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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 is equal to 30. SOLUTION First note ( ) sin 13.2 N sin30 6.60 N x P P = = = ( ) cos 13.2 N cos30 11.4315 N = = = 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 = = = = 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 knowing that the moment of P about A is counterclockwise and has a magnitude of 1.95 N m. SOLUTION By definition / sin = A B A M r P where ( ) 90 = + and 1 122 mm tan 54.819 86 mm = = 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 = + or ( ) sin 144.819 0.99725 = or 144.819 85.752 = and 144.819 94.248 = 50.6 , 59.1 = W PROBLEM 3.4 A foot valve for a pneumatic system is hinged at B . Knowing that = 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 = = = 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 = 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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