∗a=T − 300

 Σ Fx =ma  R= ma ="x 1 + "x 2− F 1 − F 2−T 

300

g

∗

3

3

0.2 4

5

5

5

∗a= ∗100 + ∗200 −

∗100 −

∗

0.3 4 5

∗200 −T 

T = '(8 N & = 1! ,/sec '

.

1. hat is the tension in the card+ (. Acceleration of the %locks ). =elocity of B after ( seconds+

B 1-*.( E A -1 E Consider B

Consider A

 #

#

#

1-*.( a

a

 -1 E

 R= ma1= T − "  196.2 9

'or

 R= ma2 =981−2 T  981

a1=T −196.2

a1

9

" 1

1

1 2

2

2

= a t  1

1

!olve 1 and ( simultaneously"

=

2

a1 a2

a1= 2 a 2

T6ere-ore:

2

s =V o t + a2 t 

2

5

a2 "

'or

s =V o t + a1 t 2

a2= 981−2 T 

1

2

= a t 

2.5

2

2

(

a1

T = !'7N

m  = ".5

s

a2= 3.27

2

m 2

s

5

10

&A

&C

&$

3etermine0 theEtension 100 and acceleration of each %locks. Consider %lock A"

#1 aA

 R= ma # R=T 1−" 

10 E

150 9

Consider %lock B" #(

a # =T 1−150

1

 R= ma $  R=" −T 2 100

aB

9

a $=100 −T 1

100 E

Consider %lock C" #(

 R= ma% 

aC

 R=T 2−"  50

0E

9

a% =T 2−50

'or #1 and #("

#1

−T  =ma

2 T 2

1

=T 

2 T 2

1

)

(

A" 150 9

!A

a # =2 T 2−150

B"

10E 100 9

( a # +a $ )=100 −T 

2

Friginal C" 50

100E

0E

9

( a # + a $ & )=12 −50

!A !B !BG

'or !A"

'or !BG" 1

S # =V o t + a # t 2 2

1

S $ & = a $ &  2

'or !B"

t 2

1

S $ = a $ t 2 2

1

S # = a # t 2 2

But

S $ =S $

G

1

+1

2

2

S $ = a $ & t 2

1 2

+ S #

1

2

¿ t  (a # + a $ & )

t 2

a$

a # t 2

2

a $=( a # + a$ & )

Iuation 100 9.81

(

will now %e"

( a # +a $ & )=100−T 

2

1 150 9.81

a # =2 T 2−150

#( $ 1(0

aA $ .)*

aBG $ ?,.4

) 50 9.81

( a # −a $ & ) =T  −50 2

m sec2 m sec2

50 9.81

( a $ & − a # ) =T  −50

T = 7( N

2

&A = 1(.58

m sec2

&$ = !."

m sec2

ASSINMENT NO.  -?).* How fast must an automo%ile of the previous pro%lem move in the last  minutes to o%tain an average speed of ) mph+ From the previous problem:

s = vt  s 1=( 30 m'h ) ( 12 min )= 6 miles

s 2=( 40 m'h) ( 20 min )=13.33 miles s 3=( 30 m'h ) ( 8 min )=6.67 miles s T =26 miles

v=

 

t =40 min

26 miles 40 min(

a=

60 sec

=39 m'h= v o

1 min

v − v o 35 m'h−39 m'h −9 m'h = = =−30 m'h2 t  40 min 40 min

-?). Fn a certain stretch of track2 trains run at *0 mph. How far %ack of a stopped train should a warning torpedo %e placed to signal an oncoming train+ Assume that the %rakes are applied at once and retard the train at the uniform rate of 4 fps (.

v o =60 m'h=88 f's v 2= v o2 + 2 as 0

=88 + 2 ( 4 ) s 2

s = 468 ft .

-?).10A ship %eing launched slides down the ways with a constant acceleration. !he takes 4 seconds to slide the 9rst foot. How long will she take to slide down the ways if their length is -00 feet+ 1

s = v o t + a t 2   2

v o =0

1

s = a t 2 2

1

2

= a( 4 )

1

900

1

2



a =0.125 f's2

= ( 0.125 ) t 

2

2

t =120 sec ¿ 2 min)tes -?).1(A stone is dropped down a well and  seconds later the sound of the splash is heard. @f the velocity of sound is 11(0 fps2 what is the depth of the well+ 1

s = v o t + a t 2 2

For the stone,

For the sound,

1

s =( 0 ) t + (9.81 ) t 2

s = v t 2

s = 4.905 t 2

s =( 341.376 ) t 2

2

t 12=

s

t 2 =

4.905

s 341.376

t 1 +t 2=5 sec

(

√

s

)  & 7 4.905

√

s 341.376

¿=5 sec

s =103.65 meters -?).14. A train moving with constant acceleration travels (4 ft during the 10 th sec of its motion and 1 ft during the 1( th sec of its motion. 'ind its initial velocity. !olution" S1oth $ (4 ft J t $ - sec to 10 sec S12th $ 1 ft J t  $ 11 sec to 1( sec J S1oth" 1

s = v o t + a t 2 2

1

2

24

=v o ( 9 )+ (9 )t 

24

=v o ( 10)+ a ( 10)



2

1 2

2



1

S9-10

24

¿ v o + a (19) 2

=v o + 9.5 a

7eI. 16

1

24

=9 v o + a (81 )

24

=10 v o + a ( 100)

2

1 2

J S12th: 1

s = v o t + a t 2 2

1

2

= v o (11)+ (11) t 

18



2

1

2

= v o (12 )+ a ( 12 )

18

2

1

=12 v o + a (144)

18



2

1

=11 v o + a (121 )

18

2

1

S11-12

¿ v o + a (23 ) 2

= v o + 11.5 a

18

7eI. (6

sing the eI. 1 K ("

v o =52.5 f's



a =−3 f' s2

-?).1*. An auto A is moving at (0 fps and accelerating at  fps ( to overtake an auto B which is )( ft ahead. @t auto B is moving at *0 fps and decelerating at ) fps(2 how soon will A pass B+ !olution" J Auto A"

J Auto B"

1

1

s = v o t + a t 2

s = v o t + a t 2

2

2

1

s =( 20 ) t + ( 5 ) t 2 2

1

7eI. (6

(6 !u%tract eI. ( from eI. 1" 384

2

=−40 t + 4 t 

2

s −384 =( 60 ) t + (−3 ) t  2

7eI.

t =16 sec

-?).1. sin

#he rectilinear motion of a particle is governed %y the eIuation s = r

* t  where r and * are constants. !how that the acceleration is a = - *

2

s.

!olution"

s =r sin *t 

)= r



v =sin *t 

d) =0 dv = cos *t 



vd) + )d v v =*r cos *t 

)= *r



v =cos *t 

d) =0 dv =−sin *t 



vd) + )dv 2

a =−* r sin *t 

!ince

s =r sin *t  2

 #herefore"

a =−* s

-?).(0 A ladder of length L moves with its ends in contact with a vertical wall and a hori8ontal Loor. @f a ladder starts from a vertical position and its lower end  A moves along the Loor with a constant velocity v   A2 show that the velocity of the upper end B is v B = – v   A tan  where  is the angle %etween the ladder and the wall. hat does the minus sign mean+ @s it physically possi%le for the upper end B to remain in contact with the wall throughout the entire motion+ 5plain.

!olution"

g= √  + −   2

2

dy 1 1 dy V b=  = (  (−2 , ) dt  2 -  dt 

But

  =V a t +

 #herefore"

hen M $ -0N2

dx  = V a dt 

V b= V b= .

− ,  - 

V a =−V a tan

2 which is impossi%le.

-?).(( #he velocity of a particle moving along the 5?a5is is de9ned %y v$k5O ?45 & *52 where v is in fps2 5 is in feet2 and k is a constant. @f 5 $ 12 compute the value of the acceleration when 5 $ ( feet.

So49on: At 5 $ ( feet v $ 7167(6O ? 47(6 & *7(6 $ 4fps

dv  dv  dv  dv a =  =3 x ²  − 8 x  + 6 dt  dt  dt  dt 

v=

dv dt 

a =3 x 2 v − 8 xv + 6 v !u%stituting v $ 4 fps.

a =( 3 ) (2 )2 ( 4 )−8 ( 2 ) ( 4 ) + 6 ( 4 ) Answer:

a =8 f's ²

ASSINMENT NO. ' 3etermine the acceleration of the ( %locks after touching each other. 3etermine the time at which the

300 ! 

 %lock will touch the

100 ! 

 %lock.

 R= ma100="  sin / −f 100="  sin / − 0 ! 1

"  a ="  sin / − 0 ! 1    ! 1=100 cos 30 1 g 100 30 1 100 cos

¿

¿ 30 1 −0.2 ¿ 100 9.81

 

a100=3.206

m sec

2

a100=100 sin ¿

1

2

s = v o t + a100 t  2

1

s = ( 3.206 ) t 2

2  716

2

 R= ma300="  sin / −f 300="  sin / − 0 ! 2 "  a ="  sin / − 0 ! 2    ! 2=300cos30 1 g 300 30 1

¿ ¿ 30 1 −0.1 ¿ 300 cos

300 9.81

 

a300= 4.055

a300=300 sin ¿

1

2

s + 1= v o t + a300 t  2

s=

1 2

( 4.055 ) t  −1 2

2  7(6

m sec2

Iuating 716 and 7(6 3.206 2

t 2=

2

0.4245 t 

4.055 2

t 2−1

=1

2

t  =2.356  

t =1 . 535 seconds

Acceleration after touch"

m at =a 100+ a300=3.206 + 4.055   at =7 . 261 sec 2 ASSINMENT NO.  1044. An elevator weighing )((0 l% starts from rest and acIuired an upward velocity of *00 ft per min in a distance of (0 ft. @f the acceleration is constant. hat is the tension in the elevator ca%le+ Piven"

T

 $ )((0 l% !olGn" v $ *00 ft/min $ 10 ft/sec

v 2= 2 as

<

( 10 ) =2 a ( 20 ) 2

s $ (0 ft

;

a =2.5

QeIGd" #

ft  sec2

"  a =T −"  g 3220 32.2

( 2.5 )=T −3220 T =3470 lb

104. A man weighing 1*1 l% is in an elevator moving upward with an acceleration of  ft per sec (. 7a6 hat pressure does he e5ert on the Loor of the elevator+ 7%6 hat will the pressure %e if the elevator is descending with the same acceleration+

Piven"!olGn" man $ 1*1 l%

161

a $  ft/sec

(

QeIGd"

"  a =T −"  g

7a6

32.2

( 8 )=T −161

T =201 lb

−"  7a6 ressure he e5ert

7%6

g

a=T −" 

−161 7%6 ressure if the elevator

32.2

( 8 )=T −161

T =121 lb

3escends with the same acceleration

104*. #he %lock in 'ig. ?104* reaches a velocity of 40 ft per sec in 100 ft2 starting from rest. Compute the coeRcient of kinetic friction %etween the %lock and the ground.

$ *0 l%

1*1 l% Piven"!olGn"

v 2= 2 as

v $ 40 ft/sec 2

s $ 100 ft

40

= 2 a (100 )

a =8

QeIGd" CoeRcient of kinetic

"  a = − 0!  g

'riction2 D

161 32.2

ft  sec2

( 8 )=60 − 0 ( 161 )

0=0.124 104,. 3etermine the force  that will give the %ody in 'ig. ?104, an acceleration of * ft per sec (. #he coeRcient of kinetic friction is 0.(0.



)(( l%

Piven"!olGn" a $ * ft/sec

(

"  a =  x − 0!  g 322

D $ 0.(

32.2

( 6 )= 4  − ( 0.2 ) !  5

4

=   + 60

0.2 ! 

QeIGd" force2

  y + ! −" =0 3

 ! =322 −   5

5

) 4

322 32.2

( 6 )= 4  − ( 0.2 ) 5

(

3 −  

322

5

)

 =135.22

10). Qeferring to 'ig. ?10(2 assume A weighs (00l% and B weighs 100l%. 3etermine the acceleration of the %odies if the coeRcient of kinetics friction is 0.10 %etween the ca%le and the 95ed drum.

B

A

'ig. ?10(

Piven"

"  # =200 lb

0=0.10

" $=100 lb

!olution"

−T  # =

200

T $−100 =

 200 32.2

 100 32.2

a

1

a

(

T  # = e 0/ T $ T  # T $

0.1 ( 3 )

=e

T  # =1.37 T $

)

!u%stitute ) to 1"

−1.37 T $ =

200

200 32.2

a

4

'rom ("

T $−100 =

 100 32.2

a

!u%stitute ( to 4" 200

−1.37 ( 100 + 3.11 a )=6.21 a

200

−1.37 − 4.261 a =6.21 a

a =6.02

ft  sec

2

10. @f the pulleys in 'ig. ?10 are weightless and frictionless2 9nd the acceleration of the %ody A.

(00

A

)00 B.'ig. ?10 'or A"

−T = ma

200

−T =

200

aA

T =200−

(00 l%

 200 32.2

 200 32.2

a #

a #

1

'or B" #

−300=

2 T 

#

aB

)00 l% Iuate 1 and ("

300

+

 300 32.2

(

a # 2

)=2 T 

300 32.2

(

a$

−

200

200 32.2

a # =5 . 85

300

a #=

+ 300 (

a #

32.2

2

)

2

ft  sec2

10,. #he coeRcient of kinetic friction under %lock  A in 'ig. ?10, is 0.)0 and under %lock B it is 0.(0. 'ind the acceleration of the system and the tension in each cord.

B (0 0l

 A 10 0l

)00l%

)0o

At C2 300

)00  #( $

32.2

a  ?????1

At B2 200 o

o

 #(  #1  (00sin)0   (00cos)0 70.(6 $  #(  #1  1)4.*4 $ *.(1a  ?????(

At A2

32.2

a

100 o

o

 #1  100sin)0   100cos)0  70.)6 $

32.2

a

 #1  ,.- $ ).11a  ?????)  #1 $ ,.- & ).11a

!u%stitute # 1 to (2  #(  7,.- & ).11a6  1)4.*4 $ *.(1a  #(  (10.*( $ -.)(a  #( $ (10.*( & -.)(a  ?????4

!u%stitute 4 to 1 300

)00  (10.*( & -.)(a $

32.2

a

-.) $ 1.*4a

a$ 4. !t"se#2



ans$

T = 75.28  !. >.8? = 2(.2

lb

T' = '(."'  2.!' >.8? = '55.!"



lb

ans. 

ans.

10-. Compute the acceleration of %ody B and the tension in the cord supporting %ody A in 'ig. ?10-.

)00l% ! h $ 0.(0

(00l%

 A )

4 @n %lock A2 200

(00  # $

32.2

aA

@n %lock B2 3

(# ?

5

4

 7)006 ?

300

 7)006 70.(06 $

5

300

(#  (( $

32.2

 aB

@n getting the acceleration for B2 !ince aA $ ( aB

200

( S (00  # $

 aA T

32.2

300

&

?(( & (# $

32.2

70.6 aA

400  (( $ 0aA

1,( $ 00 7(aB6

aB / )(.( $ 1,(/1100

a$

= 5.(! -/sec '

 &ns.



32.2

aB

10*1. Compute the time reIuired for the 100?l% %ody in 'ig. ?10*1 to move 10 ft starting from rest.

100 l% ! h$0.(0

0l% ) 4

'or 100?l% %lock2 3

 #1 ?

5

100

 71006 $

32.2

a1

'or 00?l% %lock2 4 5

3

 706 ?

5

800

 706 70.(06  # ( $

(#(  #1 $ ma (#(  #1 $ 0 (#( $ #1

!ince a( $ ( a1

a( $ (.( ft/sec( !olving for t 1 7100l%62

32.2

a(

1

!$

 a1t(

2

1

10 $

2

 7(.(6 t(

 = '.""! sec.



ans.

10*). 3etermine the acceleration of each weight in 'ig. ?10*)2 assuming the pulleys to %e weightless and frictionless.

 A

%

10 l% B

B 40 l%

'or A2 150

 #  10 $

32.2

aA

'or B2 480

(# ? 40 $

32.2

aB

)00 l%

'or %2 300

)00  # $

32.2

aC

1

!ince aB$

2

1

 aC ?

2

 aA

'or #ension2 150 32.2

aA $ # ?10

1

7406 7

2

1

aC ?

2

aA6 $ (# ? 40

300 32.2

aC $ )00  #

 # $ (1., l%

!olving for acceleration2 150

(1.,  10 $

32.2

 aA

 aA $ 14.,4,* ft/sec(

 aC

 aC $ .,(*( ft/sec(

 ans.



300

)00  (1., $

32.2

(7(1.,6  40 $ 40/)(.(a B

 ans.



&$ = 1'.85775 -/sec ' or &$ = '.85775 -/sec ' >downw&rd?

 &ns.



10*. 3etermine the ma5imum and minimum weights of the %ody C on @llustration ro%lem 104) that will keep C stationary. All other data remain unchanged. !olution"

1000  #

(B '$ 00 E$

(# B

1

aB $

2

 00

 

'

'$

E

4  ΣFx =  a g

'or A 1000

*00  1*0  # $

g

a#

'or B 800 1

(#  00 $

g

( a# ) 2

!olving for #  # $ 40, l% 'or up plane impending motion of C"

 ΣFx =0  # $ 40, $ * & 1*

; = 5!5 4 'or down plane impending motion of C

 ΣFx =0  # $ 40, $ * ? 1*

; = 2' 4

*

10*,. @n the system of connected %locks in 'ig 10*,2 the coeRcient of kinetic friction is 0.(0 under %odies B and C. determine the acceleration of each %ody and the tension in the cord. !olution"

1000l% C B ! h $ 0.(0

! h $

00l% 400l%

 A

4

) 4

3irection of motion" Assuming at rest  #$ 400 Fn B2 Eet force $ (# $ 00  40 U ' $ 1( 7B rises6 Fn B2 Eet force $ *00  7#$4006 U ' $ 1*0 7 C falls6 ith C at rest2 !aG

$ (!%

ith B at rest !a V

$ !c

Eet motion $ !aG ? !aW $ !a 3i:erentiating "

$ (!%

)

? !c

aA $ (aB  aC 1

aB $

2

(a# + a% )

4  ΣFx =  a g 'or  400

400  # $

g

a#

'or B 1000

(#  40  1( $

g

a% 

!olving

T = !8.'4 &A = .8 -@s ' &$ = !.57 -@s ' &C = '.2" -@s '

10*-. #wo %locks A and B each weighing -*.* l% and connected %y a rigid %ar of negligi%le weight move along the smooth surfaces shown in 'ig 10*-. #hey start from rest at the given position. 3etermine the acceleration of B at this instant. Hint"  #o relate aA to aB2 use the method developed in @llus ro% on (.

!olution" =$

ds , dt 

 +xVa + -Vb=0

a$

dv dt 

=a & 5aA & =% & Xa% $ 0 At start2 =a $ =% $ 0

− y aA $

 x

ab

or if A% is down & down aA $

 y  ab  $  x

8 6

ab

4  ΣFx =  a g 'or A * $

96.6

8

32.2

6

ab $ 4a%

'or B 96.6

-*.*   $

32.2

!olving

&$ = ." -@s '

 a% $ )a%

&A = 5.7 -@s '

10,1. #he pulleys in the preceding pro%lem have %een assumed to %e frictionless and weightless. hat changes would there %e in the solutions of these pro%lems if the pulleys 7a6 had friction 7%6 had apprecia%le weight+ !olution"

>&?w6 -rcon 6e ensons on 6e o@@ose sdes o- 6e @944e0 wo94d e 9neB9&4. >?;6 &@@ro@r&e we+6 6e s9@@orn+ enson wo94d no eB9&4 wce 6e o9sde ensons.

SEAT;OR3  A %all is dropped from the tower of 0 ft. high at the same instant that a second %all is thrown upward from the ground with an initial velocity of 40 ft/sec. hen and where do they pass2 and with what velocities+

!FY#@FE"

1

h =v 0 t + g t 2 2

h =40 t −

1 2

1

2

− h= ( 32.2 ) t 

80

2

( 32.2) t  (1)

2

2

h =80− 16.1 t  ( 2)

Iuate 7(6 and 716"

1 − 16.1 t  =40 t − ( 32.2 ) t  2

80

2

2

v f  =−24.4 2

ft  sec

t =2 sec. v f  =32.2 ( 2 ) 1

!u%stitute t to 716 and 7(6"

v f  =64.4 1

h =80−16.1 ( 2 )2=15.6 meters ¿ thebottom

ft  sec

& 

s =80 −15 . 6 =64 . 4 meters ¿ theto'

v =v f  + v f   1

v f  − v o =at  2

 2

v =64.4 −24.4

2

v f  = 40−32.2 ( 2 )

v =40

2

ft  sec

An automo%ile starting from rest speeds up to 40 ft/sec with a constant acceleration of 4 ft/sec ( run at this speed for a time and 9nally comes to rest with a deceleration of  ft/sec (. @f the total distance travelled is 1000 ft2 9nd the total time reIuired.

!FY#@FE"

v −v o =a t 1

1

s 1= ( 4 ) (10 )2=200 ft . 2

40

=4 t 

1

t 1 =10 sec.

v=

s2 t 2 s 1 + s 2+ s 3=1000

s 2= 40 t 2

200

+ s + 160=1000 2

s 2=640 ft .

= 40 t 

640

v −v o =−at 3

2

t 2 =16 sec.

−40=−5 t 

3

t 3 =8 sec. s 3=( 40 ) ( 8 )−

1 2

( 5) ( 8)

2

T t =t 1+ t 2 + t 3=10 + 16 + 8

s 3=160 ft . T t =34 sec.

 #he velocity of a particle moving along the 5?a5is is de9ned %y v=&' (-)' 2*+' where v  is in m/s and ' = meter and & $ 1. Compute the acceleration when '  $ (m.

!FY#@FE"

hen k $1 

v =( 1 ) x 3− 4 x 2+ 6 x v = x3 −4 x 2 + 6 x

ads = vdv a =v

dv ds

a =( x 3−4 x2 + 6 x ) ( 2 x 2−8 x + 6 )

hen 5 $ ( m

a =[ ( 2 )3−( 4 )2 + 6 ( 2)][ 2 ( 2 )2−8 ( 2 )+ 6 ] a =[ 2 ] [ 4 ]

a =8 m / s

a$

6 √ v

2

  when t $ ( sec v $ )* m/sec s $ )0 m. 3etermine s at t $ ) sec.

!FY#@FE"

dv dt 

a=

6 √ v

dv

√ v

=

dv dt 

=6 dt 

−1

v

dv =6 dt 

2

∫v

−1 2

dv =∫ 6 dt 

1

v2 1

= 6 t + c

2

1

2v

2

=6 t + c

hen v $ )* m/sec t $ ( sec 1

( ) =6 ( 2 ) + c

2 36

2

c =0  #herefore2 1

2v

Fr

2

=6 t 

v =18 t 2

v=

ds dt 

ds  =18 t 2 dt  ds =18 t 2 dt  ds =¿ ∫ 18 t 2 dt 

∫¿ 3

s=

18 t  3

+c

s =6 t 3 + c

 hen s $ )0 m t $ ( sec

=6 ( 2 ) + c

30

3

c =−18

 #herefore2

s =6 t 3−18

hen t $ ) sec

s =6 ( 3 )3 −18

s =144 meters