Mecanica vectorial

COSMOS: Complete Online Solutions Manual Organization System

Chapter 3, Solution 1.

Resolve 90 N force into vector components P and Q where Q = ( 90 N ) sin 40° = 57.851 N Then M B = − rA/BQ

= − (0.225 m )(57.851 N ) = −13.0165 N ⋅ m
M B = 13.02 N ⋅ m

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

COSMOS: Complete Online Solutions Manual Organization System

Chapter 3, Solution 2.

Fx = ( 90 N ) cos 25°

= 81.568 N
Fy = ( 90 N ) sin 25° = 38.036 N x = ( 0.225 m ) cos 65° = 0.095089 m

y = (0.225 m ) sin 65°
= 0.20392 m

M B = xFy − yFx
= ( 0.095089 m )( 38.036 N ) − ( 0.20392 m )( 81.568 N ) =−13.0165 N ⋅ m

M B = 13.02 N ⋅ 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 3, Solution 3.

Px = ( 3 lb ) sin 30°
= 1.5 lb

Py = ( 3 lb ) cos 30°= 2.5981 lb

M A = xB/ A Py + yB/ A Px
= ( 3.4 in.)( 2.5981 lb ) + ( 4.8 in.)(1.5 lb ) = 16.0335 lb ⋅ in.

M A = 16.03 lb ⋅ in.

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 ManualOrganization System

Chapter 3, Solution 4.

For P to be a minimum, it must be perpendicular to the line joining points A and B with rAB =

( 3.4 in.)2 + ( 4.8 in.)2

= 5.8822 in.

α = θ = tan −1   x
 4.8 in.  = tan −1    3.4 in. 

 y  

= 54.689°
Then or

M A = rAB Pmin Pmin = M A 19.5 lb ⋅ in. = rAB 5.8822 in.

= 3.3151 lb ∴ Pmin = 3.32 lb

54.7°
or

Pmin = 3.32lb

35.3°

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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 3, Solution 5.

By definition where and

M A = rB/ A P sin θ

θ = φ + ( 90° − α ) φ =tan −1 
 4.8 in.    3.4 in. 

= 54.689° Also

rB/ A =

( 3.4 in.)2 + ( 4.8 in.)2

= 5.8822 in.

Then or or

(17 lb ⋅ in.) = ( 5.8822 in.)( 2.9 lb ) sin ( 54.689° + 90° − α )
sin (144.689° − α ) = 0.99658 144.689° − α = 85.260°; 94.740° ∴ α = 49.9°, 59.4°

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 3, Solution 6.

(a)

(a) M A = rB/ A × TBF

M A = xTBFy + yTBFx
= ( 2 m )( 200 N ) sin 60° + ( 0.4 m )( 200 N ) cos 60° = 386.41 N ⋅ m or M A = 386 N ⋅ m (b) (b) For FC to be a minimum, it must be perpendicular to the linejoining A and C.

∴ M A = d ( FC )min
with

d =

( 2 m )2 + (1.35 m )2

= 2.4130 m
Then 386.41 N ⋅ m = ( 2.4130 m ) ( FC )min

( FC )min
and

= 160.137 N  1.35 m   = 34.019°  2m 

φ = tan −1 

θ = 90 − φ = 90° − 34.019° = 55.981°
∴ ( FC )min = 160.1 N
56.0°

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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 3, Solution 7.

(a)

M A = xTBFy + yTBFx

= ( 2 m )( 200 N ) sin 60° + ( 0.4 m )( 200 N ) cos 60° = 386.41 N ⋅ m
or M A = 386 N ⋅ m

(b)

Have or

M A = xFC

FC =

MA 386.41 N ⋅ m = 2m x
∴ FC = 193.2 N...