# Solucionario "mecacina vectorial para ingenieros beer & johnston"

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COSMOS: Complete Online Solutions Manual Organization System
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
Chapter 3, Solution 1.
Resolve 90 N force into vector components P and Q
where Q = (90 N)sin 40°
=57.851 N
Then MB=−rA/BQ
= − (0.225 m)(57.851 N)
= −13.0165 N⋅m
MB=13.02 N⋅m
COSMOS: Complete Online Solutions Manual Organization System
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
Chapter 3, Solution 2.
Fx = (90 N)cos 25°
=81.568 N
Fy =(90 N)sin 25°
=38.036 N
x = (0.225 m)cos65°
=0.095089 m
y = (0.225 m)sin 65°
=0.20392 m
MB = xFy− yFx
=(0.095089 m)(38.036 N)−(0.20392 m)(81.568 N)
= −13.0165 N⋅m
MB=13.02 N⋅m
COSMOS: Complete Online Solutions Manual Organization System
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
Chapter 3, Solution 3.
Px = (3 lb)sin30°
=1.5 lb
Py = (3 lb)cos30°
= 2.5981 lb
MA=xB/APy+yB/APx
=(3.4 in.)(2.5981 lb)+(4.8 in.)(1.5 lb)
=16.0335 lb⋅in.
MA=16.03 lb⋅in.
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. RussellJohnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
Chapter 3, Solution 4.
For P to be a minimum, it must be perpendicular to the line joining points
A and B
with ( )2 ( )2 rAB = 3.4 in. + 4.8 in.
=5.8822 in.
tan 1 y
x
α =θ = −  
 
tan 1 4.8 in.
3.4 in.
−  
=  
 
= 54.689°
Then MA=rABPmin
or min19.5 lb in.
5.8822 in.
A
AB
P M
r

= =
=3.3151 lb
∴Pmin =3.32 lb 54.7°
or Pmin = 3.32 lb 35.3°
COSMOS: Complete Online Solutions Manual Organization System
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
Chapter 3,Solution 5.
By definition MA=rB/APsinθ
where θ =φ + (90° −α)
and tan 1 4.8 in.
3.4 in.
φ −  
=  
 
=54.689°
Also ( )2 ( )2
rB/A = 3.4 in. + 4.8 in.
=5.8822 in.
Then (17 lb⋅in.) = (5.8822 in.)(2.9 lb)sin (54.689°+ 90°−α )
or sin (144.689° − α ) = 0.99658
or 144.689° − α = 85.260°; 94.740°
∴α = 49.9°, 59.4°
COSMOS: Complete Online Solutions Manual Organization System
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
Chapter 3, Solution 6.
(a)
(b)
(a) MA=rB/A×TBF
MA xTBFy yTBFx = +
= (2 m)(200 N)sin 60° + (0.4 m)(200 N)cos60°
=386.41 N⋅m
or MA = 386 N⋅m
(b) For FC to be a minimum, it must beperpendicular to the line
joining A and C.
( )∴MA=dFC min
with ( )2 ( )2 d = 2 m + 1.35 m
=2.4130 m
Then ( )( )min 386.41 N⋅m = 2.4130 m FC
( )min FC = 160.137 N
and tan 1 1.35 m 34.019
2 m
φ −  
=  = °
 
θ =90−φ = 90° − 34.019° = 55.981°
( )min ∴FC =160.1 N 56.0°
COSMOS: Complete Online Solutions Manual Organization System
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
Chapter 3, Solution 7.
(a)
(b)
(c)
y x
MA=xTBF+yTBF
=(2 m)(200 N)sin 60°+(0.4 m)(200 N)cos60°
=386.41 N⋅m
or 386 N m
AM = ⋅ 
Have A C
M =xF
or
386.41 N m
2 m
A
C
M
F
x
= = ⋅
= 193.205 N
193.2 N
C ∴F = 
For B
F to...