Solution Manual - Vector Mechanics Engineers Dynamics 8th Beer Chapter 11 PDF

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COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 2. 2 x t3 (t 2) m (a) dx 3t 2 2 ( t 2 ) dt dv 6t 2 2 dt Time at a 0. 0 6t0 2 0 t0 (b) 1 3 t0 s W Corresponding position and velocity. 3 2 x 2 2 m x 2 m W 2 v 3 2 2 Vector Mechanics for Engineers: Statics and Dynamics, ...

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Chapter 11, Solution 2. 2

x = t3 − ( t − 2) m

(a)

v=

dx = 3t 2 − 2 ( t − 2) m/s dt

a=

dv = 6t − 2 m/s 2 dt

Time at a = 0. 0 = 6t0 − 2 = 0 t0 =

(b)

1 3

t0 = 0.333 s W

Corresponding position and velocity. 3

2

⎛1 ⎞ ⎛1 ⎞ x = ⎜ ⎟ − ⎜ − 2⎟ = − 2.741 m ⎝3⎠ ⎝3 ⎠

x = − 2.74 m W

2

⎛1 ⎞ ⎛1 ⎞ v = 3 ⎜ ⎟ − 2 ⎜ − 2 ⎟ = 3.666 m/s ⎝3 ⎠ ⎝3 ⎠

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

v = 3.67 m/s W

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Chapter 11, Solution 3. Position:

x = 5t 4 − 4t 3 + 3t − 2 ft

Velocity:

v =

dx = 20t 3 − 12t 2 + 3 ft/s dt

Acceleration:

a=

dv = 60t 2 − 24t ft/s2 dt

When t = 2 s, 4

3

x = ( 5)( 2 ) − ( 4 )( 2 ) − ( 3 )( 2 ) − 2 3

x = 52 ft W

2

v = 115 ft/s W

a = ( 60 )( 2 ) − ( 24 )( 2 )

a = 192 ft/s2 W

v = ( 20 )( 2 ) − (12 )( 2 ) + 3 2

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 11, Solution 4. Position:

x = 6t 4 + 8 t 3 − 14t 2 − 10 t + 16 in.

Velocity:

v=

dx = 24t 3 + 24 t 2 − 28t − 10 in./s dt

Acceleration:

a=

dv = 72 t 2+ 48 t − 28 in./s 2 dt

When t = 3 s, 4

3

2

x = ( 6 )( 3 ) + (8 )( 3 ) − (14 )(3 ) − (10 )(3 ) + 16 3

2

v = ( 24 )(3 ) + ( 24 )( 3 ) − ( 28 )( 3) − 10 2

a = ( 72 )( 3) + ( 48 )( 3) − 28

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

x = 562 in. ! v = 770 in./s ! a = 764 in./s 2 !

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Chapter 11, Solution 5. Position:

x = 500sin kt mm

Velocity:

v=

dx = 500k coskt mm/s dt

Acceleration:

a =

dv = − 500k 2sin kt mm /s 2 dt

When t = 0.05 s,

and

k = 10 rad/s kt = (10 )( 0.05 ) = 0.5 rad x = 500sin ( 0.5)

v = (500 )(10 )cos (0.5 ) 2

a = − (500 )(10 ) sin (0.5 )

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

x = 240 mm ! v = 4390 mm/s ! a = − 24.0 × 103 mm/s2 !

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Chapter 11, Solution 6.

(

)

Position:

x = 50sin k1t − k2t 2 mm

Where

k1 =1 rad/s

Let

θ = k1t − k2 t2 = t − 0.5 t2 rad dθ = (1− t ) rad/s dt x = 50sin θ mm

Position:

k2 = 0.5 rad/s2

and

d 2θ = − 1 rad/s2 dt 2

and

dx dθ = 50 cosθ mm/s dt dt dv a= dt

v=

Velocity:

Acceleration:

a = 50 cosθ

2

d 2θ

When v = 0,

⎛dθ ⎞ 2 − 50 sinθ ⎜ ⎟ mm/s 2 dt ⎝ dt ⎠ either cosθ = 0

dθ =1 − t = 0 dt Over 0 ≤ t ≤ 2 s, values of cosθ are:

t = 1s

or

t (s )

0

0.5

1.0

1.5

2.0

θ ( rad )

0

0.375

0.5

0.375

0

cosθ

1.0

0.931

0.878

0.981

1.0

No solutions cosθ = 0 in this range.

For t = 1 s,

2 θ = 1 − ( 0.5)(1) = 0.5 rad

x = 50sin ( 0.5 ) a = 50 cos (0.5 )( −1) − 50sin (0.5 )( 0 )

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

x = 24.0 mm W a = − 43.9 mm/s 2 W

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Chapter 11, Solution 7. Given:

x = t 3 − 6t 2 + 9t + 5

Differentiate twice.

v =

dx = 3t 2− 12t + 9 dt

a =

dv = 6t − 12 dt

(a)

v=0

When velocity is zero.

3t2 − 12t + 9 = 3 ( t − 1 )( t − 3 ) = 0 t = 1 s and t = 3 s W (b)

Position at t = 5 s. 3

2

x5 = ( 5) − ( 6)( 5) + ( 9 )( 5) + 5

x5 = 25 ft W

Acceleration at t = 5 s. a5 = 18 ft/s 2 W

a5 = (6 )(5 ) − 12 Position at t = 0. x0 = 5 ft Over 0 ≤ t < 1 s

x is increasing.

Over 1 s < t < 3 s

x is decreasing.

Over 3 s < t ≤ 5 s

x is increasing.

Position at t = 1 s. 3

2

x1 = (1) − ( 6 )(1) + ( 9 )(1) + 5 = 9 ft

Position at t = 3 s. 3

2

x3 = ( 3) − ( 6)( 3) + ( 9 )( 3) + 5 = 5 ft

Distance traveled. At t = 1 s

d1 = x1 − x0 = 9 − 5 = 4 ft

At t = 3 s

d 3 = d 1 + x3 − x1 = 4 + 5 − 9 = 8 ft

At t = 5 s

d5 = d3 + x5 − x3 = 8 + 25 − 5 = 28 ft d5 = 28 ft W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 11, Solution 8. 3

x = t 2 − ( t − 2 ) ft

v= (a)

dx 2 = 2t − 3( t − 2 ) ft/s dt

Positions at v = 0. 2

2t − 3( t − 2) = − 3t 2 + 14t − 12 = 0

t=

−14 ± (14) 2 − (4)( − 3)( −12) (2)(− 3)

t1 = 1.1315 s and t 2 = 3.535 s

(b)

At t1 = 1.1315 s,

x1 = 1.935 ft

x1 = 1.935 ft W

At t 2 = 3.535 s,

x2 = 8.879 ft

x2 = 8.879 ft W

Total distance traveled. At t = t0 = 0,

x0 = 8 ft

At t = t4 = 4 s,

x4 = 8 ft

Distances traveled. 0 to t1:

d1 = 1.935 − 8 = 6.065 ft

t1 to t2:

d2 = 8.879 − 1.935 = 6.944 ft

t2 to t4 :

d3 = 8 − 8.879 = 0.879 ft

Adding,

d = d1 + d 2 + d3

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

d = 13.89 ft W

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Chapter 11, Solution 9. a = 3e − 0.2 t v

t

∫ 0 dv = ∫ 0 a dt v − 0= ∫

t − 0.2t 3e dt 0

3 e− 0.2t = − 0.2

(

)

(

t

0

v = − 15 e − 0.2t − 1 = 15 1 − e − 0.2t

At t = 0.5 s,

(

v = 15 1 − e − 0.1 x

)

)

v = 1.427 ft/s W

t

∫ 0 dx = ∫ 0 v dt t

t 0

(

x − 0 = 15 ∫ 1 − e

(

−0.2t

)

1 −0.2t ⎞ ⎛ dt = 15 ⎜ t + e ⎟ 0.2 ⎝ ⎠0

x = 15 t + 5e − 0.2t − 5 At t = 0.5 s,

(

)

)

x = 15 0.5 + 5e− 0.1 − 5

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

x = 0.363 ft W

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Chapter 11, Solution 10. Given:

a = −5.4sin kt ft/s2 , t

v0 =1.8 ft/s,

t

v − v0 = ∫ 0 a dt = − 5.4 ∫ 0 sin kt dt = v − 1.8 = Velocity:

t

0

v = 1.8cos kt ft/s t

x −0 =

When t = 0.5 s,

5.4 cos kt k

k = 3 rad/s

5.4 ( cos kt − 1) = 1.8cos kt − 1.8 3

t

x − x0 = ∫ 0 v dt = 1.8 ∫ 0 cos kt dt =

Position:

x0 = 0,

t 1.8 sin kt 0 k

1.8 ( sin kt − 0 ) = 0.6sin kt 3

x = 0.6sinkt ft kt = ( 3)( 0.5) = 1.5 rad v = 1.8cos1.5 = 0.1273 ft/s x = 0.6sin1.5 = 0.5985 ft

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

v = 0.1273 ft/s W x = 0.598 ft W

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Chapter 11, Solution 11. Given:

a = −3.24sin kt − 4.32 cos kt ft/s2 , x 0 = 0.48 ft,

k = 3 rad/s

v 0 = 1.08 ft/s

t

t

t

v − v0 = ∫0 a dt = − 3.24 ∫0 sin kt dt − 4.32 ∫0 cos kt dt v − 1.08 = =

t

3.24 cos kt k

− 0

t

4.32 sin kt k

0

3.24 4.32 ( coskt − 1) − ( sin kt − 0) 3 3

= 1.08cos kt − 1.08 − 1.44 sin kt Velocity:

v = 1.08cos kt − 1.44sin kt ft/s t

t

t

x − x0 = ∫ 0 v dt = 1.08 ∫ 0 cos kt dt − 1.44 ∫ 0 sin kt dt x − 0.48 =

1.08 sin kt k

t

+ 0

1.44 cos kt k

t 0

1.08 ( sin kt − 0 ) + 1.44 ( cos kt − 1) 3 3 = 0.36sin kt + 0.48cos kt − 0.48 =

P osition:

x = 0.36sin kt + 0.48cos kt ft

When t = 0.5 s,

kt = ( 3)(0.5 ) = 1.5 rad v = 1.08cos1.5 − 1.44sin1.5 = −1.360 ft/s x = 0.36sin1.5 + 0.48 cos1.5 = 0.393 ft

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

v = − 1.360 ft/s ! x = 0.393 ft !

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Chapter 11, Solution 12. a = kt mm/s 2

Given: At t = 0,

v = 400 mm/s;

at t =1 s,

where k is a constant.

v = 370 mm/s,

x = 500 mm

1 2 v t t ∫400 dv = ∫0 a dt = ∫0 kt dt = 2 kt v − 400 =

1 2 kt 2

or

v = 400 +

1 2 k (1) = 370, 2

At t = 1 s,

v = 400 +

Thus

v = 400 − 30t2 mm/s v7 = 400 − ( 30)( 7)

At t = 7 s, When v = 0,

400 − 30 2t = 0.

Then 2t = 13.333 s2 ,

1 2 kt 2

k = −60 mm/s3

2

v7 = −1070 mm/s W t = 3.651 s

For 0 ≤ t ≤ 3.651 s,

v >0

and

x is increasing.

For t > 3.651 s,

v 0 and reaches x = 2 m. At x = 2 m, v = 0 and 2 a < 0, so that v becomes negative and x decreases. Thus, x = 2 m is never reached. 3 xmax = 2 m !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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(b) Velocity when total distance traveled is 3 m.

The particle will have traveled total distance d = 3 m when d − xmax = xmax − x or 3 − 2 = 2 − x or x = 1 m. 2

7 4  Using v = − 12  x −  − , which applies when x is decreasing, we get 3 3  2

7 4  v = − 12 1 −  − = − 20 3 3 

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

v = − 4.47 m/s !

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Chapter 11, Solution 21.

(

a = k 1 − e −x

Note that a is a function of x.

(

)

)

Use v dv = a dx = k 1 − e−x dx with the limits v = 9 m/s when x = − 3 m, and v = 0 when x = 0. 0 0 −x ∫ 9 v dv = ∫ − 3 k (1 − e ) dx

⎛ v2 ⎞ ⎜⎜ ⎟⎟ ⎝2 ⎠

0− (a)

0

(

= k x + e− x 9

)

0

−3

92 = k ⎡⎣ 0 + 1 − ( − 3) − e3 ⎤⎦ = − 16.0855k 2 k = 2.52 m/s 2 W

k = 2.5178

(

)

(

)

Use v dv = a dx = k 1 − e− x dx = 2.5178 1 − e− x dx with the limit v = 0 when x = 0. v x −x ∫ 0 v dv = ∫ 0 2.5178( 1 − e ) dx

v2 = 2.5178 x + e − x 2

(

(

)

x

(

)

= 2.5178 x + e− x − 1

0

)

(

v 2 = 5.0356 x + e − x − 1

1/2

)

v = ± 2.2440 x + e− x − 1

(b) Letting x = −2 m,

(

1/ 2

)

v = ± 2.2440 − 2 + e2 − 1

= ± 4.70 m/s

Since x begins at x = −2 m and ends at x = 0, v > 0. Reject the minus sign. v = 4.70 m/s W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 11, Solution 22. a= v

dv = 6.8 e−0.00057 x dx

v x − 0.00057x dx ∫ 0 v dv = ∫ 0 6.8 e

v2 6.8 e − 0.00057x −0 = 2 − 0.00057

(

= 11930 1 − e −0.00057x

x 0

)

When v = 30 m/s.

( 30 ) 2 2

(

= 11930 1 − e − 0.00057x

)

1 − e − 0.00057x = 0.03772 e − 0.00057x = 0.96228 − 0.00057 x = ln (0.96228) = − 0.03845 x = 67.5 m W

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 11, Solution 23. a= v

Given:

dv = − 0.4v dx dv = − 0.4 dx

or

Separate variables and integrate using v = 75 mm/s when x = 0. v

x

∫ 75dv = − 0.4 ∫ 0

v − 75 = − 0.4 x

(a) Distance traveled when v = 0 x = 187.5 mm W

0 − 75 = − 0.4x

(b) Time to reduce velocity to 1% of initial value. v = (0.01)(75) = 0.75 t = − 2.5ln

0.75 75

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

t = 11.51 s W

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Chapter 11, Solution 24. a =v

Given:

dv = − kv2 dx

Separate variables and integrate using v = 9 m/s when x = 0. dv

v x ∫ 9 v = − k ∫ 0 dx

ln

v = − kx 9

Calculate k using v = 7 m/s when x = 13 m. ln Solve for x .

7 = − ( k )(13 ) 9 x=−

k = 19.332 × 10− 3 m − 1

1 v v ln = −51.728 ln 9 9 k

(a) Distance when v = 3 m/s. ⎛ 3⎞ x = −51.728 ln ⎜ ⎟ ⎝9 ⎠

x = 56.8 m W

(b) Distance when v = 0. x = − 51.728 ln (0 )

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

x=∞W

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Chapter 11, Solution 25. v dv = a dx = −k vdx, dx = −

∫

x − x0 =

x dx x0

2 3/2 v − v3/2 3k 0

(

)

x0 = 0,

v0 = 25 ft/s

1 1/2 v dv k

1 v 2 = − ∫ v vdv = − v3/2 0 3k k or

x=

v v0

2⎡ ( 25)3/2 − v3/2 ⎤⎥⎦ = 2 ⎡⎣ 125 − v3/2 ⎤⎦ 3k ⎢⎣ 3k

Noting that x = 6 ft when v = 12 ft/s,

Then,

6 =

2 ⎡ 55.62 125 − 123/2 ⎤ = ⎣ ⎦ k 3k

x =

2 ⎡125 − v 3/2⎤ = 0.071916 125 − v 3/2 ⎦ ( 3)( 9.27 ) ⎣

3 k = 9.27 ft/s

or

(

)

v3/2 = 125 − 13.905x (a) When x = 8 ft,

v3/2 = 125 − (13.905)( 8) = 13.759 ( ft/s)

3/2

v = 5.74 ft/s W (b)

dv = a dt = − k vdt dt = −

t =−

At rest, v = 0

t=

1 dv k v1/ 2 v 1 2 1/2 v − v 1/2 ⋅ 2 ⎡⎣v 1/2⎤⎦ = v0 k k 0

( 2)( 25) 2v1/2 0 = 9.27 k

(

)

1/2

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

t = 1.079 s W

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Chapter 11, Solution 29. x as a function of v. v = 1 − e −0.00057 x 154 2

⎛ v ⎞ e −0.00057 x = 1 − ⎜ ⎟ ⎝ 154 ⎠ ⎡ − 0.00057 x = ln ⎢1 − ⎣⎢

2 ⎛ v ⎞ ⎤ ⎜ ⎟ ⎥ ⎝ 154 ⎠ ⎥⎦

⎡ ⎛ v ⎞2 ⎤ x = −1754.4 ln ⎢1 − ⎜ ⎟ ⎥ ⎢⎣ ⎝ 154 ⎠ ⎦⎥

(1)

a as a function of x.

(

v2 = 23716 1 − e−0.00057 a =v

)

dv d ⎛ v2 ⎞ = ⎜ ⎟ = (11858 )( 0.00057 ) e−0.0005x dx dx ⎜⎝ 2 ⎟⎠

⎡ a = 6.75906 e −0.00057 x = 6.75906⎢ 1 − ⎢⎣

⎛ v ⎞ ⎜ ⎟ ⎝ 154 ⎠

2⎤

⎥ ⎥⎦

(2)

(a) v = 20 m/s. x = 29.8 m e

From (1),

x = 29.843

From (2),

a = 6.64506

a = 6.65 m/s2 e

From (1),

x = 122.54

x = 122.5 m e

From (2),

a = 6.30306

a = 6.30 m/s2 e

(b) v = 40 m/s.

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 11, Solution 30. 0.3

Given: v = 7.5 (1 − 0.04 x)<...