Solucionario Libro Estatica Cap 4

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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)

or A y = 271 lb

ΣM A = 0 :

By (3.6 ft ) − (146 lb)(5.04 ft ) − (63 lb)(6.84 ft ) − (90 lb)(9.84 ft ) = 0 By = 570.10 lb
or B y = 570 lb

VectorMechanics 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 4, Solution 2.
Free-Body Diagram:

(a)

ΣM C = 0:

    ( 3.5 kips ) (1.6 + 1.3 + 19.5cos15o ) ft  − 2FB(1.6 + 1.3 + 14 ) ft  + ( 9.5 kips )(1.6 ft ) = 0 2FB = 5.4009 kips
or FB = 2.70 kips

(b)
ΣM B = 0:

( 3.5 kips ) (19.5cos15o − 14 ) ft  − ( 9.5 kips ) (14 + 1.3) ft  + 2 FC (14 + 1.3 + 1.6 ) ft  = 0      
2FC = 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.

COSMOS: Complete Online Solutions Manual Organization System

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
or FH = 34.0 kN

(b)

ΣM H = 0:

( 25 kN )( 2.9 m ) + ( 3 kN )( 0.9 m ) − ( 50kN )( 2.0 m ) + 2FK ( 2.5 m ) = 0
2FK = 9.9200 kN
or FK = 4.96 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.

COSMOS: Complete Online Solutions Manual Organization System

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
TCD = 81.143 kN
or TCD = 81.1 kN

(b)

ΣFx = 0:

Bx = 0 so that B = By

ΣFy = 0:

( −25 − 3 − 25 − 81.143) kN + B = 0
B = 134.143 kN
or B = 134.1 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 4, Solution 5.
Free-Body Diagram:

a1 = ( 20 in.) sin α − ( 8 in.) cos α a2 = ( 32 in.) cos α − ( 20 in.) sin α b = ( 64 in.) cos α
From free-body diagram of hand truck

ΣM B = 0: P ( b ) − W ( a2 ) + W ( a1 ) = 0 ΣFy = 0: P − 2w +2B = 0
For

(1) (2)

α = 35°

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

b = 64cos 35° = 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) From Equation (2) or P = 14.99 lb

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

B = 72.5 lb

Vector Mechanics forEngineers: 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 4, Solution 6.

Free-Body Diagram:

a1 = ( 20 in.) sin α − ( 8 in.) cos α a2 = ( 32 in.) cos α − ( 20 in.) sin α b = ( 64 in.) cos αFrom free-body diagram of hand truck

ΣM B = 0: P ( b ) − W ( a2 ) + W ( a1 ) = 0 (1) ΣFy = 0: P − 2w + 2B = 0
For (2)

α = 40°

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

b = 64cos 40° = 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)...