vector mechanics for engineers - 9th edition

RMP - calculus

Cornelia demers: WIT

demersc1

Sticky Note

These problem solutions do not show a timeline. I highly recommend that you show a timeline to help visualize what is happening when.

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Chapter 11, Solution 1

Given: 4 2

3

2

1.5 30 5 10

6 60 5

18 60

x t t t

dxv t t

dtdv

a tdt

Evaluate expressions at 4 s.t

4 21.5(4) 30(4) 5(4) 10 66 mx 66.0 mx

36(4) 60(4) 5 149 m/sv 149.0 m/sv

2 218(4) 60 228 m/sa 2228.0 m/sa



Given: 3 2

2

12 18 2 5

36 36 2

72 36

x t t t

dxv t t

dtdv

a tdt

Find the time for 0.a

72 36 0 0.5 st t

Substitute into above expressions.

3 212(0.5) 18(0.5) 2(0.5) 5 3x 3.00 mx

236(0.5) 36(0.5) 2

7 m/s

v 7.00 m/sv



We have 3 25 530 8

3 2x t t t

Then 25 5 30dx

v t tdt

and 10 5dv

a tdt

When 0:v 2 25 5 30 5( 6) 0t t t t

or 3 s and 2 s (Reject)t t 3.00 st

At 3 s:t 3 23

5 5(3) (3) 30(3) 8

3 2x 3or 59.5 ftx

3 10(3) 5a 23or 25.0 ft/sa


Chapter 11 Solution 4

We have 26 8 40 cosx t t

Then 12 40 sindx

v t tdt

and 212 40 cosdv

a tdt

At 6 s:t 26 6(6) 8 40 cos 6x 6or 248 in.x

6 12(6) 40 sin 6v 6or 72.0 in./sv

26 12 40 cos 6a 2

6or 383 in./sa



We have 4 3 26 2 12 3 3x t t t t

Then 3 224 6 24 3dx

v t t tdt

and 272 12 24dv

a t tdt

When 0:a 2 272 12 24 12(6 2) 0t t t t

or (3 2)(2 1) 0t t

or2 1

s and s (Reject)3 2

t t 0.667 st

At2

s:3

t 4 3 2

2/32 2 2 2

6 2 12 3 33 3 3 3

x

2/3or 0.259 mx

3 2

2/32 2 2

24 6 24 33 3 3

v

2/3or 8.56 m/sv



3 2

2

2 15 24 4

6 30 24

12 30

x t t t

dxv t t

dtdv

a tdt

(a) 0 whenv 26 30 24 0t t

6( 1)( 4) 0 1.000 s or 4.00 st t t t

(b) 0 whena 12 30 0 2.5 st t

For 2.5 s:t 3 22.5 2(2.5) 15(2.5) 24(2.5) 4x

2.5 1.500 mx

To find total distance traveled, we note that

0 when 1s:v t 3 21

1

2(1) 15(1) 24(1) 4

15 m

x

x

For 0,t 0 4 mx

Distance traveled

From 0 to 1s:t t 1 0 15 4 11 mx x

From 1s to 2.5 s:t t 2.5 1 1.5 15 13.5 mx x

Total distance traveled 11 m 13.5 m 24.5 m



We have 3 26 36 40x t t t

Then 23 12 36dx

v t tdt

and 6 12dv

a tdt

(a) When 0:v 2 23 12 36 3( 4 12) 0t t t t

or ( 2)( 6) 0t t

or 2 s (Reject) and 6 st t 6.00 st

(b) When 0:x 3 26 36 40 0t t t

Factoring ( 10)( 2)( 2) 0 or 10 st t t t

Now observe that 0 6 s:t# , 0v ,

6 s 10 s:t, # 0v .

and at 0:t 0 40 ftx

6 s:t 3 26 (6) 6(6) 36(6) 40

256 ft

x

10 s:t 210 3(10) 12(10) 36v 10or 144.0 ft/sv

10 6(10) 12a 20or 48.0 ft/sa

Then 6 0

10 6

| | | 256 ( 40)| 216 ft

0 ( 256) 256 ft

x x

x x

Total distance traveled (216 256) ft 472 ft



We have 3 29 24 8x t t t

Then 23 18 24dx

v t tdt

and 6 18dv

a tdt

(a) When 0:v 2 23 18 24 3( 6 8) 0t t t t

or ( 2)( 4) 0t t

or 2.00 s and 4.00 st t

(b) When 0:a 6 18 0 or 3 st t

At 3 s:t 3 23 (3) 9(3) 24(3) 8x 3or 10.00 in.x

First observe that 0 2 s:t# , 0v .

2 s 3 s:t, # 0v ,

Now

At 0:t 0 8 in.x

At 2 s:t 3 22 (2) 9(2) 24(2) 8 12 in.x

Then 2 0

3 2

12 ( 8) 20 in.

| | |10 12| 2 in.

x x

x x

Total distance traveled (20 2) in. 22.0 in.



We have 28 m/sdv

adt

Then 8 constantdv dt C C or 8 (m/s)v t C

Also 8dx

v t Cdt

At 4 s, 20 m:t x 20 4

( 8 )x t

dx t C dt or 2

420 [ 4 ]tx t Ct

or 24 ( 4) 84 (m)x t C t

When 16 m/s, 4 m:v x 16 8 16 8t C C t

24 4 ( 4) 84t C t

Combining 20 4 (16 8 )( 4) 80t t t

Simplifying 2 4 4 0t t

or 2 st

and 32 m/s

8 32 (m/s)

C

v t

24 32 44 (m)x t t

(a) When 0:v 8 32 0t or 4.00 st

(b) Velocity and distance at 11 s.

11 (8)(11) 32v 11 56.0 m/sv

At 0:t 0 44 mx

4 s:t 4 20 mx

11 s:t 211 4(11) 32(11) 44 176 mx


PROBLEM 11.9 (Continued)

Now observe that 0 4 s: 0t v# , .

4 s 11 s:t, # 0v ,

Then 4 0

11 4

20 ( 44) 64 m

| | | 176 20| 196 m

x x

x x

Total distance traveled (64 196) m 260 m



We have 2 constanta kt k

Now 2dva kt

dt

At 6 s, 18 m/s:t v 2

18 6

v tdv kt dt

or 3118 ( 216)

3v k t

or 3118 ( 216)(m/s)

3v k t

Also 3118 ( 216)

3

dxv k t

dt

At 0, 24 m:t x 3

24 0

118 ( 216)

3

x tdx k t dt

or 41 124 18 216

3 4x t k t t

Now

At 6 s, 96 m:t x 41 196 24 18(6) (6) 216(6)

3 4k

or 41 m/s

9k

Then 41 1 124 18 216

3 9 4x t t t

or 41( ) 10 24

108x t t t

and 31 118 ( 216)

3 9v t

or 31( ) 10

27v t t



We have constanta kt k

Nowdv

a ktdt

At 0, 16 in./s:t v 16 0

v tdv kt dt

or 2116

2v kt

or 2116 (in./s)

2v kt

At 1 s, 15 in./s:t v 2115 in./s 16 in./s (1s)

2k

or 3 22 in./s and 16k v t

Also 216dx

v tdt

At 1 s, 20 in.:t x 2

20 1(16 )

x tdx t dt

or 3

1

120 16

3

t

x t t

or 31 1316 (in.)

3 3x t t

Then

At 7 s:t 27 16 (7)v 7or 33.0 in./sv

37

1 13(7) 16(7)

3 3x or 7 2.00 in.x

When 0:v 216 0 or 4 st t



At 0:t 013

3x

4 s:t 34

1 13(4) 16(4) 47 in.

3 3x

Now observe that

0 4 s:t# , 0v .

4 s 7 s:t, # 0v ,

Then 4 0

7 4

1347 42.67 in.

3| | |2 47| 45 in.

x x

x x

Total distance traveled (42.67 45) in. 87.7 in.


Chapter 11, Solution 11.12

2

2

a kt

dva kt

dt

(1)

0, 32 ft/st v and 4 s, 32 ft/st v

(a)32 4

2

32 0dv kt dt

3132 ( 32) (4)

3k 43.00 ft/sk

(b) Substituting 43 ft/s into (1)k

23dv

a tdt

23a t

0, 32 ft/s:t v 2

32 03

v tdv t dt

31( 32) 3( )

3v t 3 32v t

3 32dx

v tdt

4 s, 0:t x 3 4

0 44

1( 32) ; 32

4

tx tdx t dt x t t

4 4

4

1 132 (4) 32(4)

4 4

132 64 128

4

x t t

x t t

4132 64

4x t t



We have 26 constanta A t A

Now 26dv

a A tdt

At 0, 0:t v 2

0 0( 6 )

v tdv A t dt

or 32 (m/s)v At t

At 1 s, 30 m/s:t v 330 (1) 2(1)A

or 2 332 m/s and 32 2A v t t

Also 332 2dx

v t tdt

At 0, 8 m:t x 3

8 0(32 2 )

x tdx t t dt

or 2 418 16 (m)

2x t t

(a) When 0:v 3 232 2 2 (16 ) 0t t t t

or 0 and 4.00 st t

(b) At 4 s:t 2 44

18 16(4) (4) 136 m

2x

5 s:t 2 45

18 16(5) (5) 95.5 m

2x

Now observe that

0 4 s:t, , 0v .

4 s 5 s:t, # 0v ,

Then 4 0

5 4

136 8 128 m

| | |95.5 136| 40.5 m

x x

x x

Total distance traveled (128 40.5) m 168.5 m

Radial and Transverse Components




2 2

22 2 2 2

10 10(9.81 m/s ) 98.1 m/s

(8 m)(98.1 m/s ) 784.8 m /s

n

n n

a g

va v a

28.0 m/sv



(a) First note 72 km/h 20 m/sv

Now2

nv

a

or2

2

(20 m/s)

2 3.2 m/s

d

or 250 md

(b) We have2

nv

a

Then 2 2 1(0.6 9.81 m/s ) 180 m

2v

or 23.016 m/sv

or 82.9 km/hv



222.4 ft/sn n

va a

45 mi/h 66 ft/sv

22 (66 ft/s)

2.4 ft/s

1815 ft



We have2

nv

a

Then 2 2max( ) (3 32.2 ft/s )(80 ft)ABv

or max( ) 87.909 ft/sABv

or max( ) 59.9 mi/hABv



(a) At 0, 0,At v which implies ( ) 0A na

( )A Aa a t

or 220 mm/sAa

(b) We have uniformly accelerated motion

0 ( )A A tv a t

At 2 s:t 2 (20 mm/s )(2 s) = 40 mm/sAv

Now2 2

2(40 mm/s)( ) 17.7778 mm/s

90 mmA

A nA

va

Finally, 2 2 2

2 2

( ) ( )

(20) (17.7778)

A A t A na a a

or 226.8 mm/sAa



When 72 km/h 20 m/sv and 400 m,

2 22(20)

1.000 m/s400n

va

But 2 2n ta a a

2 2 2 2 2(1.5) (1.000) 1.11803 m/st na a a

Since the train is accelerating, reject the negative value.

(a) Distance to reach the speed.

0 0v

Let 0 0x

2 21 0 1 0 12 ( ) 2t tv v a x x a x

2 21

1(20)

2 (2)(1.11803)t

vx

a 1 178.9 mx

(b) Corresponding tangential acceleration.

21.118 m/sta



We have uniformly accelerated motion

2 22 1 122 tv v a s

Substituting 2 2(24 ft/s) (14 ft/s) 2 (95 ft)ta

or 22 ft/sta

Also 1 tv v a t

At 2 s:t 214 ft/s (2 ft/s )(2 s) 18 ft/sv

Now2

nv

a

At 2 s:t 2

2(18 ft/s)1.54286 ft/s

210 ftna

Finally 2 2 2t na a a

At 2 s:t 2 2 22 1.54286a

or 22.53 ft/sa

RMP - alternate accel form




(a) We have 3/20.6dv

v a vdx

When 0,x 9 m/s:v (3/2)

9 00.6

v xv dv dx

or 1/292[ ] 0.6vv x

or 1/21(3 )

0.3x v (1)

When 4 m/s:v 1/21(3 4 )

0.3x

or 3.33 mx

(b) We have 3/20.6dv

a vdt

When 0,t 9 m/s:v (3/2)

9 00.6

v tv dv dt

or (1/2)92[ ] 0.6vv t

or1 1

0.33

tv

When 1 m/s:v 1 1

0.331

t

or 2.22 st

(c) We have1 1

0.33

tv

or2

2

3 9

1 0.9 (1 0.9 )v

t t

Now2

9

(1 0.9 )

dxv

dt t



At 0,t 0:x 20 0

9

(1 0.9 )

x tdx dt

t

or0

1 19

0.9 1 0.9

t

xt

110 1

1 0.9

9

1 0.9

t

t

t

When 6 m:x 9

61 0.9

t

t

or 1.667 st

An alternative solution is to begin with Eq. (1).

1/21(3 )

0.3x v

Then 2(3 0.3 )dx

v xdt

Now

At 0, 0:t x 20 0(3 0.3 )

x tdxdt

x

or0

1 1

0.3 3 0.3 9 0.9

xx

tx x

Which leads to the same equation as above.



(a) We have 0.37.5(1 0.04 )dx

v xdt

At 0,t 0:x 0.30 0

7.5(1 0.04 )

x tdxdt

x

or 0.70

1 1[(1 0.04 ) ] 7.5

0.7 0.04xx t

or 0.71 (1 0.04 ) 0.21x t (1)

or 1/0.71[1 (1 0.21 ) ]

0.04x t

At 1 h:t 1/0.71{1 [1 0.21(1)] }

0.04x

or 7.15 mix

(b) We have

0.3 0.3

2 0.3 0.7

0.4

7.5(1 0.04 ) [7.5(1 0.04 ) ]

7.5 (1 0.04 ) [(0.3)( 0.04)(1 0.04 ) ]

0.675(1 0.04 )

dva v

dxd

x xdx

x x

x

At 0,t 0:x 2

20

5280 ft 1 h0.675 mi/h

1 mi 3600 sa

or 6 20 275 10 ft/sa

(c) From Eq. (1) 0.71[1 (1 0.04 ) ]

0.21t x

When 6 mi:x 0.71{1 [1 0.04(6)] }

0.210.83229 h

t

or 49.9 mint



(a) We have

0 0

20

3

0.18 0.18

0.0324

dva v

dx

v vd

x dx x

v

x

When 2 m:x 2

3

0.0324(3.6)

(2)a

or 20.0525 m/sa

(b) We have 00.18vdxv

dt x

From 1 m to 3 m:x x 3

1

3

01

0.18t

txdx v dt

or3

20 3 1

1

10.18 ( )

2x v t t

or12

3 1

(9 1)( )

0.18(3.6)t t

or 3 1 6.17 st t

RMP - special cases


demersc1

Sticky Note

These problems do not show a timeline. I highly recommend that you include a timeline to help visualize what is happening when.



(a) Acceleration of the car.

2 21 0 1 0

2 21 0

1 0

2 ( )

2( )

v v a x x

v va

x x

Data: 0

1

45 km/h 12.5 m/s

99 km/h 27.5 m/s

v

v

0

1

0

0.2 km 200 m

x

x

2 2(27.5) (12.5)

(2)(200 0)a

21.500 m/sa

(b) Time to reach 99 km/h.

1 0 1 0

1 01 0

( )

27.5 12.5

1.50010.00 s

v v a t t

v vt t

a

1 0 10.00 st t



(a) Initial velocity. 20 0

1

2x x v t at

00

1

2220 1

( 0.6)(10)10 2

x xv at

t

0 25.9 m/sv

(b) Final velocity. 0v v at

25.0 ( 0.6)(10)v 19.00 m/sfv

(c) Distance traveled during first 1.5 s.

20 0

2

1

21

0 (25.0)(1.5) ( 0.6)(1.5)2

x x v t at

36.8 mx



(a) Time required to reach B.

2

2

2

30 mi/h 44 ft/s, 0, 160 ft, 11 ft/s

1

21

160 0 44 (11)2

A A B

B A A

v x x a

x x v t at

t t

25.5 44 160 0t t

244 (44) (4)(5.5)( 160)

(2)(5.5)

4 6.7150

t

Rejecting the negative root. 2.7150 st 2.71st

(b) Speed at B.

44 (11)(2.7150) 73.865 ft/sB Av v at

50.4 mi/hBv



(a) We have 21 1

1

2y y v t at

At land,t 0y

Then 1

2 2

0 89.6 ft (16 s)

1( 32.2 ft/s )(16 s)

2

v

or 1 2.52 ft/sv

(b) We have 2 21 12 ( )v v a y y

At max, 0y y v

Then 2 2max0 (252 ft/s) 2( 32.2 ft/s )( 89.6) fty

or max 1076 fty



Given: 0 35 m,# # constanta

35 m 100 m, constantv , #

At 0, 0 when 35 m, 5.4 st v x t

Find:

(a) a

(b) v when 100 mx

(c) t when 100 mx

(a) We have 210 0

2x t at for 0 35 mx# #

At 5.4 s:t 2135 m (5.4 s)

2a

or 22.4005 m/sa

22.40 m/sa

(b) First note that maxv v for 35 m 100 m.x# #

Now 2 0 2 ( 0)v a x for 0 35 mx# #

When 35 m:x 2 2max 2(2.4005 m/s )(35 m)v

or max 12.9628 m/sv max 12.96 m/sv

(c) We have 1 0 1( )x x v t t for 35 m 100 mx, #

When 100 m:x 2100 m 35 m (12.9628 m/s)( 5.4) st

or 2 10.41 st



(a) For A B

and C D

we have 2 20 02 ( )v v a x x

Then,

at B 2 2

2 2

0 2(4.8 m/s )(3 0) m

28.8 m /s ( 5.3666 m/s)

BC

BC

v

v

and at D 2 2 2 ( )D BC CD D Cv v a x x d Cd x x

or 2 2 2 2(7.2 m/s) (28.8 m /s ) 2(4.8 m/s )d

or 2.40 md

(b) For A B

and ,C D

we have 0v v at

Then A B25.3666 m/s 0 (4.8 m/s ) ABt

or 1.11804 sABt

and C D 27.2 m/s 5.3666 m/s (4.8 m/s ) CDt

or 0.38196 sCDt

Now,

for ,B C

we have C B BC BCx x v t

or 3 m (5.3666 m/s) BCt

or 0.55901 sBCt

Finally, (1.11804 0.55901 0.38196) sD AB BC CDt t t t

or 2.06 sDt

Graphical Methods




(a)

Initial conditions: 0 00, 18 ft/s, 0t v x

Change in equals area under curve:v a t

0 18 ft/sv

0 4 s:t, , 24 0 (3 ft/s )(4 s) 12 ft/sv v 4 6 ft/sv

4 s 10 s:t, , 210 4 (6 ft/s )(6 s) 36 ft/sv v 10 30 ft/sv

10 s 12 s:t, , 212 10 ( 5 ft/s )(2 s) 10 ft/sv v 12 20 ft/sv


Change in equals area under curve:x v t

0 0x

0 4 s:t, , 4 01

( 18 6)(4) 48 ft2

x x 4 48 ftx

4 s 5 s:t, , 5 41

( 6)(1) 3 ft2

x x 5 51 ftx



5 s 10 s:t, , 10 51

( 30)(5) 75 ft2

x x 10 24 ftx

12 s 10 s:t, , 12 101

( 30 20)(2) 50 ft2

x x 12 74 ftx

16 s 12 s:t, , 16 121

( 20)(4) 40 ft2

x x 16 114 ftx

20 s 16 s:t, , 20 161

( 20)(4) 40 ft2

x x 20 74 ftx

(b) From above curves, we read

12 12For 12 s: 20 ft/s, 74 ftt v x

Distance traveled 0t to 12 st

From 0 st to 5 s :t Distance traveled 51 ft

From 5 st to 12 s :t Distance traveled (51 74) 125 ft

Total distance traveled 176 ft



(a)

Initial conditions: 0 00, 18 ft/s, 0t v x

Change in equals area under curve:v a t

0 18 ft/sv

0 4 s:t, , 24 0 (3 ft/s )(4 s) 12 ft/sv v 4 6 ft/sv

4 s 10 s:t, , 210 4 (6 ft/s )(6 s) 36 ft/sv v 10 30 ft/sv



Change in equals area under curve:x v t

0 0x

0 4 s:t, , 4 01

( 18 6)(4) 48 ft2

x x 4 48 ftx

4 s 5 s:t, , 5 41

( 6)(1) 3 ft2

x x 5 51 ftx

5 s 10 s:t, , 10 51

( 30)(5) 75 ft2

x x 10 24 ftx

12 s 10 s:t, , 12 101

( 30 20)(2) 50 ft2

x x 12 74 ftx



16 s 12 s:t, , 16 121

( 20)(4) 40 ft2

x x 16 114 ftx

20 s 16 s:t, , 20 161

( 20)(4) 40 ft2

x x 20 74 ftx

From v t and x t curves, we read

(a) At 10 s:t max 30 ft/sv

(b) At 16 s:t max 114 ftx


Problem 11, Solution 63

(a) slope ofta v t curve at time t

From 0t to 10 s:t constant 0v a

10 st to 26 s:t 220 605 ft/s

26 10a

26 st to 41s:t constant 0v a

41st to 46 s:t 25 ( 20)3 ft/s

46 41a

46 s:t constant 0v a

2 1x x (area under v t curve from 1t to 2 )t

At 10 s:t 10 540 10(60) 60 ftx

Next, find time at which 0.v Using similar triangles

00

10 26 10or 22 s

60 80v

v

tt

At 22

26

41

46

50

122 s: 60 (12)(60) 420 ft

21

26 s: 420 (4)(20) 380 ft2

41s: 380 15(20) 80 ft

20 546 s: 80 5 17.5 ft

2

50 s: 17.5 4(5) 2.5 ft

t x

t x

t x

t x

t x



(b) From 0t to 22 s:t Distance traveled 420 ( 540)

960 ft

t 22 s to 50 s:t Distance traveled | 2.5 420|

422.5 ft

Total distance traveled (960 422.5) ft 1382.5 ft

1383 ft

(c) Using similar triangles

Between 0 and 10 s: 0 1( ) 0 10

540 600xt

or 0 1( ) 9.00 sxt

Between 46 s and 50 s: 0 2( ) 46 4

17.5 20xt

or 0 2( ) 49.5 sxt



SOLUTION

(a) slope ofta v t curve at time t

From 0t to 10 s:t constant 0v a

10 st to 26 s:t 220 605 ft/s

26 10a

26 st to 41s:t constant 0v a

41st to 46 s:t 25 ( 20)3 ft/s

46 41a

46 s:t constant 0v a

2 1x x (area under v t curve from 1t to 2 )t

At 10 s:t 10 540 10(60) 60 ftx

Next, find time at which 0.v Using similar triangles

00

10 26 10or 22 s

60 80v

v

tt

At 22

26

41

46

50

122 s: 60 (12)(60) 420 ft

21

26 s: 420 (4)(20) 380 ft2

41s: 380 15(20) 80 ft

20 546 s: 80 5 17.5 ft

2

50 s: 17.5 4(5) 2.5 ft

t x

t x

t x

t x

t x



(b) Reading from the x t curve max 420 ftx

(c) Between 10 s and 22 s

100 ft 420 ft (area under curve from , to 22 s) ftv t t

or 1 11

100 420 (22 )( )2

t v

or 1 1(22 )( ) 640t v

Using similar triangles

11 1

1

60or 5(22 )

22 12

vv t

t

Then 1 1(22 )[5(22 )] 640t t

or 1 110.69 s and 33.3 st t

We have 110 s 22 st , , 1 10.69 st

Between 26 s and 41s:

Using similar triangles

241 15

20 300

t

or

2 40 st



Assume second deceleration is constant. Also, note that

200 km/h 55.555 m/s,

50 km/h 13.888 m/s

(a) Now area under curve for given time intervalx v t

Then 155.555 13.888

(586 600) m m/s2

t

or 1

2

0.4032 s

(30 586) m (13.888 m/s)

t

t

or 2

3

40.0346 s

1(0 30) m ( )(13.888 m/s)

2

t

t

or 3 4.3203 st

total (0.4032 40.0346 4.3203) st

or total 44.8 st

(b) We have initialinitial

1

2

[ 13.888 ( 55.555)] m/s

0.4032 s

103.3 m/s

va

t

or 2initial 103.3 m/s a



First note that (80 mm/s) (20 s) 4000 mm,, so that the speed of the pallet must be increased. Sincevpaint constant, it follows that paint maxv v and then 1 .5 s.t # From the curve,v t 1 2 4000 mmA A and it is seen that max min( )v occurs when

max1

1

80va

t

is maximum.

Thus,

max

1

( 80)mm/s60 mm/s

(s)

v

t

or max1

( 80)

60

vt

and max1 1 max

80(20 ) ( ) 4000

2

vt t v

Substituting for 1t

max max maxmax

( 80) 80 8020 4000

60 2 60

v v vv

Simplifying 2max max2560 486400 0v v

Solving max max207 mm/s and 2353 mm/sv v

For max 1207 mm/s, 5 sv t ,

max 12353 mm/s, 5 sv t .

max min( ) 207 mm/sv



The v t curve is first drawn as shown. Then

right

2right

left

left

12 in./s2 s

6 in./s

30 in./s

20 in./s

1.5 s

a

d

vt

a

vt

a

Now 1 60 in.A

or 1[( 2) s](12 in./s) 60 in.t

or 1 7 st

and 2 60 in.A

or 2{[( 7) 1.5] s}(30 in./s) 60 in.t

or 2 10.5 st

Now cycle 2t t cycle 10.5 st

We have (area under curve from to )ii i i iix x v t t t

At 2 s:t 21

(2) (12) 12 in.2

x

5 s:t 5 12 (5 2)(12)

48 in.

x

7 s:t 7 60 in.x

8.5 s:t 8.51

60 (1.5)(30)2

37.5 in.

x

9 s:t 9 37.5 (0.5)(30)

22.5 in.

x

10.5 s:t 10.5 0x



Given: At 0, 40 mi/h 0; when 2.5 mi,t v x x 20 mi/h;v at 7.5 min, 3 mi; constantt x decelerations

The v t curve is first drawn is shown.

(a) We have 1 2.5 miA

or 140 20 1 h

( min) mi/h 2.5 mi2 60 min

t

or 1 5 mint

(b) We have 2 0.5 miA

or 220 1 h(7.5 5) min mi/h 0.5 mi

2 60min

v

or 2 4 mi/hv

(c) We have final 12

(4 20) mi/h 5280 ft 1 min 1 h

(7.5 5)min mi 60 s 3600 s

a a

or 2final 0.1564 ft/sa



Given: max max

max

( ) 100 km/h, ( ) 70 km/h;

0; min, 20 s;

| | constant; as such as possible

AC CB

A B AB

v v

v v t B

a v v

The v t curve is first drawn as shown, where the magnitudes of the slopes (accelerations) of the threeinclined lines are equal.

Note:5

8 min 20 s h36

At 1, 8 kmt x

5h, 12 km

36x

Denoting the magnitude of the accelerations by a, we have

100

30

70

a

b

c

at

at

at

where a is in km/h2 and the times are in h.

Now 1 8 km:A 11 1

( )(100) ( )(100) ( )(30) 82 2a bt t t



Substituting 11 100 1 30

100 (100) (30) 82 2

ta a

or 154.5

0.08ta

Also 2 4 km:A 15 1

(70) ( )(70) 436 2 ct t

Substituting 15 1 70

(70) (70) 436 2

ta

or 1103 35

1260t

a

Then54.5 103 35

0.081260a a

or2

2 1000 m 1 h51.259 km/h

km 3600 sa

or 23.96 m/sa



SOLUTION

Given: 0 6 m/s;v for 0 , ;t t a t# #

for 21 1.4 s 2 m/s ;t t a # #

at 210 12 m/s ; att a t t

2 22 m/s , 1.8 m/sa v

The a t and v t curves are first drawn as shown.

(a) We have1 0 1tv v A

or 21

12 21.8 m/s 6 m/s ( s) m/s

2t

or 1 0.6 st

(b) We have11.4 2tv v A

or 21.4 1.8 m/s (1.4 0.6)5 2 m/sv

or 1.4 0.20 m/sv

Now 1.4 3 4 ,x A A where 3A is most easily determined usingintegrating. Thus,

for 10 :t t# # 2 ( 12) 5012 12

0.6 3a t t

Now

5012

3

dva

dt

t



At 0, 6 m/s:t v 6 0

5012

3

v tdv t dt

or 2256 12

3v t t

We have 2256 12

3

dxv t t

dt

Then 10.6

23

0 0

0.62 3

0

25(6 12 )

3

256 6 2.04 m

9

txA dx t t dt

t t t

Also 41.8 0.2

(1.4 0.6) 0.8 m2

A

Then 1.4 (2.4 0.8) mx

or 1.4 2.84 mx


= x = 6 m


Geometry of “bell-shaped” portion of v t curve

The parabolic spandrels marked by * are of equal area. Thus, total area of shaded portion of v t diagram is:

20 50 m/sv

(a) When 20 s:t

20 (60 m/s) (20 s) (shaded area)x

1200 m 6 m 20 1194 mx

(b) From 6 s to 14 s: 8 st t t

(60 m/s) (14 s 6 s) (shaded area)

(60 m/s)(8 s) 6 m 480 m 6 m 474 m

x

average474 m

8 s

xv

t

average 59.25 m/sv

Curvilinear Motion




First note 0 315 km/h

87.5 m/s

v

Vertical motion. (Uniformly accelerated motion)

210 (0)

2y t gt

At B: 2 2180 m (9.81 m/s )

2t

or 4.03855 sBt

Horizontal motion. (Uniform)

00 ( )xx v t

At B: (87.5 m/s)(4.03855 s)d

or 353 md



(a) Vertical motion. (Uniformly accelerated motion)

210 (0)

2y t gt

At B: 2 211 m (9.81 m/s ) or 0.451524 s

2 Bt t

Horizontal motion (Uniform)

00 ( )xx v t

At B: 07 m (0.451524 s)v

or 0 15.5031 m/sv 0 15.50 m/sv

(b) Vertical motion: At C: 2 213 m (9.81 m/s )

2t

or 0.782062 sCt

Horizontal motion.

At C: (7 )m (15.5031 m/s)(0.782062 s)d

or 5.12 md




210 (0)

2y t gt


0 00 ( )xx v t v t

At B: y: 2 21 13 ft (32.2 ft/s )

3 2t

or 0.455016 sBt

Then x: 07 ft ( ) (0.455016 s)Bv

or 0( ) 15.38 ft/sBv

At C: y: 2 212 ft (32.2 ft/s )

2t

or 0.352454 sCt

Then x: 01

12 ft ( ) (0.352454 s)3 Cv

or 0( ) 35.0 ft/sCv

015.38 ft/s 35.0 ft/sv# #



(a) Vertical motion. (Uniformly accelerated motion)

210 (0)

2y t gt


0 00 ( )xx v t v t

When5

31 in., 2 ft:12

h y 2 25 12 ft (32.2 ft/s )

12 2t

or 31 0.387432 st

Then 0 3140 ft ( ) (0.387432 s)v

or 0 31( ) 103.244 ft/s 70.4 mi/hv

When 42 in., 1.5 ft :h y 2 211.5 ft (32.2 ft/s )

2t

or 42 0.305234 st

Then 0 4240 ft ( ) (0.305234 s)v

or 0 42( ) 131.047 ft/s 89.4 mi/hv

070.4 mi/h 89.4 mi/hv# #

(b) For the vertical motion

(0)yv gt

Now0

|( ) |tan

( )y B

x B

v gt

v v

When 31 in.:h 2(32.2 ft/s )(0.387432 s)

tan103.244 ft/s

0.120833

or 31 6.89

When 42 in.:h 2(32.2 ft/s )(0.305234 s)

tan131.047 ft/s

0.075000

or 42 4.29



First note 0

0

( ) (13.40 m/s) cos 20 12.5919 m/s

( ) (13.40 m/s) sin 20 4.5831 m/sx

y

v

v

(a) Horizontal motion. (Uniform)

00 ( )xx v t

At C 9 m (12.5919 m/s) or 0.71475 sCt t


20 0

1( )

2yy y v t gt

At C:

2 2

2.1 m (4.5831 m/s)(0.71475 s)

1(9.81 m/s )(0.71475 s)

22.87 m

cy

2.43 mCy . (height of net) ball clears net

(b) At B, 0:y 2 210 2.1 m (4.5831 m/s) (9.81 m/s )

2t t

Solving 1.271175 s (the other root is negative)Bt

Then 0( ) (12.5919 m/s)(1.271175 s)

16.01 mx Bd v t

The ball lands (16.01 9.00) m 7.01 mb from the net



First note

0

0

( ) (1.2 m/s) cos 40 0.91925 m/s

( ) (1.2 m/s) sin 40 0.77135 m/sx

y

v

v


00 ( )xx v t


20 0

1( )

2yy y v t gt

Milk enters glass at B

2 2

: 0.08 m (0.91925 m/s) or 0.087028 s

: 0.140 m ( 0.77135 m/s)(0.087028 s)

1(9.81 m/s )(0.087028 s)

2

B

B

x t t

y h

or 0.244 mBh

Milk enters glass at C

2 2

: 0.146 m (0.91925 m/s) or 0.158825 s

: 0.140 m ( 0.77135 m/s)(0.158825 s)

1(9.81 m/s )(0.158825 s)

2

C

C

x t t

y h

or 0.386 mCh

0.244 m 0.386 mh# #



First note 0

0

( ) (160 ft/s)cos 25

( ) (160 ft/s)sin 25x

y

v

v

and at B cos 5 sin 5B Bx d y d Now Horizontal motion. (Uniform)

00 ( )xx v t

At Bcos 5

cos 5 (160 cos 25 ) or160 cos 25Bd t t d


2 20

10 ( ) ( 32.2 ft/s )

2yy v t gt g

At B: 21sin 5 (160 sin 25 )

2B Bd t gt

Substituting for Bt2

2cos 5 1 cos 5sin 5 (160 sin 25 )

160 cos 25 2 160 cos 25d d g d

or 22(160 cos 25 ) (tan 5 tan 25 )

32.2 cos 5

726.06 ft

d

or 242 ydd



First note 0

0

( ) (2.5 ft/s) cos 15 2.4148 ft/s

( ) (2.5 ft/s) sin 15 0.64705 ft/sx

y

v

v


20

10 ( )

2yy v t gt

At the top of the trough

2 218.8 ft ( 0.64705 ft/s) (32.2 ft/s )

2t t

or 0.719491 sBCt (the other root is negative)


00 ( )xx v t

In time BCt (2.4148 ft/s)(0.719491 s) 1.737 ftBCx

Thus, the trough must be placed so that

1.737 ft 1.737 ftB Cx x# $

Since the trough is 2 ft wide, it then follows that 0 1.737 ftd# #



0 0

0

0 0

0

( ) cos 20

0.9397

( ) sin 20

0.3420

x

y

v v

v

v v

v

Horizontal motion. 0 0( ) (0.9397 )xx v t v t

At B: 0 030 ft (0.9397 ) 31.925/v t t v (1)

Vertical motion. 20

1( )

2yy v t gt

At B: 20

118 ft (0.3420) 32.2

2v t t

Using Eq. (1):2

0

2

0

31.92518 (0.3420)(31.925) (16.1)

31.92528.918 (16.1)

v

v

22 2 20

(16.1)(31.925)567.44 ft /s

28.918v

0 23.82 ft/sv 0 23.8 ft/sv



First note

0 0

0 0

( ) cos 30

( ) sin 30x

y

v v

v v


00 ( )xx v t

At B: 00

16(16 ) ( cos 30 ) or

cos 30Bd

d v t tv


2 20

10 ( ) ( 32.2 ft/s )

2yy v t gt g

At B: 20

13.2 ( sin 30 )

2B Bv t gt

Substituting for tB

2

00 0

16 1 163.2 ( sin 30 )

cos 30 2 cos 30

d dv g

v v

or2

20

2 (16 )~ ft

13 (16 ) 3.2

3

g dv d

d

(a) 9 in.:d

29122

091

123

2(32.2) 16

3 16 3.2v

or 0 29.8 ft/sv

(b) 17 in.:d

217122

0171123

2(32.2) 16

3 16 3.2v

or 0 29.6 ft/sv



First note

0 0

0 0

( ) cos 3

( ) sin 3x

y

v v

v v


00 ( )xx v t

When 6 m:x 0 60

66 ( cos 3 ) or

cos 3v t t

v


2 20

10 ( ) ( 9.81 m/s )

2yy v t gt g

When the ball reaches the tire, 6t t

, 00

2

0

6( sin 3 )

cos 3

1 6

2 cos 3

B Cy vv

gv

or 20 2

,

18(9.81)

cos 3 (6 tan 3 )B C

vy

or 20

,

177.065

0.314447 B C

vy

At B, 0.53 m:y 20

177.065

0.314447 ( 0.53)v

or 0( ) 14.48 m/sBv

At C, 1.25 m:y 20

177.06 s

0.314447 ( 1.25)v

or 0( ) 10.64 m/sCv

010.64 m/s 14.48 m/sv# #



First note

0 0

0 0

( ) cos 65

( ) sin 65x

y

v v

v v


00 ( )xx v t

At either B or C, 5 mx

0 ,( cos 65 ) B Cs v t

or ,0

5

( cos 65 )B Ctv


2 20

10 ( ) ( 9.81 m/s )

2yy v t gt g

At the tree limbs, ,B Ct t

2

, 00 0

5 1 5( sin 65 )

cos 65 2 cos 65B Cy v gv v

or1

2 20 2

,

,

(9.81)(25)

cos 65 (5 tan 65 )

686.566

5 tan 65

B C

B C

vy

y

At Point B: 20 0

686.566or ( ) 10.95 m/s

5 tan 65 5 Bv v

At Point C : 20 0

686.566or ( ) 11.93 m/s

5 tan 65 5.9 Cv v

010.95 m/s 11.93 m/sv# #



First note 0 0

0 0

( ) sin 15

( ) cos 15x

y

v v

v v


0 0( ) sin 15x xv v v


20 0

20 0

1( ) 0 ( )

21

cos 15 ( cos 15 )2

y y yv v gt y v t gt

v gt v t gt

At Point B, 0yv ,

Then 0

0

( ) sin 15tan 12

|( ) | cos 15x B

y B B

v v

v gt v

or 20

0

sin 15cos 15 9.81 m/s

tan 12

0.22259

B

vt g

g

v

Noting that 0.2 m,By

We have 20 0 0

10.2 ( cos 15 )(0.22259 ) (9.81)(0.22259 )

2v v v

or 0 2.67 m/sv



First note 0 105 mi/h 154 ft/sv

and 0 0

0 0

( ) cos (154 ft/s) cos

( ) sin (154 ft/s) sinx

y

v v

v v


00 ( ) (154 cos )xx v t t

At the front of the net, 16 ftx

Then 16 (154 cos ) t

or enter8

77 cost


20

2 2

10 ( )

21

(154 sin ) ( 32.2 ft/s )2

yy v t gt

t gt g

At the front of the net,

2front enter enter

2

2

1(154 sin )

2

8 1 8(154 sin )

77 cos 2 77 cos

3216 tan

5929 cos

y t gt

g

g

Now 2 22

1sec 1 tan

cos

Then 2front

3216 tan (1 tan )

5929

gy

or 2front

5929 5929tan tan 1 0

2 32y

g g

Then

1/225929 5929 5929

front2 2 324 1tan

2

g g g y

or

1/22

front5929 5929 5929

tan 14 32.2 4 32.2 32 32.2

y

or 2 1/2fronttan 46.0326 [(46.0326) (1 5.7541 )]y



Now front0 4 fty# # so that the positive root will yield values of 45 . for all values of yfront.When the negative root is selected, increases as yfront is increased. Therefore, for max , set

front 4 ftCy y

Then 2 1/2tan 46.0326 [(46.0326) (1 5.7541 4)]

or max 14.6604 max 14.66

(b) We had found enter8

77 cos

8

77 cos 14.6604

t

or enter 0.1074 st



First note

0 72 km/h 20 m/sv

and 0 0

0 0

( ) cos (20 m/s) cos

( ) sin (20 m/s) sinx

y

v v

v v


00 ( ) (20 cos )xx v t t

At Point B:7

14 (20 cos ) or10 cosBt t


2 2 20

1 10 ( ) (20 sin ) ( 9.81 m/s )

2 2yy v t gt t gt g

At Point B: 210.08 (20 sin )

2B Bt gt

Substituting for Bt

27 1 7

0.08 (20 sin )10 cos 2 10 cos

g

or2

1 498 1400 tan

2 cosg

Now 2 22

1sec 1 tan

cos

Then 28 1400 tan 24.5 (1 tan )g

or 2240.345 tan 1400 tan 248.345 0

Solving 10.3786 and 79.949

Rejecting the second root because it is not physically reasonable, we have

10.38

(b) We have 0( ) 20 cosx xv v

and 0( ) 20 siny yv v gt gt

At Point B: ( ) 20 sin

720 sin

10 cos

y B Bv gt

g



Noting that at Point B, 0,yv , we have

710 cos

7 9.81200 cos10.3786

|( ) |tan

20 sin

20 cos

sin 10.3786

cos 10.3786

y B

x

g

v

v

or 9.74



First note 0 0

0 0

( ) cos

( ) sinx

y

v v

v v


0 00 ( ) ( cos )xx v t v t

At Point B: 01.8 ( cos )v t

or0

1.8

cosBt v


20

2 20

10 ( )

21

( sin ) ( 9.81 m/s )2

yy v t gt

v t gt g

At Point B: 20

11.4 ( sin )

2B Bv t gt

Substituting for Bt

2

00 0

1.8 1 1.81.4 ( sin )

cos 2 cosv g

v v

or 20 2

2

1.62

cos (1.8 tan 1.4)

1.62

0.9 sin 2 1.4 cos

gv

g

Now minimize 20v with respect to .

We have20

2 2

(1.8 cos 2 2.8 cos sin )1.62 0

(0.9 sin 2 1.4 cos )

dvg

d

or 1.8 cos 2 1.4 sin 2 0

or18

tan 214

or 26.0625 and 206.06

Rejecting the second value because it is not physically possible, we have

26.1

Finally, 20 2

1.62 9.81

cos 26.0625 (1.8 tan 26.0625 1.4)v

or 0 min( ) 2.94 m/sv



First note 0 0

0 0

( ) cos (8 m/s) cos

( ) sin (8 m/s) sinx

y

v v

v v


00 ( ) (8 cos )xx v t t

At Point B: : (8 cos )x d d t

or8 cosB

dt


20

2 2

10 ( )

21

(8 sin ) ( 9.81 m/s )2

yy v t gt

t gt g

At Point B: 210 (8 sin )

2B Bt gt

Simplifying and substituting for Bt

10 8 sin

2 8 cos

dg

or64

sin 2dg

(1)

(a) When 0,h the water can follow any physically possible trajectory. It then follows fromEq. (1) that d is maximum when 2 90

or 45

Then64

sin (2 45 )9.81

d

or max 6.52 md

(b) Based on Eq. (1) and the results of Part a, it can be concluded that d increases in value as increasesin value from 0 to 45° and then d decreases as is further increased. Thus, maxd occurs for the valueof closest to 45° and for which the water just passes over the first row of corn plants. At this row,xcom 1.5 m

so that corn1.5

8 cost

Also, with corn ,y h we have

2corn corn

1(8 sin )

2h t gt



Substituting for cornt and noting 1.8 m,h

21.5 1 1.5

1.8 (8 sin )8 cos 2 8 cos

g

or2

2.251.8 1.5 tan

128 cos

g

Now 2 22

1sec 1 tan

cos

Then 22.25(9.81)1.8 1.5 tan (1 tan )

128

or 20.172441 tan 1.5 tan 1.972441 0

Solving 58.229 and 81.965

From the above discussion, it follows that maxd d when

58.2

Finally, using Eq. (1)

64sin (2 58.229 )

9.81d

or max 5.84 md



First note 0 0

0 0

( ) cos (11.5 m/s) cos

( ) sin (11.5 m/s) sinx

y

v v

v v

By observation, maxd occurs when max 1.1 m.y


20 0

2

1( ) 0 ( )

21

(11.5 sin ) (11.5 sin )2

y y yv v gt y v t gt

gt t gt

When max at , ( ) 0y By y B v

Then ( ) 0 (11.5 sin )y Bv gt

or 211.5 sin( 9.81 m/s )Bt g

g

and 21(11.5 sin )

2B B By t gt

Substituting for Bt and noting 1.1 mBy

2

2 2

11.5 sin 1 11.5 sin1.1 (11.5 sin )

2

1(11.5) sin

2

gg g

g

or 22

2.2 9.81sin 23.8265

11.5


00 ( ) (11.5 cos )xx v t t

At Point B: max and Bx d t t

where11.5

sin 23.8265 0.47356 s9.81Bt

Then max (11.5)(cos 23.8265 )(0.47356)d

or max 4.98 md

(b) From above 23.8



First determine the velocity of the bag as it lands on the belt. Now

0 0

0 0

[( ) ] ( ) cos 30

(2.5 ft/s)cos 30

[( ) ] ( ) sin 30

(2.5 ft/s)sin 30

B x B

B y B

v v

v v


00 [( ) ]B xx v t 0( ) [( ) ]B x B xv v

(2.5 cos 30 ) t 2.5 cos 30


20 0

1[( ) ]

2B yy y v t gt 0( ) [( ) ]B y B yv v gt

211.5 (2.5 sin 30 )

2t gt 2.5 sin 30 gt

The equation of the line collinear with the top surface of the belt is

tan 20y x

Thus, when the bag reaches the belt

211.5 (2.5 sin 30 ) [(2.5 cos 30 ) ]tan 20

2t gt t

or 21(32.2) 2.5(cos 30 tan 20 sin 30 ) 1.5 0

2t t

or 216.1 0.46198 1.5 0t t

Solving 0.31992 s and 0.29122 st t (Reject)

The velocity Bv of the bag as it lands on the belt is then

(2.5 cos 30 ) [2.5 sin 30 32.2(0.319 92)]

(2.1651 ft/s) (9.0514 ft/s)B

v i j

i j

Finally /B A B A v v v

or / (2.1651 9.0514 ) 4(cos 20 sin 20 )

(1.59367 ft/s) (10.4195 ft/s)B A

v i j i j

i j

or / 10.54 ft/sB A v 81.3°

impulse and momentum



1

1 1 1

0k

k k k

mv Wt

mv mv vt

W mg g

(a) For 0.75k

2

25 m/s

(0.75)(9.81 m/s )t 3.40 st

(b) For 0.10k

2

25 m/s

(0.10)(9.81 m/s )t 25.5 st


1

2

0

(40,000 tons)(2000 lb/tons)(3.667 ft/s) (35,000 lb) 0

32.2 ft/s

mv Ft

t

260.3 st 4 min 20 st


(a) 0

2

30 ft/s0

(32.2 ft/s )(0.30)A

A kk

vW v Wt tg g

3.11st

(b) 20

Impulse-momentum in x direction

cos 20 sin 20 0A kW

v Wt Wtg

2

( cos 20 sin 20 )

30 ft/s

(32.2 ft/s )(0.30cos 20 sin 20 )

A

k

vt

g

1.493 st


0m dt m v F v (1)

Where 2

0

2 3 2

[(8 6 ) (4 ) (4 ) ]

1 1(8 3 ) 4 4

3 2

tdt t t t

t t t t t t

F i j k

i j k

Substituting 2 kg,m 0 150 100 250v i j k into (1):

2 3 21 1(2 kg)(150 100 250 ) (8 3 ) 4 4 (2 kg)

3 2t t t t t t

i j k i j k v

2 3 23 1 1150 4 100 2 250 2

2 6 4t t t t t t

v i j k

(a) v is parallel to yz plane when 0,xv that is, when

23150 4 0 11.422 s

2t t t 11.42t s

(b) 3

2

1100 2(11.422) (11.422)

6

1250 2(11.422) (11.422)

4

v j

k

(125.5 m/s) (194.5 m/s) v j k


1

2

90 km/h 25 m/s

30 km/h 8.33 m/s

v

v

N mg

1 2smv Nt mv

2(25 m/s) (0.65) (9.81 m/s ) (8.33 m/s)m m t m

25 8.33

(0.65)(9.81)t

2.61 st


1 4tan 2.29

100

1 1 2 2 impm m v v

1 2Nmv Wt Ft mv #

1

2

60 mi/h 88 ft/s cos

20 mi/h 29.33 ft/s coss s

v N W W mg

v F N W

( m )(88 ft/s) ( m 2)(32.2 ft/s )( )(sin 2.29 ) (0.60)(t m 2)(32.2 ft/s )(cos 2.29 )( ) (t m )(29.33 ft/s)

88.0 29.33

32.2[(0.60)cos 2.29 sin 2.29 ]t

3.26 st


90 km/h 25 m/sv

(a) The shortest time for the rig to come to a stop will be when the friction force on the wheels is maximum.The downward force exerted by the trailer on the cab is assumed to be zero. Since the trailer brakes fail,all of the braking force is supplied by the wheels of the cab, which is maximum when the wheels of thecab are at impending sliding.

1 2 1 2

1 2

(2000)

(0.65)(2000)s C C CFt N t N m g g

Ft gt

2 1[( ) ] [( ) ]C T C Tm m v Ft m m v

21 20 (0.65)(2000 kg)(9.81 m/s )( ) 10,000 kg(25 m/s)t

1 2 19.60 st

(b) For the trailer:

2 1 2 1[ ] [ ]T Tm v Qt m v

From (a), 1 2 19.60 s

0 (19.60 s) (8000 kg)(25 m/s)

10,204 N

t

Q

Q

10.20 kN (compression)Q


(a) Blocks A and C

1 1 2 1 2 2[( ) ] ( ) ( ) [( ) ]A C A C A Cm m v T t m m gt m m v

0 (12 )(0.8) 12g T v (1)

Block B

1 1 2 1 2 2[ ] ( ) ( )B B Bm v T t m gt m v

0 ( 4 )(0.8) 4T g v (2)

Adding (1) and (2), (eliminating)

(12 4 )(0.8) (12 4)g g 2(8 kg)(9.81 m/s )(0.8 s)

16 kgv 3.92 m/sv

(b) Collar A

1( ) 0 0 ( )A C Am v F m g (3)

From Eq. (2) with 3.92 m/sv

2

44

0.8(4 kg)(3.92 m/s)

(4 kg)(9.81 m/s )(0.8 s)

58.84 N

vT g

T

T

Solving for CF in (3)

2(4 kg)(3.92 m/s)(4 kg)(9.81 m/s ) 58.84 N

(0.8 s)CF

39.2 NCF


Kinematics: Dependent motion

Cable length: constantA A BL x x x

2 0A BdL

v vdt

Here velocities are defined as positive if downward.

2B Av v (1)

Let T be the tension in the cable.

Use the principle of impulse and momentum.

Collar A, components : 2 A A ATt m gt m v (2)

Block B, components : B B BTt m gt m v (3)

Subtract twice Eq. (3) from Eq. (2) to eliminate Tt.

(2 ) 2

2

(2 )

B A B B A A

B B A A

B A

m m gt m v m v

m v m vt

m m g

Data: 20 kg 30 kg 2 10 kgA B B Am m m m

1.0 m/s, i.e., 1.0 m/sAv

From Eq. (1) 2.0 m/sBv

[(2)(15)(2.0) (20)( 1.0)]

(10)(9.81)t

0.815 st


0 Pdt Fdt mv

At any time:1

v Pdt Fdtm (1)

(a) Block starts moving at t.

(0.50)(125 lb) 62.5 lbs sP F W

1 18 s 8 s;

100 lb 62.5 lb 100 lbs

t t

F 1 5.00 st

(b) Maximum velocity: At mt t

where 0.4(125) 50 lbk kP F W

Block moves at 5 s.t

Shaded area is maximum net impulse RPdt F dt when 1m mt t v v

Eq. (1):shaded1 1 1 1 1

(12.5 50)(3) (50)(4) (193.75) area 2 2mv

m m m

125 lb

32.2

1[193.75] 49.91 ft/smv 49.9 ft/sm v


(c) Block stops moving when 0; orPdt Fdt Qdt Fdt Assume 16 s.mt

1(100)(16) 800 lb s

21

(62.5)(5) (50)( 5)2 m

Pdt

Fdt t

800 [156.25 50( 5)] 0mPdt Fdt t 17.875 smt

16 s OKmt 17.88 smt


1 2( ) 0.18 sm t m t v P W v

Vertical components

0 ( )(0.18) (12)(sin 50 )

(12)(sin 50 )

(9.81)(0.18)

v

v

WP W

g

P W W

6.21vP W


185 lb

0.22 s

W

t

1 2( )m t m v P W v

Horizontal components

2

(30)(cos 35 ) (0.22) 0

(185 lb)(30 ft/s)(cos 35 )641.7 lb

(32.2 ft/s )(0.22 s)

H

H

WP

g

P

642 lbHP

Newton's 2nd Law - blocks



(a) Assume uniformly decelerated motion.

Then 0v v at

At 9 s:t 00 (9)v a

or 0

9

va

Also 2 20 2 ( 0)v v a x

At 9 s:t 200 2 (30)v a

Substituting for a 2 000 2 (30) 0

9

vv

or 0 6.6667 m/sv 0or 6.67 m/sv

and 26.66670.74074 m/s

9a

(b) We have

0: 0yF N W or N W mg

Sliding: k kF N mg

:xF ma F ma or k mg ma

or

2

2

0.74074 m/s

9.81 m/sk

a

g

or 0.0755k

demerssborovc

Sticky Note

It's useful to draw a timeline for RMP.

demerssborovc

Sticky Note

This is Newton's 2nd Law. Notice the puck is moving to the right and decelerating therefore the decel vector points to the left. The friction vector opposes the direction of motion and therefore properly points to the left.


Kinematics: Uniformly accelerated motion. 0 0( 0, 0)x v

20 0

1,

2x x v t at

or 22 2

2 (2)(5)0.100 m/s

(10)

xa

t

0: sin 50 cos 20 0

sin 50 cos 20

yF N P mg

N P mg

: cos 50 sin 20xF ma P mg N ma

or cos 50 sin 20 ( sin 50 cos 20 )P mg P mg ma

(sin 20 cos 20 )

cos 50 sin 50

ma mgP

For motion impending, set 0 and 0.4sa

(20)(0) (20)(9.81)(sin 20 0.4 cos 20 )

cos 50 0.4 sin 50

419 N

P

For motion with 20.100 m/s , use 0.3.ka

(20)(0.100) (20)(9.81)(sin 20 0.3 cos 20 )

cos 50 0.3 sin 50P

301 NP

demerssborovc

Sticky Note

The author starts with using RMP. A timeline would be helpful for you. I would suggest to add that to this problem to assist in writing the RMP equations.


For any angle .

Use x and y coordinates as shown.

0ya

: cos 0

cos

y yF ma N mg

N mg

: sinx x k xF ma mg N ma

(sin cos )x ka g

At Point A. 230 , m/s

sin 30

cos 30

9.81sin 30 3

9.81 cos 30

0.22423

x

xk

a

g a

g

At Point B. 15 , 0.22423

9.81(sin 15° 0.22423 cos 15°)k

xa

0.414 m/s 20.414 m/sa 15

demerssborovc

Sticky Note

No RMP here, just Newton's 2nd Law.


Load: We assume that sliding of load relative to trailer is impending:

m

s

F F

N

Deceleration of load is same as deceleration of trailer, which is the maximum allowable deceleration max .a

0: 0yF N W N W

0.40m sF N W

max:x mF ma F ma

max0.40W

W ag

2max 12.88 ft/sa

2max 12.88 ft/sa

Uniformly accelerated motion.

2 20 2 with 0v v ax v 0 45 mi/h 66ft/sv

2max

2

12.88 ft/s

0 (66) 2( 12.88)

a a

x

169.1 ftx

demerssborovc

Sticky Note

Since the load is not shifting (not moving), it has no acceleration of the load with respect to the truck. But since the truck is moving, the load moves with the truck and therefore has the same acceleration as the truck.


(a) Kinematics of the belt. 0ov

1. Acceleration phase with 21 2 m/sa

1 1 1

2 21 1 1 1

0 (2)(1.3) 2.6 m/s

1 10 0 (2)(1.3) 1.69 m

2 2

o

o o

v v a t

x x v t a t

2. Deceleration phase: 2 0v since the belt stops.

2 22 1 2 2 1

2 2 22 1

22 1

2 ( )

0 (2.6)6.63

2( ) 2(2.2 1.69)

v v a x x

v va

x x

2

2 6.63 m/sa

2 12 1

2

0 2.60.3923 s

6.63

v vt t

a

(b) Motion of the package

1. Acceleration phase. Assume no slip. 21( ) 2 m/spa

0: 0 oryF N W N W mg

1: ( )x f pF ma F m a

The required friction force is Ff.

The available friction force is 0.35 0.35 s N W mg

21( ) (0.35)(9.81) 3.43 m/sf s

p s

F Na g

m m

,

Since 2 22.0 m/s 3.43 m/s ,, the package does not slip.

1 1 1( ) 2.6 m/s and ( ) 1.69 m.p pv v x

2. Deceleration phase. Assume no slip. 22( ) 11.52 m/spa

2: ( )x f pF ma F m a

22( ) 6.63 m/sf

p

Fa

m

2 23.43 m/s 6.63 m/ss ss

N mgg

m m

,

Since the available friction force s N is less than the required friction force Ff for no slip, thepackage does slip.

demerssborovc

Sticky Note

Start with RMP. There are two segments of motion: the acceleration phase and the deceleration phase. A timeline would help in setting up the RMP equations.

demerssborovc

Sticky Note

Newton's 2nd Law. If the package does not slip then it's only acceleration is the same as the conveyor belt. Use the static coefficient of friction because the package does not slip on the belt. BEST APPROACH: What the author is doing is checking what the maximum acceleration of the belt would be for the package not to slip. The max accel is 3.43 m/sec2 and the belt is moving at 2 m/sec2. Therefore the package will not slip. You can always calculate the friction force the package experiences and compare it to that required for no slip. The normal force will not change between the two scenarios but the coefficient of friction will. The static coefficient of friction is that for motion impending.

demerssborovc

Sticky Note

Since the package slips we can now calculate the position of the package with respect to the belt.

demerssborovc

Sticky Note

This is a challenging problem. I will not be including anything like this on the exam. However, I have included it for your learning.


22( ) 6.63 m/s ,p f ka F N,

2 2( ) : ( )x p k pF m a N m a

2

2

2 1 2 2 1

2

2 22 1 1 2 1 2 2 1

2

( )

(0.25)(9.81)

2.4525 m/s

( ) ( ) ( ) ( )

2.6 ( 2.4525)(0.3923)

1.638 m/s

1( ) ( ) ( ) ( ) ( ) ( )

21

1.69 (2.6)(0.3923) ( 2.4525)(0.3923)2

2.521 m

kp k

p p p

p p p p

Na g

m

v v a t t

x x v t t a t t

Position of package relative to the belt

2 2( ) 2.521 2.2 0.321px x

/belt 0.321 mpx


Acceleration 1:a Impending slip. 1 1 10.30sF N N

: y A yF m a 1 1 sin 65 A AN W m a

1 1

1

sin 65

( sin 65 )

A A

A

N W m a

m g a

1 1: cos 65x A x AF m a F m a

1 sF N

or 1 1cos 65 0.30 ( sin 65 )A Am a m g a

1

2

0.30

cos 65 0.30 sin 65

(1.990)(9.81)

19.53 m/s

ga

21 19.53 m/sa 65

Deceleration 2:a Impending slip. 2 2 20.30sF N N

1 2: sin 65y y A AF ma N W m a

1 2 sin 65A AN W m a

: x xF ma 2 2 cos 65 AF m a

2 2sF N

or 2 2cos 65 0.30 ( cos 65 )A Am a m g a

20.30

cos 65 0.30 sin 65

(0.432)(9.81)

ga

24.24 m/s 22 4.24 m/sa 65

demerssborovc

Sticky Note

This is similar to the previous problem but it limits the motion for the shingles to slip impending. So this is a great problem to study for the exam.


Let Pa be the acceleration of the plywood, Ta be the acceleration of the truck, and /P Ta be the accelerationof the plywood relative to the truck.

(a) Find the value of Ta so that the relative motion of the plywood with respect to the truck is impending.

P Ta a and 1 1 10.40sF N N

1

1

: cos 20 sin 20

( cos 20 sin 20 )

y P y P P T

P T

F m a N W m a

N m g a

1

1

: sin 20 cos 20

( sin 20 cos 20 )x x P P T

P T

F ma F W m a

F m g a

( sin 20 cos 20 ) 0.40 ( cos 20 sin 20 )P T P Tm g a m g a

(0.40cos 20 sin 20 )

cos 20 0.40sin 20(0.03145)(9.81)

0.309

Ta g

20.309 m/sT a

(b) 2 2/ / / / /

1 1( ) ( ) 0 0

2 2P T P T o P T P T P Tx x v t a t a t

2// 2 2

2 (2)(2)4.94 m/s

(0.9)P T

P T

xa

t

2/ 4.94 m/sP T a 20

2/ ( ) (4.94 m/sP T P T Ta a a a 20 )

2

2

: cos 20 sin 20

( cos 20 sin 20 )

y P y P P T

P T

F m a N W m a

N m g a

2 /: sin 20 cos 20x x P P T P P TF ma F W m a m a

2 /( sin 20 cos 20 )P T P TF m g a a

For sliding with friction 2 2 20.30kF N N

/

/

( sin 20 cos 20 ) 0.30 ( cos 20 sin 20 )

(0.30cos 20 sin 20 )

cos 20 0.30sin 20

( 0.05767)(9.81) (0.9594)(4.94)

P T P T P T

P TT

m g a a m g a

g aa

4.17 24.17 m/sT a

demerssborovc

Sticky Note

If the plywood is not slipping on the truck (there is no acceleration of the plywood with respect to the truck), then the plywood only experiences the acceleration of the truck. Since the plywood is just about to slip (but hasn't), then use the static coefficient of friction.

demerssborovc

Sticky Note

Now the truck is accelerating and the plywood is slidding. Therefore, the plywood experiences both the acceleration of the truck AND the acceleration of the plywood with respect to the truck.

demerssborovc

Sticky Note

We can not find the acceleration of the plywood directly. We must use vector addition and resolve all accelerations into x and y coordinates. The author does not show you the solution to this final equation. You can only solve this last part after you find the acceleration of the truck.


(a) : sin 30 cos 40x A A A AB A AF m a W N m a

or

122

2cos 40

A

AB

a gN

Now we note: / , B A B Aa a a where /B Aa is directed along the top surface of A.

: cos 20 sin 50 y B y AB B B AF m a N W m a

or 10 ( cos 20 sin 50 )AB AN g a

Equating the two expressions for ABN

122

210( cos 20 sin 50 )

cos 40

A

A

a gg a

or2(9.81)(1.1 cos 20 cos 40 )

6.4061 m/s2.2 cos 40 sin 50Aa

/: sin 20 cos50 x B x B B B A B AF m a W m a m a

or /

2

2

sin 20 cos50

(9.81sin 20 6.4061cos50 ) m/s

7.4730 m/s

B A Aa g a


Finally /B A B A a a a

We have 2 2 26.4061 7.4730 2(6.4061 7.4730)cos50Ba

or 25.9447 m/sBa

and7.4730 5.9447

sin sin 50

A:

B:

or 74.4

25.94 m/sB a 75.6

(b) Note: We have uniformly accelerated motion, so that

0v at

Now / /B A B A B A B At t t v v v a a a

At 0.5 s:t 2/ 7.4730 m/s 0.5 sB Av

or / 3.74 m/sB A v 20

cables - newton's 2nd law



A:

B:

From the diagram

3 constantA Bx y

Then 3 0A Bv v

and 3 0A Ba a

or 3A Ba a (1)

(a) A: : x A AF m a A BT m a

Using Eq. (1) 3 A BT m a

B: : 3y B B B B BF m a W T m a

Substituting for T

3(3 )B A B B Bm g m a m a

or2

29.81 m/s0.83136 m/s

30 kg1 91 9

25 kg

BA

B

ga

m

m

Then 22.49 m/sA a

and 20.831 m/sB a

(b) We have 23 30 kg 0.83136 m/sT

or 74.8 NT

demerssborovc

Sticky Note

A cable relationship is needed because the cable is directly attached to both block A and B.


A:

B:

From the diagram

3 constantA Bx y

Then 3 0A Bv v

and 3 0A Ba a

or 3A Ba a (1)

First determine if the blocks will move with 0. A Ba a We have

A:1

0: 3 0 or3y B BF W T T m g

B: 0:xF 0AF T

Then 2125 kg 9.81 m/s 81.75 N

3AF

0: 0 ory A A A AF W N N m g

Also, max

2

( ) ( ) ( )

0.25 30 kg 9.81 m/s

73.575 N

A s A A s A AF N m g

max( ) ,A AF F. which implies that the blocks will move.

(a) A: 0: 0 ory A A A AF W N N m g

Sliding: ( ) 0.20A k A A AF N m g

:x A A A A AF m a F T m a

Using Eq. (1) 0.20 3A A BT m g m a

B: : 3y B B B B BF m a W T m a

or 3(0.20 3 )B A A B B Bm g m g m a m a


Referring to solution of Problem 11.51 we note that

1

2B Aa a (1)

where minus sign indicates that Aa and Ba have opposite sense.

Block B.

: 25 2 7.5 AF ma T a (2)

Block A.

: 2 10 AF ma T T a

10 AT a (3)

(a) Substituting for T from (3) into (2):

25 2(10 ) 7.5

25 27.5A A

A

a a

a

20.909 m/sA a

0( ) 0 0.909(1.2),A A Av v a t 1.091 m/sA v

(b) Substituting 0.909a into Eq. (1): 20.455 m/sB a

0( ) 0 0.455(1.2)B B Bv v a t 0.545 m/sB v


Let y be positive downward for both blocks.

Constraint of cable: constantA By y

0A Ba a or B Aa a

For blocks A and B, :F ma

Block A: AA A

WW T a

g or A

A AW

T W ag

Block B: B BB B A

W WP W T a a

g g

A BB A A A

W WP W W a a

g g

Solving for aA, A BA

A B

W W Pa g

W W

(1)

2 20 0 0( ) 2 [ ( ) ] with ( ) 0A A A A A Av v a y y v

02 [ ( ) ]A A A Av a y y (2)

0 0( ) with ( ) 0A A A Av v a t v

A

A

vt

a (3)

(a) Acceleration of block A.

System (1): 200 lb, 100 lb, 0A BW W P

By formula (1), 1200 100

( ) (32.2)200 100Aa

2

1( ) 10.73 ft/sA a

System (2): 200 lb, 0, 50 lbA BW W P

By formula (1), 2200 100

( ) (32.2)200Aa

22( ) 16.10 ft/sA a

System (3): 2200 lb, 2100 lb, 0A BW W P

By formula (1), 32200 2100

( ) (32.2)2200 2100Aa

2

3( ) 0.749 ft/sA a


(b) 0 at ( ) 10 ft. Use formula (2).A A Av y y

System (1): 1( ) (2)(10.73)(10)Av 1( ) 14.65 ft/sAv

System (2): 2( ) (2)(16.10)(10)Av 2 17.94 ft/sAv

System (3): 3( ) (2)(0.749)(10)Av 3( ) 3.87 ft/sAv

(c) Time at 20 ft/s. Use formula (3).Av

System (1): 120

10.73t 1 1.864 st

System (2): 220

16.10t 2 1.242 st

System (3): 320

0.749t 3 26.7 st

Chapter 12, Solution 28We first check that static equilibrium is not maintained:

( ) ( ) ( )

0.24(5 10)

3.6

A m C m s A CF F m m g

g

g

Since 10g 3.6g,B BW m g . equilibrium is not maintained.

Block A: :y A AF N m g

0.2A k A AF N m g

: 0.2A A A A AF m a T m g m a (1)

Block C: :y C CF N m g

0.2C k C CF N m g

: 0.2x C C C C CF m a T m g m a (2)

Block B: y B BF m a

2B B Bm g T m a (3)

From kinematics:1

( )2B A Ca a a (4)

(a) Tension in cord. Given data: 5 kg

10 kgA

B C

m

m m

Eq. (1): 0.2(5) 5 AT g a 0.2 0.2Aa T g (5)

Eq. (2): 0.2(10) 10 CT g a 0.1 0.2Ca T g (6)

Eq. (3): 10 2 10 Bg T a 0.2Ba g T (7)

Substitute into (4):

2

10.2 (0.2 0.2 0.1 0.2 )

224 24

1.2 0.35 (9.81 m/s )7 7

g T T g T g

g T T g

33.6 NT

(b) Substitute for T into (5), (7), and (6):

2240.2 0.2 0.4857(9.81 m/s )

7Aa g g

24.76 m/sA a

2240.2 0.3143(9.81 m/s )

7Ba g g

23.08 m/sB a

2240.1 0.2 0.14286(9.81 m/s )

7Ca g g

21.401 m/sC a


A:

B:

C:

D:

Note: As shown, the system is in equilibrium.

From the diagram:

Cord 1: 2 2 constantA B Cy y y

Then 2 2 0A B Cv v v

and 2 2 0A B Ca a a (1)

Cord 2: ( ) ( ) constantD A D By y y y

Then 2 0D A Bv v v

and 2 0D A Ba a a (2)

(a) 1 2: 2 Ay A A A A

WF m a W T T a

g

or 1 220

20 2 AT T ag

(3)

1 2: 2 By B B B B

WF m a W T T a

g

or 1 220

20 2 BT T ag

(4)

Note: Eqs. (3) and (4) A B a a

Then Eq. (1) 4C Aa a

Eq. (2) D Aa a

1: Cy C C C C

WF m a W T a

g

or 1 14 1 CaT

g

414 1 Aa

g

(5)

2 ext: 2 ( ) Dy D D D D D

WF m a W T F a

g

demerssborovc

Sticky Note

There will not be anything this complicated on the exam. I include it for your learning only.


or 21

16 1 242

20 8

D

A

aT

g

a

g

Substituting for T1 [Eq. (5)] and T2 [Eq. (6)] in Eq. (3)

4 2020 2 14 1 20 8A A

Aa a

ag g g

or 2 23 332.2 ft/s 2.76 ft/s

35 35Aa g

22.76 ft/sA B D a a a

and 24(2.76 ft/s )Ca 2or 11.04 ft/sC a

(b) Substituting into Eq. (5)

14 2.76

14 132.2

T

1or 18.80 lbT


A:

B:

Note: As shown, the system is in equilibrium.From the diagram:Cord 1: 2 2 constantA B Cy y y

Then 2 2 0A B Cv v v

and 2 2 0A B Ca a a (1)Cord 2: ( ) ( ) constantD A D By y y y

Then 2 0D A Bv v v

and 2 0D A Ba a a (2)We determine the accelerations of blocks A, C, and D, using theblocks as free bodies.

1 2: 2 Ay A A A A

WF m a W T T a

g

or 1 220

20 2 AT T ag

(3)

1 2 ext: 2 ( ) By B B B B B

WF m a W T T F a

g

or 1 220

20 2 10 BT T ag

(4)

Forming20

(3) (4) 10 ( )A Ba ag

or1

2B Aa a g

C:

Then Eq. (1):1

2 2 02A A Ca a g a

or 4C Aa a g

Eq. (2):1

2 02D A Aa a a g

or1

4D Aa a g

1: Cy C C C C

WF m a W T a

g

or 11

14 1 14 1 ( 4 ) 28 1 2C AA

a aT a g

g g g

(5)

2: 2 Dy D D D D

WF m a W T a

g

demerssborovc

Sticky Note

I will not include anything this complicated on the exam. This is included for your learning only.

D: PROBLEM 12.31 (Continued)

or 21 1 1 3

16 1 8 1 82 4 4

D AA

a aT a g

g g g

(6)

Substituting for 1T [Eq. (5)] and 2T [Eq. (6)] in Eq. (3)

3 2020 2 28 1 2 8

4A A

Aa a

ag g g

or 2 23 3(32.2 ft/s ) 6.90 ft/s

14 14Aa g

Then 2 2 2

2 2 2

4( 6.90 ft/s ) 32.2 ft/s 4.60 ft/s

16.90 ft/s (32.2 ft/s ) 1.15 ft/s

4

C

D

a

a

Note: We have uniformly accelerated motion, so that

0v a t (a) We have /D A D A v v v

or 2/ [1.15 ( 6.90)] ft/s 3 sD A D Aa t a t v

or / 24.2 ft/sD A v

(b) And /C D C D v v v

or 2/ ( 4.60 1.15) ft/s 3 sC D C Da t a t v

or / 17.25 ft/sC D v


(a) First we note / ,B A B A a a a where /B Aa is directed along the inclined surface of A.

B: /: sin 25 cos 25 x B x B B A B B AF m a P W m a m a

or /225 15 sin 25 15( cos 25 )A B Ag a a

or /15 sin 25 cos 25A B Ag a a (1)

: cos 25 sin 25y B y AB B B AF m a N W m a

or 15( cos 25 sin 25 )AB AN g a

A: : cos 25 sin 25 x A A AB A AF m a P P N m a

or [25 225(1 cos 25 )] / sin 25AB AN a

Equating the two expressions for ABN

25 225(1 cos 25 )15( cos 25 sin 25 )

sin 25A

Aa

g a

or2

2

3(9.81)cos 25 sin 25 45(1 cos 25 )

5 3sin 25

2.7979 m/s

Aa

22.80 m/sA a

(b) From Eq. (1)

/ 15 (9.81)sin 25 2.7979cos 25B Aa

or 2/ 8.32 m/sB A a 25

B:

A:

demerssborovc

Sticky Note

Make sure you can calculate the Accel of block B (aB), just like we did in class with this problem.


(a) Panel: Force exerted by counterweightF

:xF ma 40

T F ag

(1)

Counterweight A: Its acceleration has two components

/A P A P a a a a a

25:x xF ma F a

g (2)

25: 25g gF ma T a

g (3)

Adding (1), (2), and (3):

T 25F F T 40 25 25

ag

25 25(32.2)

90 90a g 28.94 ft/s a

Substituting for a into (3):25 25 625

25 2590 90

T g Tg

18.06 lbT


(b) Panel:

:yF ma 40

T ag

(1)

Counterweight A:

:yF ma 25

25 T ag

(2)

Adding (1) and (2): T 25 T 40 25

ag

25

65a g 212.38 ft/s a

Substituting for a into (1):

40 25 1000

65 65T g

g

15.38 lbT

(c) Since panel is accelerated to the left, there is no force exerted by panel on counterweight and viceversa.

Panel:

:xF ma 40

T ag

(1)

Counterweight A: Same free body as in Part (b):

:yF ma 25

25 T ag

(2)

Since Eqs. (1) and (2) are the same as in (b), we get the same answers:

212.38 ft/s ; 15.38 lbT a

wheels



(a) For maximum acceleration

max 0.8(0.62 )

0.496 0.496 mgF s FF F N W

W

Now :x FF ma F ma

or 0.496 mg ma

Then 2 20.496(9.81 m/s ) 4.86576 m/sa

Since a is constant, we have

2 0 2 ( 0)v a x

When 2 2max400 m: 2(4.86576 m/s )(400 m)x v

or max 62.391 m/sv

or max 225 km/hv

(b) For maximum acceleration

max 0.8(0.43 )

0.344 0.344 mgR s RF F N W

W

Now :x RF ma F ma

or 0.344 mg ma

Then 2 20.344(9.81 m/s ) 3.37464 m/sa

Since a is constant, we have

2 0 2 ( 0)v a x

When 2 2max400 m: 2(3.37464 m/s )(400 m)x v

or max 51.959 m/sv

or max 187.1 km/hv


First consider when the bus is on the level section of the highway.

2level 3 ft/sa

We have level:xW

F ma P ag

Now consider when the bus is on the upgrade.

We have : sin 7xW

F ma P W ag

Substituting for P level sin 7W W

a W ag g

or level

2

2

sin 7

(3 32.2 sin 7 ) ft/s

0.92419 ft/s

a a g

For the uniformly decelerated motion

2 20 upgrade upgrade( ) 2 ( 0)v v a x

Noting that 60 mi/h 88 ft/s, then when 05

50 mi/h ,6

v v

we have

22 2

upgrade5

88 ft/s (88 ft/s) 2( 0.92419 ft/s )6

x

or upgrade 1280.16 ftx

or upgrade 0.242 mix


Assume uniformly decelerated motion in all cases.

For braking on the level surface,

0

0

2 20 0

2 20

0

2

2

60 mph 88 ft/s, 0

150 ft

2 ( )

2( )

0 (88)

(2)(150)

25.813 ft/s

f

f

f f

f

f

v v

x x

v v a x x

v va

x x

Braking force.

25.813

32.20.80166

bF maW

ag

W

W

(a) Going up a 5° incline.

F ma

2

2 20

0

sin 5

sin 5

(0.80166 sin 5 )(32.2)

28.62 ft/s

2

b

b

ff

WF W a

g

F Wa g

W

v vx x

a

20 (88)

(2)( 28.62)

0 135.3 ftfx x


(b) Going down a 3 percent incline.

3tan 1.71835

100

sin

(0.80166 sin )(32.2)

24.848 ft/s

bW

F W ag

a

2

00 (88)

(2)( 24.848)fx x

0 155.8 ftfx x


(a) totalbr trac br trl: ( ) ( )x

WF ma F F a

g

or2

2

32.2 ft/s(3600 13,700) lb

(15,000 17,400) lb

17.1932 ft/s

a

For uniformly decelerated motion

2 20 0 02 ( ) 60 mi/h 88 ft/sv v a x x v

When 0:v 2 20 (88 ft/s) 2( 17.1932 ft/s )( )x

or 225 ftx

(b) trltrl br trl hitch: ( )x

WF m a F P a

g

Then 2hitch 2

17,400 lb13,700 lb ( 17.1932 ft/s )

32.2 ft/sP

or hitch 44.10 lb (tension)P

work & energy - cables



(a) Kinematics: 2

2B A

B A

x x

v v

A and B. Assume B moves down.

1

1

2 22

22

22

1 2

1 2

1 2

0

0

1 1

2 2

1(2 kg)

2 4

5

4(cos30 )( ) (cos30 )

2 m

1 m

3(2)(9.81) [ 1 2]

2

16.99 J

A A B B

BB

B

A A B B

B

A

v

T

T m v m v

vv

T v

U m g x m g x

x

x

U

U

Since work is positive, block B does move down.

1 1 2 2

2

2

50 16.99

4

13.59

B

B

T U T

v

v

3.69 m/sB v 60°

(b) B alone.

1

1

2

0

0

3.69 m/s, (from ( ))

v

T

v a

2 22 2

1 2

21 2

1 2

1 1(2)(3.69) 13.59 J

2 2( )(cos30 )( ) ( )( )

3(2 kg)(9.81 m/s ) ( ) (2 m)

2

33.98 2

B

B B B

T m v

U m g x T x

U T

U T


1 1 2 2 0 33.98 2 13.59

2 33.98 13.59 20.39

T U T T

T

10.19 NT


Check at to see if blocks move. With motion impending at B downward, determine required friction forceat A for equilibrium.

Block B:

( )(sin 30 ) 0

1(2 )

2

B B

B

F N m g

N g g

( )(cos30 ) ( ) 0B B fF T m g F

( ) (0.30)( )

(2 ) 3/2 (0.30)

( )3 0.30

B f s BF N g

T g g

T g

Block A:

( )(sin 30 ) 0

1(2 )

2

A A

A

F N m g

N g g

2 ( )(cos30 ) ( ) 0A A fF T m g F

( ) 2 (2 ) 3/2A fF T g

Substituting T from into

( ) (2) ( ) 33 0.30A fF g g

Requirement for equilibrium ( ) 1.1323 0.60A fF g g

Maximum friction that can be developed at 0.3s AA N g

Since 0.3 1.132 ,g g, blocks move.


(a) A and B.( ) (0.20)

( ) (0.20)

A f k B

A f k A

F N g

F N g

Kinematics: 2

2B A

B A

x x

v v

1

1

22 2 2

2

22

0

0

1 1 1(2 kg)

2 2 2 4

5

4

BA A B B B

B

v

T

vT m v m v v

T v

1 2

1 2

2

1 2

1 1 2 2

(cos30 )( ) (cos30 )

( ) ( ) ( ) ( )

2 m, 1 m

3 3[ (2kg) (1 m) (2 kg) (2m)2 2

(0.20)(1 m) (0.20)(2 m)][9.81 m/s ]

[(1.732) (0.6)][9.81] 11.105 J

A A B B

A f A B f B

B A

U m g x m g x

F x F x

x x

U

U

T U T

2

2

0 11.105 1.25

8.88

B

B

v

v

2.98 m/s B v 60

(b) B alone.

1 1

2

2 22

2

1 2

21 2

1 2

1 1 2 2

0 02.98 m/s (from ( ))

1 1(2)(2.98)

2 28.88 J

sin 30 9 N(cos30 )(2) ( )(2) ( ) (2)

3(2 kg)(9.81 m/s ) (2m) 2 (0.2)(9 N)(2 m)

2

2 3 2 0.6

0 2 3

B B

B B

B B f

v T

v a

T m v

T

N m g

U m g T F

U T

U g T g

T U T g

2 0.4 8.88

2 (2 3 0.4)( ) 8.88 21.179

T g

T g

10.59 NT


Kinematics: 2A Bv v23 kg (3 kg) (9.81 m/s )

29.43 NA A A

A

m W m g

W

(a) 150–N force acts through entire 0.6 m motion of B.

1 2

2 21 2 2 2

(150 N)(0.6 m) (29.43 N)(1.2 m) 54.68 J

1 10 ( ) ( )

2 2A A B B

U

T T m v m v

2 2 22 2 2

1 1(3 kg) 2( ) (8 kg)( ) 10( )

2 2B B Bv v v

21 1 2 2 2: 0 54.68 10( )BT U T v

2( ) 2.338 m/sBv 2.34 m/sB v

(b) Initial and final velocities are zero.

1 20 0T T

Remove 150-N force after B moves distance d.

1 2 (150 N) (29.43 N)(1.2 m)

150 35.316 J

U d

d

1 1 2 2: 0 150 31.316 0T U T d 0.2354 md

235 mmd


Energy dissipated.

(a) 1 1

2

22 2

22

0 0

3 m / s

1( )

216

kg (3 m/s) 72 J2

A B

A B

v T

v v v

T m m v

T

1 2

21 2

1 2

(2) (2)

(6 kg)(9.81 m/s (2 m)

117.72

A B p

p

p

U m g m g E

U E

U E

1 1 2 2

0 117.72 72p

T U T

E

117.72 72pE 45.7 JpE

(b) Block A:

2 21 2 2

21 2

1 2

1 110 kg (3 m/s) 49.5 J

2 2

( )(2) [(11 kg)(9.81 m/s ) ][2 m]

215.82 2

A

A A A

A

T T m v

U m g T T

U T

1 1 2 2 0 215.82 2 49.5AT U T T 83.2 NAT

Block B:

1 1 2 2T U T

2 21 2 2

21 2

1 2

1 50 kg (3 m/s) 22.5 J

2 2

(2) (2) (5 kg)(9.81 m/s )(2 m) 2

98.1 2

B

B B B

B

T T m v

U m g T T

U T

0 98.1 2 22.5BT 60.3 NBT


Kinematics:

2

2B A

B A

x x

v v

(a) Blocks A and B.

1

2 22

2 22

2 2

22

0

1 1

2 21

(20 lb/32.2 ft /s )(2 )2

1 (40 lb/32.2 ft/s )( )

260

( )32.2

B B A A

A

A

A

T

T m v m v

T v

v

T v

1 2

1 2

1 2

(100)( ) ( )( ) ( )( )

(100 lb)(2 ft) (40 lb)(2 ft) (20 lb)(4 ft)

200 80 80 200 lb ft

A A A B BU x W x W x

U

U

21 1 2 2

2

600 200

32.2

107.33

A

A

T U T v

v

10.36 ft /sA v

(b) Since the 20-lb weight at B is replaced by a 20-lb force, the kinetic energy at is

2 22

1

1 1 40

2 2

0

A A AT m v vg

T

The work done is the same as in Part (a).

1 2 200 lb ftU

1 1 2 2T U T

2

2

200 200

322

A

A

vg

v

17.94 ft /sA v


Position to Position .

1 10 0v T

At before C is removed from the system

22 2

2 22 2 2

1 2

1 2

2

1 2

1 1 2 2

1( )

21

(12 kg) 62( ) (0.9 m)

(4 3 5)( )(0.9 m)

(2 kg)(9.81 m/s )(0.9 m)

17.658 J

A B C

A C B

T m m m v

T v v

U m m m g

U g

U

T U T

22

22

0 17.658 6

2.943

v

v

At Position , collar C is removed from the system.

Position to Position .

22 2

2

2 23 3 3

1 9( ) kg (2.943)

2 2

13.244 J

1 9( )( )

2 2

A B

A B

T m m v

T

T m m v v

2 3

2

2 3

2 2 3 3

23

23

3

( )( )(0.7 m)

( 1 kg)(9.81 m/s )(0.7 m)

6.867 J

13.244 6.867 4.5

1.417

1.190 m/s

A B

A

U m m g

U

T U T

v

v

v v

1.190 m/sAv


(a) At the initial Position , the force in the spring equals the weight of both blocks, i.e., 5g N, thusat a distance x, the force in the spring is

5

5 40s

s

F g kx

F g x

Maximum velocity of the 2 kg block occurs while the spring is still in contact with the block.

2 2 21 2

21 2

0

1 10 (2 kg)( )

2 2

(5 40 ) 2 3 20x

T T mv v v

U g x dx gx gx x

2 2

1 1 2 2 0 3 20T U T gk x v (1)

Maximum v when 0

3 40

3 (max ) m

40

dv

dxg x

gx v

Substituting in (1) 2 (max ) 0.7358 mx v

2 2max (3)(9.81)(0.7358) (20)(0.7358)

21.65 10.83

10.82

v

max 3.29 m/sv


(b) 0

0

1 3

initial compression of the spring

(2 3 )m

40 85 40

0 0s

x

g g gx

F g x

T T

/8

1 30

2 2

1 3

2

1 1 3 3

(5 40 ) 2

5 202

8 64

200 2 0

6410 (10)(9.81)

64 64

gU g x dx gh

g gU gh

gT U T gh

gh

1.533 mh

work & energy



(a) Up the plane, from A to C, 0.Cv

21, 0

2( sin15 )(10 m)

A A C

A C

T mv T

U W F

0: cos1 5 0F N W

cos1 5N W

2

2

2

0.12 cos1 5

(sin15 0.12cos15 )(10 m)

1(sin15 0.12cos15 )(10 m)

2

(2)(9.81)(sin15 0.12cos15 )(10 m)

73.5

k

A C

A A C C A

A

A

F N W

U W

WT U T v W

g

v

v

8.57 m/sA v 15

(b) Down the plane from C to A.

2

2

2

2

10 ( sin15 )10

2( reverses direction.)

10 (sin15 0.12cos15 )(10 m)

2

(2)(9.81)(sin15 0.12cos15 )(10 m)

28.039

C A A C A

C C A A A

A

A

T T mv U W F

F

T U T W mv

v

v

5.30 m/sA v 15°


(a) Up the plane from A to B.

2 21 1(8 m/s) 32 0

2 2

( sin15 ) 0.12 N

A A B

A B k

W WT mv T

g g

U W F d F N

0 cos15 0 cos15F N W N W

(sin15 0.12cos15 ) (0.3747)

32 (0.3743) 0

A B

A A B B

U W d Wd

WT U T Wd

g

32

(9.81)(0.3747)d 8.70 md

(b) Down the plane from B to A (F reverses direction).

2

2

10 8.72 m/s

2

( sin15 )

(sin15 0.12cos15 )(8.70 m/s)

1.245

10 1.245

2

A A B

B A

B A

B B A A A

WT v T d

g

U W F d

W

U W

WT U T W v

g

2 (2)(9.81)(1.245)

253.9

4.94 m/s

A

A

v

v

4.94 m/sA v 15


20

2 2

1

21 1

(8 ft/s)2 232

( sin15 )(20 ft)

A

B B

B

A B k

T mv

T mv m

T m

U W N

0 cos15 0

cos15

F N W

N W

20

(sin15 0.40cos15 )(20 ft)

(2.551)( ) 2.551

12.551 32

2

A B

A B

A A B B

U W

U W mg

T U T

mv mg m

2 20

20

(2)(32 (2.551)(32.2 ft/s ))

228.29

v

v

0 15.11 ft/sv 15


20

10

2( sin15 )(20 ft)

A B

A B k

T mv T

U W N

0 cos15 0F N

cos15N W

(sin15 0.40cos15 )(20 ft)

(2.551)( ) 2.551A B

A B

A A B B

U W

U W mg

T U T

20

12.551 0

2mv mg

2 20

20

(2)(2.551)(32.2 ft/s )

164.28

v

v

0 12.82 ft/sv 15°


On incline AB: cos30

0.25 cos 30

sin 30

(sin 30 cos 30 )

AB

AB k AB

A B AB

k

N mg

F N mg

U mg d F d

mg d

On level surface BC: 7 m

BC BC

BC k

B C k BC

N mg x

F mg

U mg x

At A, 21and 1 m/s

2A A AT mv v

At C, 21and 2 m/s

2C C CT mv v

Assume that no energy is lost at the corner B.

Work and energy. A A B B C CT U U T

2 20

1 1(sin 30 cos30 )

2 2A k k BCmv mg d mg x mv

Dividing by m and solving for d,

2 2

2 2

/2 /2

(sin 30 cos30 )

(2) /(2)(9.81) (0.25)(7) (1) /(2)(9.81)

sin 30 0.25cos30

C k BC A

k

v g x v gd

6.71 md


(a) On incline AB:

cos 30

0.25 cos 30

sin 30

(sin 30 cos 30 )

AB

AB k AB

A B AB

k

N mg

F N mg

U mg d F d

mg d

On level surface BC: 7 m

BC BC

BC k

B C k BC

N mg x

F mg

U mg x

At A, 21and 1 m/s

2A A AT mv v

At C, 21and 2 m/s

2C C CT mv v

Assume that no energy is lost at the corner B.

Work and energy. A A B B C CT U U T

2 20

1 1(sin 30 cos30 )

2 2A k k BCmv mg d mg x mv

Solving for 2Cv ,

2 2

2

2 (sin 30 cos30 ) 2

(1) (2)(9.81)(7.5)(sin 30 0.25cos30 ) (2)(0.25)(9.81)(7)

C A k k BCv v gd g x

2 28.3811 m /s 2.90 m/sCv

(b) Box on belt: Let beltx be the distance moved by a package as it slides on the belt.

0y y

x k k

F ma N mg N mg

F N mg

At the end of sliding, belt 2 m/sv v

Principle of work and energy.

2 2belt belt

2 2belt

belt

2

1 1

2 2

2

8.3811 (2)

(2)(0.25)(9.81)

C k

C

k

mv mg x mv

v vx

g

belt 0.893 mx

work & energy - springs



(a) 0 1Find initial position of the block at .x

0 0

0

12 lb/m 144 lb/ft

1At , 20 lb 20 lb (144 lb/ft)

20/144 0.1389 fts s

k

F F kx x

x

2 21 2 2 22

22 2

1 1 10 lb0,

2 2 32.2 ft/s

0.1553

WT T v v

g

T v

0

0

0

1 2 0

02

1 2 0

21 2

1 2

( ) ( ); 144 ( )

144( ) ( ) ( ) (0.4)(10) 4 lb

2

(72 lb/ft)(0.1389 ft) (4 lb)( 0.1389 ft)

1.389 0.5556 0.8335 lb ft

s f k s f k kx

f k f k

x

U F dx F x F kx x F N

xU F x F

U

U

21 1 2 2 2

22

2

0 0.8335 0.1553

5.367

2.32 ft/s

T U T v

v

v

At original position, 2.32 ft/sv

(b) 2 2 .For any position at a distance to the right of the orignal positionx 2 2

1 2 2 21

0 ( ) 0.1553 2

WT T v v

g

0 0

0

1 2' 0

2

1 2' 0

( ) 0.1389

144( ) ( ) ( ) 4 lb

2

x x

s f kx k

x

f k f k

x

U F dx F dx x

xU F x x F

2 21 1 2' 2

22'

0 (72 lb/ft)[(0.1389) ]

(4 lb)( 0.1389) 0.1553

T U T x

x v


Max v, when2' 0

144 4 0

dv

dxx

Max v, when 0.027778 mx

2 2 2max0.1553 (72)[(0.1389) (0.02778) ] (4)(0.02778 0.1389)v

2max

2max

0.1553 1.3336 0.4445 0.8891

5.725

v

v

max 2.39 ft/sv


(a) At the initial Position , the force in the spring equals the weight of both blocks, i.e., 5g N, thusat a distance x, the force in the spring is

5

5 40s

s

F g kx

F g x

Maximum velocity of the 2 kg block occurs while the spring is still in contact with the block.

2 2 21 2

21 2

0

1 10 (2 kg)( )

2 2

(5 40 ) 2 3 20x

T T mv v v

U g x dx gx gx x

2 2

1 1 2 2 0 3 20T U T gk x v (1)

Maximum v when 0

3 40

3 (max ) m

40

dv

dxg x

gx v

Substituting in (1) 2 (max ) 0.7358 mx v

2 2max (3)(9.81)(0.7358) (20)(0.7358)

21.65 10.83

10.82

v

max 3.29 m/sv


(b) 0

0

1 3

initial compression of the spring

(2 3 )m

40 85 40

0 0s

x

g g gx

F g x

T T

/8

1 30

2 2

1 3

2

1 1 3 3

(5 40 ) 2

5 202

8 64

200 2 0

6410 (10)(9.81)

64 64

gU g x dx gh

g gU gh

gT U T gh

gh

1.533 mh


(a) See solution to (a) of Problem 13.28. max 3.29 m/sv

(b) Refer to figure in (b) of Problem 13.28.

1 3

1 30

0 0

(5 40 ) 2h

T T

U g x dx gh

Since the spring remains attached to the 2 kg block, the integration must be carried out throughout the totaldistance h.

21 1 3 2 0 5 20 2 0

3 (3)(9.81)

20 20

T U T gh h gh

gh

1.472 mh


(a) 65 mi/h 95.3 ft/s

Assume auto stops in 5 14 ft.d, ,

1

2 21 1 2

1

2

2

1 2

1 1 2 2

95.33 ft/s

1 1 2250 lb(95.3 ft/s)

2 2 32.2 ft/s

317,530 lb ft

317.63 k ft

0

0

(18 k)(5 ft) (27 k)( 5)

90 27 135

27 45 k ft

317.53 27 45

v

T mv

T

v

T

U d

d

d

T U T

d

13.43 ftd

Assumption that 14 ftd , is ok.

(b) Maximum deceleration occurs when F is largest. For 13.43 ft, 27 k.d F Thus, DF ma

2

2250 lb(27,000 lb) ( )

32.2 ft/sDa

2386 ft/sDa

work & energy wheels



0.35 (0.35)(100 kips) 35 kips

(0.35)(80 kips) 28 kipsk B

C

F

F

1 30 mi/h 44 ft/sv 2 20 0v T

(a) Entire train: 1 1 2 2T U T

22

1 (80 kips 100 kips 80 kips)(44 ft/s) (28 kips 35 kips) 0

2 32.2 ft/sx

124.07 ftx 124.1 ftx

(b) Force in each coupling: Recall that 124.07 ft.x

Car A: Assume ABF to be in tension.

1 1 2 2

21 80 kips(44) (124.07 ft) 0

2 32.219.38 kips

AB

AB

T V T

F

F

19.38 kips (tension)ABF

Car C: 1 1 2 2T U T

21 80 kips(44) ( 28 kips)(124.07 ft) 0

2 32.228 kips 19.38 kips

BC

BC

F

F

8.62 kipsBCF 8.62 kips (tension)BCF


(a) Entire train:

1

(0.35)(80 kips) 28 kips

30 mi/h 44 ft/s A AF N

v

2 20 0v T

1 1 2 2

22

1 (80 kips 100 kips 80 kips)(44 ft/s) (28 kips) 0

2 32.2 ft/s

T v T

x

279.1 ftx 279 ftx

(b) Force in each coupling:

Car A: Assume ABF to be in tension.

1 1 2 2T v T

22

1 80 kips(44 ft/s) (28 kips )(279.1 ft) 0

2 32.2 ft/sABF

28 kips 8.62 kipsABF

19.38 kipsABF 19.38 kips (compression)ABF

Car C: 1 1 2 2T v T

22

1 80 kips(44 ft/s) (279.1 ft) 0

2 32.2 ft/sBCF

8.617 kipsBCF 8.62 kips (compression)BCF

Chapter 13, Solution 17Initial speed: 1 108 km/h 30 m/sv

Final speed: 2 72 km/h 20 m/sv

Vertical drop: (0.02)(300) 6.00 mh

Braking distance: 300 md

(a) Braking force. Use cab and trailer as a free body.

3

1800 5400

7200 kg

(7200)(9.81)

70.632 10 N

m

W mg

Work and energy: 1 1 2 2T U T

2 21 2

2 21 2

2 3 2

1 1

2 21 1 1

2 2

1 1 1(7200)(30) (70.632 10 )(6.00) (7200)(20)

300 2 2

b

b

mv Wh F d v

F mv Wh mvd

37.4126 10 N 7.41 kNbF

(b) Coupling force .cF Use the trailer alone as a free body.

Braking force: 3

3

3

(0.70)(7.4126 10 )

5.1888 10 N

5400 kg

(5400)(9.81)

52.974 10 N

bF

m

W mg

Work and energy: 1 1 2 2

2 21 2

1 1

2 2b c

T U T

mv Wh F d F d mv

The plus sign before cF means that we have assumed that the coupling is in tension.

2 21 1

3 3 22

1 1 1

2 2

1 115.1888 10 (52.974 10 )(6.00) (5400)(20)(5400)(30)

300 22

c bF F mv Wh mvd

33.7068 10 N 3.71 kN (compression)cF


Initial speed: 1 72 km/h 20 m/sv

Final speed: 2 108 km/h 30 m/sv

Vertical rise: (0.02)(300) 6.00 mh

Distance traveled: 300 md

(a) Traction force. Use cab and trailer as a free body.

3

1800 5400

7200 kg

(7200)(9.81)

70.632 10 N

m

W mg


2 21 2

2 21 1

2 3 2

1 1

2 21 1

2

1 1 1(7200)(30) (70.632 10 )(6.00) (7200)(20)

300 2 2

t

t

mv Wh F d mv

F mv Wh mvd

37.4126 10 N 7.41 kNtF

(b) Coupling force .cF Use the trailer alone as a free body.

3

5400 kg

(5400)(9.81)

52.974 10 N

m

W mg

Assume that the tangential force at the trailer wheels is zero.


2 21 2

1 1

2 2cmv Wh F d mv

The plus sign before cF means that we have assumed that the coupling is in tension.

2 22 1

2 3 2

1 1 1

2 2

1 1 1(5400)(30) (52.974 10 )(6.00) (5400)(20)

300 2 2

cF mv Wh mvd

35.5595 10 N 5.56 kN (tension)cF

vector mechanics for engineers - 9th edition

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