Assignment 2

SOLUTION SOLUTION A Annss.. v v_{22 == 77.9 ft77.9 ft>>ss} v v22_{22 == (78.093)(78.093)}22 _{++ 2(2(--3.22)(53.22)(5 -- 0)0)} v v22_{22 ==} vv22_{11 ++ 22aacc(s(s22 -- ss11))} a a = = --3.22 ft3.22 ft>>ss22 ; ;++

a

a

F F  _x x == mamaxx;; --8080 == 800 800 32.2 32.2aa v v_{11 = = == 78.093 ft78.093 ft>>ss} v v22_{11 == 00 ++ 2(21.561)(1002(21.561)(100}

2 

2 

_{22 -- 0))0))} v v22_{11 ==} vv22_{00 ++ 22aacc(s(s -- ss00))} a a == 21.561 ft21.561 ft>>ss22 + + b b 

a

a

F F  _x x == mama _x x;; 800 sin 45°800 sin 45° -- 3030 == 800800 32.2 32.2aa

The water-park ride consists of an 800-lb sled which slides

The water-park ride consists of an 800-lb sled which slides

from rest do

from rest down the inclinwn the incline and then into the pooe and then into the pool.l.If theIf the

frictional

frictional resistance resistance on on the the incline incline is is and and inin

fast the sled is traveling when

fast the sled is traveling whenss _{== 5 ft.5 ft.}

t thhee FF_{rr == 80 lb,80 lb,} ddeetteerrmmiinnee F F_{rr == 30 lb,30 lb,} 100 ft 100 ft 100 ft 100 ft  s  s how how distance distance short short a a for for pool pool

*13–112.

The pi_{lot of an a}i_{rplane executes a vert}i_cal

loop whi_ch i_{n part follows the path of a card}i_oi_d,

. If his speed at  A ( ) is a

constant , determi_{ne the vert}i_{cal force the}

seat belt must exert on hi_{m to hold h}i_{m to h}i_{s seat when}

the planei_{s ups}i_{de down at} A._{He we}i_{ghs 150 lb.}

vP = 80 ft

>

s u _{= 0°} r =600(1 + cosu_{) ft} SOLUTION

 a

150

_b

au = ru $ + 2r#u# _{= 0 + 0 = 0} ar = r $ - ru#2 _{= -600(0}.06667)2 - 1200(0.06667)2 = -8 ft

>

s2 0 = 0 + 0 + 2r2 uu$ u$ = ₀ 2vpvp = 2rr $ + 2

a

ru#

b a

_r#u _{+ r}u$

b

(80)2 = 0 +

 a

1200u#

b

2 u# _{= 0}.06667 v2p = r #₂ +

 a

ru#

b

2 r $ = -600 si_nuu$ _{- 600 cos}uu#2_{u= 0° = -600}u#2 r# = -600 si_nuu# u_{= 0°} = 0 r = 600(1 + cosu_{)|u=0° = 1200 ft}  A u u r  _{600 (1 + cos ) ft}

A freight elevator, including its load, has a mass of 500 kg. It is prevented from rotating by the track and wheels mounted along its sides.When , the motorM _{draws in the cable} with a speed of 6 m s,measured relative to the elevator._{If it} starts from rest, determine the constant acceleration of the elevator and the tension in the cable. Neglect the mass of  the pulleys, motor, and cables.

> t

=

2 s SOLUTION Ans. Ans. T

=

1320 N

=

1.32 kN

+ c ©

F_y

=

ma_y; 4T

-

500(9.81)

=

500(0.75) aE

=

0.75 m>s2

c

1.5

=

0

+

aE(2)

A

+ c

_B

v

=

v₀

+

act vE

=

-6 4

= -

1.5 m>s

=

1.5 m>s

 c

-

3vE

=

vE

+

6

A

+ T

_B

vP

=

vE

+

vP>E 3vE

= -

vP 3sE

+

sP

=

l M 

*13–44.

When the blocks are released, determine their acceleration and the tension of the cable. Neglect the mass of the pulley.

SOLUTION

 Free-Body Diagram:The free-body diagram of blocks AandBareshown in Figs.b

andc, respectively. Here, a

 AandaBare assumed to be directed downwards so that they are consistent with the positive sense of position coordinates s

 A and sB of  blocks AandB,Fig.a.Since the cable passes over thesmooth pulleys, the tension in

the cable remains constant throughout.

 Equations of Motion:By referring to Figs.bandc,

; (1)

and

; (2)

 Kinematics:We can express the length of the cable in terms of s

 Aand sBby referring to Fig.a.

Thesecond derivative of the above equationgives

(3) Solving Eqs. (1), (2), and (3) yields

Ans. aB

=

7.546 m>s2

=

7.55 m>s2

T

aA

= -

3.773 m>s2

=

3.77 m>s2

c

2aA

+

aB

=

0 2sA

+

sB

=

l T

-

30(9.81)

= -

30aB

+ c ©

Fy

=

may 2T

-

10(9.81)

= -

10aA

+ c ©

Fy

=

may  A B 10 kg 30 kg

The 0.8-Mg car travels over the hill having the shape of a parabola. If the driver maintains a constant speed of 9 m s, determine both the resultant normal force and the resultant frictional force that all the wheels of the car exert

on the road at the instant it reaches point A. Neglect the

size of the car.

>

y  A  x  y 20 (1 ) 80 m  x2 6400 SOLUTION

Geometry : _{Here, and . The slope angle at point}

 Ais given by

and the radius of curvature at point Ais

 Equ_at i _{ons of Mot} i _on : _{Here, . Applying Eq. 13–8 with and}

, we have Ans. Ans. N = 6729.67 N = 6.73 kN ©Fn = man; 800(9.81) cos 26.57° - N = 800

a

92 223.61

b

F _f = 3509.73 N = 3.51 kN ©Ft = mat; 800(9.81) sin 26.57° - F f = 800(0) r = 223.61 m u = 26.57° at = 0 r = [1 + (dy

>

dx) 2_]3

>

2 |d2y

>

dx2| = [1 + (-0.00625x)2_]3

>

2 |-0.00625|

2

 x=80 m = 223.61 m tanu = dy dx

2

 x=80 m = -0.00625(80) u = -26.57° u d2y dx2 = -0.00625 dy dx = -0.00625x

13–81.

A 5-Mg airplane is flying at a constant speed of 350 km h along a horizontal circular path. If the banking angle , determine the uplift forceL _{acting on the} airplane and the radiusr of the circular path. Neglect the

size of the airplane.

u

>

= 15°

r 

L

u

SOLUTION

 Free-Body Di a gram :The free-body diagram of the airplane is shown in Fig. (a). Here,a

nmust be directed towards the center of curvature (positivenaxis).

 Equat i ons of Mot i on :The speed of the airplane isv ₌

 a

₃₅₀km h

 b a

1000 m 1 km

 b a

1 h 3600 s

b

.Realizing that and referring to Fig. (a),

Ans. Ans. r _{= 3595.92 m = 3.60 km} ;+ ©Fn = man; 50780.30 sin 15° = 5000

¢

97.222 r

≤

L _{= 50780.30 N = 50.8 kN} + c ©F_{b = 0;} L_{cos 15° - 5000(9.81) = 0} a_{n =} v 2 r  = 97.222 r = 97.22 m

>

s

The boy of mass 40 kg i_{s sl}i_di_{ng down the sp}i_{ral sl}i_{de at a} constant speed such that hi_{s pos}i_ti_{on, measured from the} top of the chute, has components

and where t i_s i_{n seconds. Determ}i_{ne the} components of force and whi_{ch the sl}i_{de exerts on} hi_{m at the} i_nstant t ₌ 2 s._{Neglect the s}i_{ze of the boy.}

F_z F_u, F_r, z ₌ 1_-0.5t2 m, u _{= 10}.7t2 rad, r = 1.5 m, SOLUTION Ans. Ans. Ans. F _{z = 392 N} ©F _{z =} ma_z; F z - 40(9.81) = 0 ©F _{u =} ma_u; F u = 0 ©F _r = ma_r ; F r = 40(-0.735) = -29.4 N a_{z =} z$ ₌ 0 a_{u =} r u$ ₊ 2r #u# ₌ 0 a_r = r $ _- r ₍u#₎2 = _{0 - 1}.5(0.7)2 = -0.735 u$ _{= 0} z$ = 0 r # ₌ r$ ₌ 0 u# _{= 0}.7 z# _{= -}0.5 r ₌ 1.5 u = 0.7t  z = -0.5t  z z r  1.5 m u

13–95. SOLUTION Thus, Ans. Since

a

Fr = mar; -T = 2(-4) a_r = r$ - r#(u₎2 _{= 0 -  0.25(4.00)}2 _{= -4 m}

>

_s2 r# = -0.2 m

>

s, r$ = 0 u# _{= 4.00 rad}

>

_s (0.5)2(1) = C = (0.25)2u# r2u# = C d(r2u#) = 0

a

Fu = mau; 0 = m[ru $ + 2r#u#_{] = m}

c

1 r d dt (r2u # )

d

= 0 The ball has a mass of 2 kg and a negligible size. It is originally traveling around the horizontal circular path of radius such that the angular rate of rotation is If the attached cord ABC   _{is drawn down} throughthe hole ata constant speed of determinethe tension the cord exerts on the ball at the instant

Also, compute the angular velocity of the ball at this instant. Neglect the effects of friction between the ball and horizontal plane.  Hint: _{First show that the equation of motion in the} direction yields

When integrated, where the constantc_{is determined} from the problem data.

r2u# = c, au = ru $ + 2r#u# =

 1

1

>

r

21

d

1

r2u#

2>

dt

2

= 0. u r = 0.25 m. 0.2 m

>

s, u#_{0 = 1 rad}

>

_s. r0 = 0.5 m C  F r  r ₀ B  A 0.2 m/ s · u u₀