Solution Manual - Vector Mechanics Engineers Dynamics 8th Beer Chapter 04 PDF

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Chapter 4, Solution 1. Free-Body Diagram:

(a)

ΣM B = 0:

− Ay ( 3.6 ft ) − ( 146 lb)( 1.44 ft) − ( 63 lb)( 3.24 ft) − ( 90 lb)( 6.24 ft) = 0

Ay = − 271.10 lb (b)

Σ M A = 0:

or A y = 271 lb ⊳

By( 3.6 ft) − (146 lb)( 5.04 ft) − (63 lb)(6.84 ft ) − (90 lb)(9.84 ft ) = 0 B y = 570.10 lb

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.

or B y = 570 lb

⊳

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Chapter 4, Solution 2. Free-Body Diagram:

(a)

ΣM C = 0:

(3.5 kips ) (1.6 + 1.3 + 19.5 cos15o ) ft −

2FB ( 1.6 + 1.3 + 14) ft + ( 9.5 kips)( 1.6 ft) = 0

2 FB = 5.4009 kips or FB = 2.70 kips

⊳

(b) ΣM B = 0:

( 3.5 kips) (19.5cos15 o − 14 )ft  − (9.5 kips )  14 (

+1.3 )ft  + 2 FC 14 ( +1.3 +1.6 )ft  = 0

2 FC = 7.5991 kips, or or FC = 3.80 kips

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 4, Solution 3. Free-Body Diagram:

(a)

ΣM K = 0:

( 25 kN)( 5.4 m) + ( 3 kN)( 3.4 m) − 2FH ( 2.5 m) + ( 50 kN )( 0.5 m) = 0 2FH = 68.080 kN

(b)

ΣM H = 0:

or FH = 34.0 kN

⊳

( 25 kN)( 2.9 m) + ( 3 kN )( 0.9 m ) − ( 50 kN )( 2.0 m) + 2FK ( 2.5 m ) = 0 2FK = 9.9200 kN

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.

or FK = 4.96 kN

⊳

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Chapter 4, Solution 4. Free-Body Diagram: (boom)

(a)

ΣM B = 0:

( 25 kN)( 2.6 m) + ( 3 kN)( 0.6 m ) − ( 25 kN)( 0.4 m) − TCD ( 0.7 m) = 0 or TCD = 81.1 kN ⊳

TCD = 81.143 kN (b)

ΣFx = 0:

Bx = 0 so that B = By

ΣFy = 0:

( −25 − 3 − 25 − 81.143) kN

+B =0

B = 134.143 kN

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.

or B = 134.1 kN

⊳

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Chapter 4, Solution 5. Free-Body Diagram:

a1 = ( 20 in. ) sin α − ( 8 in.) cos α a 2 = ( 32 in.) cosα − ( 20 in.) sin α b = ( 64 in.) cosα From free-body diagram of hand truck

ΣM B = 0: P ( b ) − W ( a2 ) + W ( a1 ) = 0

(1)

ΣFy = 0: P − 2w + 2 B = 0

(2)

α = 35°

For

a1 = 20 sin 35 ° − 8 cos 35° = 4.9183 in. a2 = 32 cos 35 ° − 20 sin 35° = 14.7413 in.

b = 64cos35 ° = 52.426 in. (a)

From Equation (1)

P ( 52.426 in.) − 80 lb (14.7413 in.) + 80 lb (4.9183 in. ) = 0 ∴ P = 14.9896 lb (b)

or P = 14.99 lb ⊳

From Equation (2)

14.9896 lb − 2 (80 lb ) + 2 B = 0 ∴ B = 72.505 lb

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.

or

B = 72.5 lb ⊳

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Chapter 4, Solution 6.

a1 = ( 20 in. ) sin α − ( 8 in.) cos α

Free-Body Diagram:

a 2 = ( 32 in.) cosα − ( 20 in.) sin α b = ( 64 in.) cosα From free-body diagram of hand truck

ΣM B = 0:

P ( b) − W ( a2 ) + W ( a1) = 0 (1)

ΣFy = 0: P − 2w + 2 B = 0

(2)

α = 40°

For

a1 = 20sin 40 ° − 8cos 40 ° = 6.7274 in. a2 = 32 cos 40 ° − 20 sin 40 ° = 11.6577 in.

b = 64cos40 ° = 49.027 in. (a)

From Equation (1)

P ( 49.027 in.) − 80 lb (11.6577 in.) + 80 lb ( 6.7274 in.) = 0 P = 8.0450 lb or P = 8.05 lb ⊳ (b)

From Equation (2)

8.0450 lb − 2( 80 lb) + 2 B = 0 B = 75.9775 lb or B = 76.0 lb ⊳

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 4, Solution 7. Free-Body Diagram:

(a) a = 2.9 m

ΣFx = 0:

ΣM B = 0:

Ax = 0

− ( 12 m) A y + ( 12 − 2.9) m ( 3.9 kN) +  ( 12 − 2.9 − 2.6) m ( 6.3 kN) +  ( 2.8 + 1.45 ) m ( 7.9 kN ) + (1.45 m )( 7.3 kN ) = 0

or

ΣF y = 0: or

Ay = 10.0500 kN

or A = 10.05 kN

⊳

or B = 15.35 kN

⊳

10.0500 kN − 3.9 kN − 6.3 kN − 7.9 kN − 7.3 kN + B y = 0 By = 15.3500 kN

(b) a = 8.1 m

ΣM B = 0:

− ( 12 m) A y +  ( 12 − 8.1) m ( 3.9 kN) +  ( 12 − 8.1− 2.6) m ( 6.3 kN) +  (2.8 + 4.05 ) m  (7.9 kN ) + (4.05 m )(7.3 kN ) = 0

or

ΣF y = 0: or

Ay = 8.9233 kN

or A = 8.92 kN

⊳

8.9233 kN − 3.9 kN − 6.3 kN − 7.9 kN − 7.3 kN + B y = 0 By = 16.4767 kN

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.

or B = 16.48 kN

⊳

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Chapter 4, Solution 8. Free-Body Diagram:

(a)

ΣFx = 0:

ΣM B = 0:

Ax = 0

− (12 m ) A y + (12 m − a )(3.9 kN ) + (12 − 2.6 ) m − a  (6.3 kN ) a a  +  2.8 m +  ( 7.9 kN ) + ( 7.3 kN) = 0 2 2  

or

( 12 m) A y = 128.14 kN⋅ m − ( 10.2 kN) a + ( 15.2 kN) a2 ( 12 m) Ay

= 128.14 kN⋅ m − ( 2.6 kN) a

Thus A y is maximum for the smallest possible value of a:

a = 0⊳ (b) The corresponding value of A y is

( Ay )max = 10.6783 kN, and ΣFy = 0:

or A = 10.68 kN

⊳

or B = 14.72 kN

⊳

10.6783 kN − 3.9 kN − 6.3 kN − 7.9 kN − 7.3 kN + By = 0 B y = 14.7217 kN

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 4, Solution 9. Free-Body Diagram:

For (TC )max , T B = 0

ΣM O = 0:

(TC )max (4.8 in. ) − (80 lb )(2.4 in. ) = 0 (TC ) = 40 lb  > [Tmax = 36 lb ] max 

(TC )max = 36.0 lb For (TC ) min , TB = Tmax = 36 lb ΣM O = 0:

(TC )min ( 4.8 in.) + ( 36 lb)(1.6 in.) − ( 80 lb)( 2.4 in.) = 0 ( TC )min

= 28.0 lb

Therefore:

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.

28.0 lb ≤ TC ≤ 36.0 lb ⊳

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Chapter 4, Solution 10. Free-Body Diagram:

For Qmin , TD = 0 ΣM B = 0:

(7.5 kN )(0.5 m ) − Q min (3 m ) = 0 Qmin = 1.250 kN

For Qmax , TB = 0 ΣM D = 0:

(7.5 kN )( 2.75 m ) − Qmax ( 0.75 m ) = 0 Qmax = 27.5 kN

Therefore:

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.

1.250 kN ≤ Q ≤ 27.5 kN ⊳

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Chapter 4, Solution 11. Free-Body Diagram:

ΣM D = 0:

(7.5 kN )( 2.75 m ) − TB (2.25 m ) + (5 kN )(1.5 m ) − Q (0.75 m ) = 0 Q = ( 37.5 − 3TB ) kN

ΣM B = 0:

(1)

(7.5 kN )(0.5 m ) − (5 kN )(0.75 m ) + TD (2.25 m ) − Q (3 m ) = 0 Q = ( 0.75 TD ) kN

(2)

For the loading to be safe, cables must not be slack and tension must not exceed 12 kN. Thus, making 0 ≤ TB ≤ 12 kN in. (1), we have 1.500 kN ≤ Q ≤ 37.5 kN

(3)

And making 0 ≤ TD ≤ 12 kN in. (2), we have 0 ≤ Q ≤ 9.00 kN (3) and (4) now give:

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.

(4) 1.500 kN ≤ Q ≤ 9.00 kN ⊳

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Chapter 4, Solution 12. Free-Body Diagram:

For (WA ) min , E = 0 ΣM F = 0:

( WA ) min ( 7.5 ft) + ( 9 lb)( 4.8 ft) + ( 28 lb)( 3 ft) − ( 90 lb)( 1.8 ft) = 0 (WA )min = 4.6400 lb

For (W A )max , F = 0 ΣM E = 0:

(WA )max (1.5 ft ) − (9 lb )(1.2 ft ) − ( 28 lb)( 3 ft) − ( 90 lb)( 7.8 ft) =

( WA ) max

0

= 531.20 lb

Thus

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.

4.64 lb ≤ WA ≤ 531 lb ⊳

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Chapter 4, Solution 13. Free-Body Diagram:

ΣM D = 0:

(750 N )(0.1 m − a ) − (750 N )( a + 0.075 m − 0.1 m ) − (125 N )(0.05 m ) + B (0.2 m ) = 0  87.5 N + 0.2B  a=  1500 N  

(1)

Using the bounds on B:

B = −250 N (i.e. 250 N downward) in (1) gives amin = 0.0250 m B = 500 N (i.e. 500 N upward) in (1) gives amax = 0.1250 m

Therefore:

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.

25.0 mm ≤ a ≤ 125.0 mm ⊳

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Chapter 4, Solution 14. Free-Body Diagram:

Note that W = mg is the weight of the crate in the free-body diagram, and that

0 ≤ E y ≤ 2.5 kN ΣFx = 0: ΣM A = 0: or ΣF y = 0: or

Ax = 0 − (1.2 m )(1.2 kN ) − ( 2.0 m )(1.6 kN ) − ( 3.8 m ) E y + ( 6 m ) W = 0

6W = 4.64 kN + 3.8 Ey

(1)

A y − 1.2 kN − 1.6 kN − E y + W = 0 Ay = 2.8 kN + Ey − W

(2)

Considering the smallest possible value of Ey : For

E y = 0, W = Wmin = 0.77333 kN

From (2) the corresponding value of A y is:

Ay = 2.02667 kN ≤ 2.5 kN, which satisfies the constraint on Ay . For the largest allowable value of E y :

E y = 2.5 kN , W = Wmax = 2.3567 kN From (2) the corresponding value of A y is:

A y = 2.9433 kN ≥ 2.5 kN which violates the constraint on Ay . Thus

( A y )max = 2.5 kN. Solving (1) and (2) for W with ( Ay )max

= 2.5 kN,

W = Wmax = 1.59091 kN Therefore:

773.33 N ≤ W ≤ 1590.91 N, or 773.33 N ≤ m(9.81 m/s2 ) ≤ 1590.91 N, and 78.8 kg ≤ m ≤ 162.2 kg ⊳

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 4, Solution 15. Free-Body Diagram:

Calculate lengths of vectors BD and CD:

BD = (11.2) 2 + (21.0) 2 ft = 23.8 ft 2

CD = (11.2) + (8.4) ft = 14.0 ft (a)

ΣM A = 0 :

 11.2 ft   11.2 ft  ( 221 lb)( 24 ft) +  TCD (11.4 ft) = 0 − (161 lb)( 24 ft) +  23 . 8 ft    14.0 ft  TCD = 150.0 lb ⊳

TCD = 150.000 lb

(b)

ΣF x = 0:

 11.2 ft   11.2 ft  161 lb −   ( 221 lb ) −   (150 lb ) + A x = 0 23.8 ft    14.0 ft  A x = 63.000 lb

ΣFy = 0:

A x = 63.000 lb

 21.0 ft   11.2 ft  Ay −   (221 lb) −  14.0 ft  (150 lb) = 0  23.8 ft   

285.00 A y = lb

A=

or

A2x + A 2y =

or

A y = 285.00 lb

2 2 (63) + (285) = 291.88 lb

( 63 )

θ = tan −1 285 = 77.535 ° Therefore

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.

A = 292 lb

77.5° ⊳

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Chapter 4, Solution 16. Free-Body Diagram:

(a)

Equilibrium for ABCD:

ΣM C = 0:

( A cos 60° )( 1.6 in.) − ( 6 lb)( 1.6 in.) + ( 4 lb)( 0.8 in.) = A = 8.0000 lb

(b)

ΣF x = 0:

A = 8.00 lb

or

C x = 8.0000 lb

C y − 6 lb + (8 lb ) sin 60° = 0

or Cy = −0.92820 lb C =

60° ⊳

C x + 4 lb + ( 8 lb) cos 60° = 0 or Cx = − 8.0000 lb

ΣF y = 0:

0

C x2 + C y2 =

or

C y = 0.92820 lb

( 8) 2 + ( 0.92820) 2 =

8.0537 lb

 − 0.92820   = 6.6182° −8  

θ = tan−1  Therefore:

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.

C = 8.05 lb

6.62° ⊳

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Chapter 4, Solution 17. Free-Body Diagram:

Equations of equilibrium:

−( 330 N)( 0.25 m) B +

ΣΜ Α = 0:

+ sinα ( 0.3 m) B

cosα ( 0.5 m) = 0

(1)

ΣFx = 0:

Ax − B sinα = 0

(2)

Σ Fy = 0:

Ay − ( 330 N ) + Bcos α = 0

(3)

(a) Substitution α = 0 into (1), (2), and (3) and solving for A and B:

B = 165.000 N, Ax = 0, Ay = 165.0 N or A = 165.0 N , B = 165.0 N

⊳

(b) Substituting α = 90° into (1), (2), and (3) and solving for A and B:

B = 275.00 N, Ax = 275.00 N, Ay = 330.00 N

A=

Ax2 + A2y = (275) 2 + (330) 2 = 429.56 N

θ = tan− 1

Ay Ax

= tan− 1

330 = 50.194° 275

∴ A = 430 N

50.2°, B = 275 N

⊳

(c) Substituting α = 30 ° into (1), (2), and (3) and solving for A and B:

B = 141.506 N, Ax = 70.753 N, A y = 207.45 N, ⇒

A=

Ax2 + Ay2 =

θ = tan −1

Ay Ax

(70.753) 2 + (207.45) 2 = 219.18 N

= tan −1

207.45 = 71.168° 70.753 ∴ A = 219 N

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.

71.2°, B = 141.5 N

60° ⊳

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Chapter 4, Solution 18. Free-Body Diagram:

Equations of equilibrium:

ΣΜ Α

−( 82.5 N ⋅ m) + B sin α ( 0.3 m) + B cos α ( 0.5 m) = 0

= 0:

(1)

Σ F x = 0:

A x − B sin α = 0

(2)

Σ Fy = 0:

Ay + Bcosα = 0

(3)

(a) Substituting α = 0 into (1), (2), and (3) and solving for A and B:

B = 165.000 N, Ax = 0, Ay = −165.0 N or A = 165.0 N , B = 165.0 N

⊳

∴ A = 275 N

⊳

(b) Substituting α = 90° into (1), (2), and (3) and solving for A and B:

B = 275.00 N, Ax = 275.00 N, Ay = 0 , B = 275 N

(c) Substituting α = 30 ° into (1), (2), and (3) and solving for A and B:

B = 141.506 N, Ax = 70.753 N, A y = −122.548 N A=

Ax2 + Ay2 = ( 70.753) 2 + (− 122.548)2 = 141.506 N

θ = tan −1

Ay Ax

− = tan 1

122.548 = 60.000° 70.753 ∴ A = 141.5 N

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.

60.0°, B = 141.5 N

60° ⊳

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Chapter 4, Solution 19. Free-Body Diagram: (a) From free-body diagram of lever BCD

ΣMC = 0: TAB (50 mm ) − 200 N (75 mm ) = 0 ∴ TAB = 300 (b) From free-body diagram of lever BCD

ΣF x = 0: 200 N + C x + 0.6 (300 N ) = 0 ∴ C x = −380 N

or

C x = 380 N

ΣF y = 0: C y + 0.8 (300 N ) = 0

∴ C y = −240 N

or

C y = 240 N

( 380)2 + ( 240)2

2

2

 Cy  Cx

 − 240   = tan −1   = 32.276 °  − 380  

Then

C =

and

θ = tan −1 

Cx + Cy =

= 449.44 N

or C = 449 N

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.

32.3° ⊳

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Chapter 4, Solution 20. Free-Body Diagram: From free-body diagram of lever BCD

ΣM C = 0: T AB ( 50 mm ) − P ( 75 mm ) = 0 ∴ TAB = 1.5P

(1)

ΣFx = 0: 0.6 TAB + P − Cx = 0 ∴ Cx = P + 0.6TAB

(2)

C x = P + 0.6 (1.5 P ) = 1.9 P

From Equation (1)

Σ Fy = 0: 0.8TAB − Cy = 0 ∴ Cy = 0.8TAB

(3)

Cy = 0.8 (1.5 P ) = 1.2 P

From Equation (1) From Equations (2) and (3)

C = C 2x + C 2y =

(1.9P )2 + (1.2P)2

= 2.2472P

Since C max = 500 N,

∴ 500 N = 2.2472Pmax or

Pmax = 222.49 lb or P = 222 lb

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-Hi...