# Estructuras

Páginas: 6 (1325 palabras) Publicado: 18 de febrero de 2013
PROBLEM 3.1
A 13.2-N force P is applied to the lever which controls the auger of a snowblower. Determine the moment of P about A when α is equal to 30°.

SOLUTION
First note Px = P sin α = (13.2 N ) sin 30° = 6.60 N

Py = P cos α = (13.2 N ) cos 30° = 11.4315 N Noting that the direction of the moment of each force component about A is counterclockwise, M A = xB/ A Py + yB/ A Px = ( 0.086 m)(11.4315 N ) + ( 0.122 m )( 6.60 N ) = 1.78831 N ⋅ m or M A = 1.788 N ⋅ m

PROBLEM 3.2
The force P is applied to the lever which controls the auger of a snowblower. Determine the magnitude and the direction of the smallest force P which has a 2.20- N ⋅ m counterclockwise moment about A.

SOLUTION
For P to be a minimum, it must be perpendicular to the line joining points A and B.
rAB =(86 mm )2 + (122 mm )2
 y

= 149.265 mm

α = θ = tan −1   = tan −1   = 54.819° x  86 mm 
Then or

 122 mm 

M A = rAB Pmin Pmin =
=

MA rAB
2.20 N ⋅ m  1000 mm    149.265 mm  1 m 

= 14.7389 N ∴ Pmin = 14.74 N 54.8° or Pmin = 14.74 N 35.2°

PROBLEM 3.3
A 13.1-N force P is applied to the lever which controls the auger of a snowblower. Determine the value of αknowing that the moment of P about A is counterclockwise and has a magnitude of 1.95 N ⋅ m.

SOLUTION
By definition where and
M A = rB/ A P sin θ

θ = φ + ( 90° − α ) φ = tan −1   = 54.819°  86 mm 
rB/ A =

 122 mm 

Also Then or or and

(86 mm )2 + (122 mm )2

= 149.265 mm

1.95 N ⋅ m = ( 0.149265 m )(13.1 N ) sin ( 54.819° + 90° − α ) sin (144.819° − α ) = 0.99725 144.819°− α = 85.752° 144.819° − α = 94.248° ∴ α = 50.6°, 59.1°

PROBLEM 3.4
A foot valve for a pneumatic system is hinged at B. Knowing that α = 28°, determine the moment of the 4-lb force about point B by resolving the force into horizontal and vertical components.

SOLUTION
Note that and

θ = α − 20° = 28° − 20° = 8°
Fx = ( 4 lb ) cos8° = 3.9611 lb Fy = ( 4 lb ) sin 8° = 0.55669 lb

Also

x= ( 6.5 in.) cos 20° = 6.1080 in. y = ( 6.5 in.) sin 20° = 2.2231 in.

Noting that the direction of the moment of each force component about B is counterclockwise,
M B = xFy + yFx

= ( 6.1080 in.)( 0.55669 lb ) + ( 2.2231 in.)( 3.9611 lb ) = 12.2062 lb ⋅ in. or M B = 12.21 lb ⋅ in.

PROBLEM 3.5
A foot valve for a pneumatic system is hinged at B. Knowing that α = 28°, determine the momentof the 4-lb force about point B by resolving the force into components along ABC and in a direction perpendicular to ABC.

SOLUTION
First resolve the 4-lb force into components P and Q, where
Q = ( 4.0 lb ) sin 28° = 1.87787 lb

Then

M B = rA/BQ

= ( 6.5 in.)(1.87787 lb ) = 12.2063 lb ⋅ in. or M B = 12.21 lb ⋅ in.

PROBLEM 3.6
It is known that a vertical force of 800 N is requiredto remove the nail at C from the board. As the nail first starts moving, determine (a) the moment about B of the force exerted on the nail, (b) the magnitude of the force P which creates the same moment about B if α = 10°, (c) the smallest force P which creates the same moment about B.

SOLUTION
(a) Have
M B = rC/B FN

= ( 0.1 m )( 800 N ) = 80.0 N ⋅ m or M B = 80.0 N ⋅ m (b) By definitionM B = rA/B P sin θ

where

θ = 90° − ( 90° − 70° ) − α
= 90° − 20° − 10° = 60° ∴ 80.0 N ⋅ m = ( 0.45 m ) P sin 60° P = 205.28 N or P = 205 N

(c) For P to be minimum, it must be perpendicular to the line joining points A and B. Thus, P must be directed as shown. Thus or M B = dPmin = rA/B Pmin 80.0 N ⋅ m = ( 0.45 m ) Pmin ∴ Pmin = 177.778 N or Pmin = 177.8 N 20°

PROBLEM 3.7
A sign issuspended from two chains AE and BF. Knowing that the tension in BF is 45 lb, determine (a) the moment about A of the force exert by the chain at B, (b) the smallest force applied at C which creates the same moment about A.

SOLUTION
(a) Have
M A = rB/ A × TBF

Noting that the direction of the moment of each force component about A is counterclockwise, M A = xTBFy + yTBFx = ( 6.5 ft )( 45...

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