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STATICS solution maunal

# STATICS solution maunal - PROBLEM 3.1 A 13.2-N force P is...

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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 ( ) sin 13.2 N sin30 6.60 N x P P α = = ° = ( ) cos 13.2 N cos30 11.4315 N α = = ° = y P P Noting that the direction of the moment of each force component about A is counterclockwise, / / A B A y B A x M x P y P = + ( )( ) ( )( ) 0.086 m 11.4315 N 0.122 m 6.60 N = + 1.78831 N m = or 1.788 N m A = M W

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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 . ( ) ( ) 2 2 86 mm 122 mm 149.265 mm = + = AB r 1 1 122 mm tan tan 54.819 86 mm α θ = = = = ° y x Then min = A AB M r P or min = A AB M P r 2.20 N m 1000 mm 149.265 mm 1 m = 14.7389 N = min 14.74 N = P 54.8 ° or min 14.74 N = P 35.2 ° W
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 α knowing that the moment of P about A is counterclockwise and has a magnitude of 1.95 N m. SOLUTION By definition / sin θ = A B A M r P where ( ) 90 θ φ α = + ° − and 1 122 mm tan 54.819 86 mm φ = = ° Also ( ) ( ) 2 2 / 86 mm 122 mm 149.265 mm = + = B A r Then ( )( ) ( ) 1.95 N m 0.149265 m 13.1 N sin 54.819 90 α = ° + ° − or ( ) sin 144.819 0.99725 α ° − = or 144.819 85.752 α ° − = ° and 144.819 94.248 α ° − = ° 50.6 , 59.1 α = ° ° W

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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 20 28 20 8 θ α = ° = ° − ° = ° and ( ) 4 lb cos8 3.9611 lb = ° = x F ( ) 4 lb sin8 0.55669 lb = ° = y F Also ( ) 6.5 in. cos20 6.1080 in. = ° = x ( ) 6.5 in. sin 20 2.2231 in. = ° = y Noting that the direction of the moment of each force component about B is counterclockwise, B y x M xF yF = + ( )( ) ( )( ) 6.1080 in. 0.55669 lb 2.2231 in. 3.9611 lb = + 12.2062 lb in. = or 12.21 lb in. B = M W
PROBLEM 3.5 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 components along ABC and in a direction perpendicular to ABC . SOLUTION First resolve the 4-lb force into components P and Q , where ( ) 4.0 lb sin 28 1.87787 lb Q = ° = Then / = B A B M r Q ( )( ) 6.5 in. 1.87787 lb = 12.2063 lb in. = or 12.21 lb in. B = M W

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PROBLEM 3.6 It is known that a vertical force of 800 N is required to 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 / = B C B N M r F ( )( ) 0.1 m 800 N = 80.0 N m = or 80.0 N m B = M W ( b ) By definition / sin θ = B A B M r P where ( ) 90 90 70 θ α = ° − ° − ° 90 20 10 = ° − ° − ° 60 = ° ( ) 80.0 N m 0.45 m sin60 P = ° 205.28 N = P or 205 N P = W ( 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.
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STATICS solution maunal - PROBLEM 3.1 A 13.2-N force P is...

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