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**Unformatted text preview: **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 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 A B 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 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....

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