Homework Chapter 12(solution) PDF

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PROBLEM 12.14 The two blocks shown are originally at rest. Neglecting the masses of the pulleys and the effect of friction in the pulleys and assuming that the coefficients of friction between block A and the horizontal surface are m s = 0.25 and mk = 0.20, determine (a) the acceleration of each block, (b) the tension in the cable.

SOLUTION From the diagram x A + 3 yB = constant Then

v A + 3v B = 0

and

aA + 3aB = 0 aA = -3aB

or

First determine if the blocks will move withaA = aB = 0. We have +

1 SFy = 0 : WB - 3T = 0 or T = mB g 3

B:

+

SFx = 0:

Then

FA = +

A:

A:

Also, B:

FA - T = 0

1 ¥ 25 kg ¥ 9.81 m/s2 = 81. 75 N 3

SFy = 0 : WA - N A = 0 or N A = mA g

(FA )max = (ms )A N A = (ms )A mA g = 0 .25 ¥ 30 kg ¥ 9.81 m/s 2 = 73. 575 N

FA ( FA ) max, which implies that the blocks will move.

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(1)

PROBLEM 12.14 (Continued) (a)

A:

+

FA = ( mk )A N A = 0.20 mA g +

Sliding:

+

B:

SF x = m Aa A: F A - T = m Aa A T = 0.20 mA g + 3mA aB

Using Eq. (1)

or

SF y = 0 : W A - N A = 0 or N A = m A g

SF y = m B aB : WB - 3T = mB aB

mB g - 3(0.20 mA g + 3mA aB ) = mB aB

or

Ê m ˆ g Á1 - 0. 6 A ˜ mB ¯ Ë aB = m 1+ 9 A mB

=

Ê 30 kg ˆ (9 .81 m/s 2 )Á 1 - 0 .6 Ë 25 kg ¯˜ kg 1 + 9 30 25 kg

= 0 .23278 m/s 2 aA = 0.698 m/s2 Æ b

Then

a B = 0.233 m/s2 Ø b

and (b)

We have

T = (30 kg)(0.20 ¥ 9.81 + 3 ¥ 0.23278) m/s2 T = 79.8 N b

or

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PROBLEM 12.38 Human centrifuges are often used to simulate different acceleration levels for pilots. When aerospace physiologists say that a pilot is pulling 9g’s, they mean that the resultant normal force on the pilot from the bottom of the seat is nine times their weight. Knowing that the centrifuge starts from rest and has a constant angular acceleration of 1.5 RPM per second until the pilot is pulling 9g’s and then continues with a constant angular velocity, determine (a) how long it will take for the pilot to reach 9g’s (b) the angle q of the normal force once the pilot reaches 9 g’s. Assume that the force parallel to the seat is zero.

SOLUTION Given:

a = 1.5 RPM/s = 0.157 rad/s2 w0 = 0 N = 9mg R =7 m

Free Body Diagram of Pilot:

Equations of Motion:

åF

y

= ma y

N sin q - mg = m ( 0) N sin q = mg (1)

åF

n

N cosq = mRw 2 w=

Substitute N = 9mg into Eqn. (1):

= ma n

N cos q mR

(2)

9mg sin q = mg æ 1ö q = sin -1 çç ÷÷÷ èç 9 ø q = 6.379 

Substitute N = 9mg and θ into (2): 9 * 9.81cos6.379 7 w = 3.540 rad/s w=

For constant angular acceleration: w = w0 + a t 3.540 = 0 + 0.157 * t (a) Solving for t:

t = 22.55 s ◀

From earlier:

q = 6.379  ◀

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PROBLEM 12.50 A 55-kg pilot flies a jet trainer in a half vertical loop of 1200-m radius so that the speed of the trainer decreases at a constant rate. Knowing that the plane has a speed of 550 km/h at point A, and the pilot experiences weightlessness at point C (i.e., the normal force from the seat bottom is zero), determine (a) the deceleration of the plane, (b) the force exerted on her by the seat of the trainer when the trainer is at point B.

SOLUTION vA = 550 km/h(1 hr/3600 s)(1000m/km)=152.78 m/s

(a) Analyze the jet at C FBD

KD

t

N F

n

mat

man

mg

 å Fn = man : N + mg = mvc 2 / r v c2 = gr vc =

(9.81 m/s2 )1200 N = 108.499 m/s2

Kinematics (constant at) vC 2 = v A 2 + 2 at s

s = rq = (1200 m)p = 3769.9 m So

at =

vC2 - v A2 (108.499)2 - (152.78) 2 = = -1.534 m/s 2 2s 2(3769.9) a t = - 1.534 m/s2 ◀

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PROBLEM 12.50 CONTINUED (b) At B

FBD

t

F

KD

mat

n N

man= mg

¬ å Fn = man : N = mvB2 / r

N=

55 kg(132.50 m/s2 )2 = 804.7 N 1200 m

 å Fy = mat : F - mg = mat F = 55(9.81) + 55(-1.534) F = 455.2 The total force is then,

F = 924.48 N æ455.2 ÷ö q = tan - 1 çç = 29.5  çè804.7 ÷÷÷ø

F = 924 N

29.5 ◀

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PR ROBLEM 12.71 Thee two blocks are released from rest when r 0.8 m and a   30 . Neg glecting the mass of the pulley and the effect of friiction in the pullley and between block A and the horizontal surfacee, determine (a) the initial tension in the cable, (b) the initial acceleration of block A, (c) ( the initial acceleration of blockB.

SOLUTION Let r and d  be polar coordinates of block A as shown, and let y B be the position coordinate (positive downward, origin at the pulley) fo or the rectilinea ar motion of block B. Constraint of cable:

r  yB  constant, r  vB  0,

For block A,

r  aB

(1)

Fx  m Aa A : T cos  m Aa A or T  m Aa A sec

(2)

For block B,

r  aB  0

or

Fy  mB aB : mB g  T  mB aB

Adding Eq. E (1) to Eq. (2) to eliminate T,

mB g  mA aA sec  mBBaBB

(3) (4)

Radial an nd transverse components of aA . Use eithe er the scalar product of vectors or the triangle construction shown, being careful to note the positive directions of the components

 r  r  2  a r  a A  er  a A cos 

(5)

Noting th hat initially   0, using Eq. (1) to eliminate r, and changing signs givves aB  aA cos

(6)

Substitutting Eq. (6) into Eq. (4) and solving for a A , aA 

mBg (25) (9.81)   5.448 m/s 2 mA sec   mB cos 20sec30   25cos30 

From Eq q. (6), a B  5.48cos30  4.75 m/s2 (a)

From Eq. (2), T  (20)(5.48) sec30  126.6

(b)

Ac cceleration of block A.

(c)

Ac cceleration of block B.

T  126.6 N 

8 m/s2 aA  5.48



a B  4.75 4 m/s 2 

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PROBLEM 12.71 The parasailing system shown uses a winch to let rope out at a constant rate so that the 70-kg rider moves away from the boat, which is travelling with a constant velocity. At the instant shown, the rope has a length of 30 m, it is increasing in length at a constant 1 m/s, the angle is increasing at a rate of 0.05 rad/s, and q is −0.01 rad/s2. Knowing that when the rope makes a 30 degree angle with respect to the water, the tension in the rope is 10 kN, determine the magnitude and direction of the force of the parasail on the parasailer.

SOLUTION Strategy: Newton’s 2nd law using polar coordinates r = 30 m, q=30 r = 1 m/s, q = 0.05 rad/s r = 0, q= - 0.01 rad/s2

Kinematics:

So,

ar = r - r q 2 = 0 - (30 m)(0.05 rad/s) 2 = -0.075 m/s 2 a = r q + 2 rq = (30 m)(- 0.01 rad/s 2) + 2(1 m/s)(0.05 rad/s) q

= - 0.3 + 0.1 = -0.2 m/s 2 Analyze the parasailer FBD

Fr

Fθ

r

T 30°

KD

θ

maθ

NA

mar

mg

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PROBLEM 12.71 (CONTINUED)  å Fr = ma r : Fr - T - mg sinq = ma r Fr = T + mgsin q + mar = 10, 000 N + (70 kg)(9.81 m/s 2) sin(30) + (70 kg)(- 0.075 rad/s2 ) = 10, 338 N  å Fq = maq : Fq - mg cos q = maq Fq = mgcos q + maq = (70 kg)(9.81 m/s 2) cos(30) + (70 kg)(- 0.2 rad/s 2) = 580.70 N

Resolve into horizontal and vertical components,

Fx = Fq sin 30 - Fr cos30 = 580.70sin(30) - 10.338cos(30) = - 8662.7 N Fy = Fq cos30 + Fr sin 30

= 580.70 cos(30) + 10,338sin(30) = 5671.95 N So,

F

r

30° F

30°

θ

F = Fx 2 + Fy 2 = 10,354 N æ 5671.95ö ÷÷ = 33.2 q = tan- 1çç èç 8662.7 ÷ø F = 10,350 N

33.2 ◀

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PROBLEM 12.88 To place a communications satellite into a geosynchronous orbit (see Problem 12.80) at an altitude of 35800 km above the surface of the earth, the satellite first is released from a space shuttle, which is in a circular orbit at an altitude of 300 km, and then is propelled by an upper-stage booster to its final altitude. As the satellite passes through A, the booster’s motor is fired to insert the satellite into an elliptic transfer orbit. The booster is again fired at B to insert the satellite into a geosynchronous orbit. Knowing that the second firing increases the speed of the satellite by 1440 m/s, determine (a) the speed of the satellite as it approaches B on the elliptic transfer orbit, (b) the increase in speed resulting from the first firing at A.

SOLUTION For earth,

R = 6370 km = 6.37 ¥ 106 m GM = gR 2 = (9 .81 m/s2 )(6 .37 ¥ 10 6 ) 2 = 3.9806 ¥ 10 14 m3 /s2 rA = 6370 + 300 = 6670 km = 6 .67 ¥ 10 6 m rB = 6370 + 35800 = 42 ,170 km = 42 .17 ¥ 106 m

Speed on circular orbit through A. (v A )circ = =

GM rA 3. 9806 ¥ 1014 6 .67 ¥10 6

= 7 .7252 ¥ 10 3 m/s Speed on circular orbit through B. ( vB ) circ = =

GM rB 3. 9806 ¥ 1014 42. 17¥ 106

= 3. 07236 ¥ 10 3 m/s

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PROBLEM 12.88 (Continued)

(a)

Speed on transfer trajectory at B. (v B )tr = 3.07236 ¥ 10 3 - 1440 = 1632.36 ¥ m/s

1632 m/s b

Conservation of angular momentum for transfer trajectory. rA (v A )tr = rB (v B )tr r (v ) (v A )tr = B B tr rA ( 42. 17 ¥ 106 )(1632. 36) ( 6. 6 7¥ 106 ) = 10, 320. 33m/s =

(b)

Change in speed at A. DvA = ( vA )tr - ( vA )circ = 10, 320. 33 - 7725. 2 = 2595.13 m/s Dv A = 2600 m/s b

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PROBLEM 12.102 A satellite describes an elliptic orbit about a planet of mass M. Denoting by r0 and r1, respectively, the minimum and maximum values of the distance r from the satellite to the center of the planet, derive the relation 1 1 2GM + = 2 r0 r1 h

whereh is the angular momentum per unit mass of the satellite.

SOLUTION Using Eq. (12.39),

GM 1 = 2 + C cos qA rA h

and

1 GM = 2 +C cos qB . rB h

But

q B = q A + 180 ,

so that

cos q A = - cos q B .

Adding,

1 1 1 1 2 GM + = + = 2 rA rB r0 r1 h

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PROBLEM 12.104 A satellite describes a circular orbit at an altitude of 19 110 km above the surface of the earth. Determine (a) the increase in speed required at point A for the satellite to achieve the escape velocity and enter a parabolic orbit, (b) the decrease in speed required at point A for the satellite to enter an elliptic orbit of minimum altitude 6370 km, (c) the eccentricity e of the elliptic orbit.

SOLUTION For earth, GM = gR2 =( 9.81)( 6.37 ´106 ) = 398.06 ´1012 m3 /s2 rA = 6370 + 19110 = 25480 km = 25.48´ 106 m

vcirc =

GM = rA

vesc =

398.06´ 1012 = 3.9525´ 103 m/s 6 25.48´ 10

2GM = 2 vcirc = 5.5897´103 m/s rA

(a) Increase in speed at A. Dv = vesc - vcirc =1.637 ´10 3 m/s

Dv = 1.637 ´10 3 m/s ◀

Elliptical orbit with rB = 6370 + 6370 =12740 km =12.74 ´10 6 m. Using Eq. (12.39), But

1 GM = 2 + C cos q A rA h

and

1 GM = 2 + Ccos q B. rB h

qB = qA + 180 , so that cosqA =- cosqB

Adding,

1 1 r + rB 2GM + = A = rA rB r ArB h2

(2 )(398.06 ´10 12 )(25.48 ´10 6 )(12.74 ´10 6 ) 2GMrArB = h= rA + rB 38.22 ´10 6 = 82.230 ´10 9 m 2 /s Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2285

PROBLEM 12.104 (Continued) vA =

h 82.230 ´10 9 = = 3.2272 ´10 3 m/s rA 25.48 ´10 6 D v = 725 m /s ◀

(b) Decrease in speed. Dv = v circ - vA = 725 m/s (c)

1 1 r -r - = A B = C cosqB - C cosA = 2C rB rA rArB C=

By Eq. (12.40),

r A - rB 12.74´ 10 6 = = 19.623 ´10 - 9 m - 1 6 6 2rArB (2) 25.48 ´10 12.74 ´10

(

(

)(

)

)(

19.623 ´10 -9 82.230 ´10 9 Ch 2 e= = GM 398.06 ´1012

2

)

e = 0.33 3 ◀

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PROBLEM 12.129 Telemetry technology is used to quantify kinematic values of a 200-kg roller coaster cart as it passes overhead. According to the system, r = 25 m, r = -10 m /s, r = -2 m/s 2, q = 90, q = - 0.4 rad/s, q = - 0.32 rad/s2 . At this instant, determine

(a) the normal force between the cart and the track, (b) the radius of curvature of the track.

SOLUTION Find the acceleration and velocity using polar coordinates.

vr = r = - 10 m/s v q = r q = (25 m)(- 0.4 rad/s)= - 10 m/s So the tangential direction is

45° and v = 10 2 m/s.

ar =  r - r q 2 = - 2 m/s - (25 m)(- 0.4 rad/s)2 = - 6 m/s 2 aq = r  q + 2rq = (25 m)(- 0.32)rad/s2 |+ 2(- 10 m/s)(- 0.4 rad/s) =0

So the acceleration is vertical and downward. (a)

To find the normal force use Newton’s second law. y-direction N - mgsin 45  = - macos45  N = m (g sin 45 - a cos 45) 2 2 = (200 kg)(9.81)m/s - 6 m/s )(0.70711) = 538.815 N

N = 539 N ◀

(b)

Radius l curvature of the track.

v2 r 2 (10 2 ) 2 v = r= a n 6cos 45

an =

r = 47.1 m ◀ Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2326...