Vector Mechanics for Engineers - Statics 8E Solutions - chapter 4

Vector Mechanics for Engineers - Statics 8E Solutions - chapter 4

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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.

Chapter 4, Solution 1. Free-Body Diagram:

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Chapter 4, Solution 2. Free-Body Diagram:

or 2.70kipsB =F or 3.80kipsC =F

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Chapter 4, Solution 3. Free-Body Diagram:

(b)

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

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Chapter 4, Solution 5. Free-Body Diagram:

()64in.cosbα= From free-body diagram of hand truck

MPbWaWaΣ=−+= (1)

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Chapter 4, Solution 6.

Free-Body Diagram:

()64in.cosbα= From free-body diagram of hand truck

MPbWaWaΣ=−+= (1) or 8.05lb=P (b) From Equation (2)

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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.

Chapter 4, Solution 7. Free-Body Diagram:

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Chapter 4, Solution 8. Free-Body Diagram:

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

(b) The corresponding value of y A is

() max

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Chapter 4, Solution 9. Free-Body Diagram:

For max 36.0 lbC

For ()maxmin , 36 lbCB min 28.0 lbC

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Chapter 4, Solution 10. Free-Body Diagram:

For min, 0D QT= min 1.250 kNQ=

For max, 0B QT=

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Chapter 4, Solution 1. Free-Body Diagram:

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 ≤ 9.0 kN (4)

(3) and (4) now give: 1.50 kN ≤ Q ≤ 9.0 kN

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Chapter 4, Solution 12. Free-Body Diagram:

For () min, 0A

For

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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.

Chapter 4, Solution 13. Free-Body Diagram:

Using the bounds on B:

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Chapter 4, Solution 14. Free-Body Diagram:

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

Considering the smallest possible value of :y E

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

For the largest allowable value of :y E

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

A= Solving (1) and (2) for Wwith ()max

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Chapter 4, Solution 15. Free-Body Diagram:

x x FA

A=or 63.000lbx
A=or 285.0 lby

285.0 lby =A

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Chapter 4, Solution 16. Free-Body Diagram:

(a) Equilibrium for ABCD:

C=−or 8.0000lbx
C=−or 0.92820lby

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Chapter 4, Solution 17. Free-Body Diagram:

Equations of equilibrium:

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

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Chapter 4, Solution 18. Free-Body Diagram:

Equations of equilibrium:

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

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

yx A

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Chapter 4, Solution 19.

Free-Body Diagram: (a) From free-body diagram of lever BCD

(b) From free-body diagram of lever BCD

380 Nor380 Nxx
N 240orN 240=−=∴

y C

240 tantan y C

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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.

Chapter 4, Solution 20.

Free-Body Diagram: From free-body diagram of lever BCD

1.5AB TP∴= (1)

From Equations (2) and (3)

() () PC yx

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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.

Chapter 4, Solution 21. Free-Body Diagram:

−=Σ spΒx FAΜ α

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Chapter 4, Solution 2. Free-Body Diagram:

cos

ΒxΜ

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Chapter 4, Solution 23. Free-Body Diagram:

From free-body diagram for (a):

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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.

Chapter 4, Solution 24. Free-Body Diagram:

From free-body diagram for (a):

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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.

Chapter 4, Solution 25. Free-Body Diagram:

Geometry:

Equilibrium for lever:

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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.

Chapter 4, Solution 26. Free-Body Diagram:

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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.

Chapter 4, Solution 27. Free-Body Diagram:

Geometry:

Equilibrium for mast:

2.5 y

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Chapter 4, Solution 28. Free-Body Diagram:

Geometry:

Equilibrium for mast:

T= or384NBC

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Chapter 4, Solution 29. Free-Body Diagram:

Geometry:

Equilibrium for bracket:

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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.

Chapter 4, Solution 30. Free-Body Diagram:

Geometry:

Equilibrium for bracket:

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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.

Chapter 4, Solution 31. Free-Body Diagram:

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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.

Chapter 4, Solution 32. Free-Body Diagram:

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Chapter 4, Solution 3. Free-Body Diagram:

For both parts (a) and (b) x CP=− (1) cos P

B θ = (2) cos y

(a) The magnitudes of the forces at B and C are equal:

xy BCC=+ cos P P θθ or continued

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cosθ θ +=, this gives cos cos θθθ =+−, or and using (1) and (3)

2 xy

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Chapter 4, Solution 34. Free-Body Diagram:

For both parts (a) and (b) x CP=− (1) cos P

B θ = (2) cos y

(a) The magnitude of the reaction at C:

xy C=+ and or P=C

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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.

Chapter 4, Solution 35. Free-Body Diagram:

Equilibrium for bracket:

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Chapter 4, Solution 36. Free-Body Diagram:

cos C ABD ABD or with 3/4 ABD TP=:

4 cos 4 a P aP P θθ x CP=

Therefore: 1

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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.

Chapter 4, Solution 37. Free-Body Diagram:

Equilibrium for bracket:

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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.

Chapter 4, Solution 38. Free-Body Diagram:

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Chapter 4, Solution 39. Free-Body Diagram:

Equilibrium for rod:

(b)() ( )0: 4.2426 lb cos45 6 lb cos60 sin45 cos45 0

xA D

A D N= (1)

Solving (1) and (2) gives:

Therefore:

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