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# Resolução do Capítulo 3 - Corpos Rígidos

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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 \u3b1 is equal to 30°. SOLUTION First note ( )sin 13.2 N sin 30 6.60 NxP P \u3b1= = ° = ( )cos 13.2 N cos30 11.4315 N\u3b1= = ° =yP P Noting that the direction of the moment of each force component about A is counterclockwise, / /A B A y B A xM x P y P= + ( )( ) ( )( )0.086 m 11.4315 N 0.122 m 6.60 N= + 1.78831 N m= \u22c5 or 1.788 N mA = \u22c5M 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\u22c5 counterclockwise moment about A. SOLUTION For P to be a minimum, it must be perpendicular to the line joining points A and B. ( ) ( )2 286 mm 122 mm 149.265 mm= + =ABr 1 1 122 mmtan tan 54.819 86 mm \u3b1 \u3b8 \u2212 \u2212\uf8eb \uf8f6 \uf8eb \uf8f6= = = = °\uf8ec \uf8f7 \uf8ec \uf8f7\uf8ed \uf8f8 \uf8ed \uf8f8 y x Then min=A ABM r P or min = A AB MP r 2.20 N m 1000 mm 149.265 mm 1 m \u22c5 \uf8eb \uf8f6= \uf8ec \uf8f7\uf8ed \uf8f8 14.7389 N= min 14.74 N\u2234 =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 \u3b1 knowing that the moment of P about A is counterclockwise and has a magnitude of 1.95 N m.\u22c5 SOLUTION By definition / sin\u3b8=A B AM r P where ( )90\u3b8 \u3c6 \u3b1= + ° \u2212 and 1 122 mmtan 54.819 86 mm \u3c6 \u2212 \uf8eb \uf8f6= = °\uf8ec \uf8f7\uf8ed \uf8f8 Also ( ) ( )2 2/ 86 mm 122 mm 149.265 mm= + =B Ar Then ( )( ) ( )1.95 N m 0.149265 m 13.1 N sin 54.819 90 \u3b1\u22c5 = ° + ° \u2212 or ( )sin 144.819 0.99725\u3b1° \u2212 = or 144.819 85.752\u3b1° \u2212 = ° and 144.819 94.248\u3b1° \u2212 = ° 50.6 , 59.1\u3b1\u2234 = ° °W PROBLEM 3.4 A foot valve for a pneumatic system is hinged at B. Knowing that \u3b1 = 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\u3b8 \u3b1= \u2212 ° = ° \u2212 ° = ° and ( )4 lb cos8 3.9611 lb= ° =xF ( )4 lb sin8 0.55669 lb= ° =yF Also ( )6.5 in. cos 20 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 xM xF yF= + ( )( ) ( )( )6.1080 in. 0.55669 lb 2.2231 in. 3.9611 lb= + 12.2062 lb in.= \u22c5 or 12.21 lb in.B = \u22c5M W PROBLEM 3.5 A foot valve for a pneumatic system is hinged at B. Knowing that \u3b1 = 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 lbQ = ° = Then /=B A BM r Q ( )( )6.5 in. 1.87787 lb= 12.2063 lb in.= \u22c5 or 12.21 lb in.B = \u22c5M W 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 \u3b1 = 10°, (c) the smallest force P which creates the same moment about B. SOLUTION (a) Have /=B C B NM r F ( )( )0.1 m 800 N= 80.0 N m= \u22c5 or 80.0 N mB = \u22c5M W (b) By definition / sin\u3b8=B A BM r P where ( )90 90 70\u3b8 \u3b1= ° \u2212 ° \u2212 ° \u2212 90 20 10= ° \u2212 ° \u2212 ° 60= ° ( ) 80.0 N m 0.45 m sin 60P\u2234 \u22c5 = ° 205.28 N=P or 205 NP = 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. Thus min / min= =B A BM dP r P or ( ) min80.0 N m 0.45 m\u22c5 = P min 177.778 N\u2234 =P or min 177.8 N=P 20°W PROBLEM 3.7 A sign is suspended 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 /A B A BF=M r T× Noting that the direction of the moment of each force component about A is counterclockwise, = +A BFy BFxM xT yT ( )( ) ( )( )6.5 ft 45 lb sin 60 4.4 ft 3.1 ft 45 lb cos60= ° + \u2212 ° 282.56 lb ft= \u22c5 or 283 lb ftA = \u22c5M W (b) Have ( )/ minA C A C=M r F× For CF to be minimum, it must be perpendicular to the line joining points A and C. ( ) minA CM d F\u2234 = where ( ) ( )2 2/ 6.5 ft 4.4 ft 7.8492 ftC Ad r= = + = ( )( ) min 282.56 lb ft 7.8492 ft CF\u2234 \u22c5 = ( ) min 35.999 lbCF = 1 4.4 fttan 34.095 6.5 ft \u3c6 \u2212 \uf8eb \uf8f6= = °\uf8ec \uf8f7\uf8ed \uf8f8 90 90 34.095 55.905\u3b8 \u3c6= ° \u2212 = ° \u2212 ° = ° or ( ) min 36.0 lbC =F 55.9°W PROBLEM 3.8 A sign is suspended from two chains AE and BF. Knowing that the tension in BF is 45 lb, determine (a) the moment about A of the force exerted by the chain at B, (b) the magnitude and sense of the vertical force applied at C which creates the same moment about A, (c) the smallest force applied at B which creates the same moment about A. SOLUTION (a) Have /A B A BF=M r T× Noting that the direction of the moment of each force component about A is counterclockwise, = +A BFy BFxM xT yT ( )( ) ( )( )6.5 ft 45 lb sin 60 4.4 ft 3.1 ft 45 lb cos60= ° + \u2212 ° 282.56 lb ft= \u22c5 or 283 lb ftA = \u22c5M W (b) Have /A C A C=M r F× or =A CM xF 282.56 lb ft 43.471 lb 6.5 ft A C MF x \u22c5\u2234 = = = or 43.5 lbC =F W (c) Have ( )/ minA B A B=M r F× For BF to be minimum, it must be perpendicular to the line joining points A and B. ( ) min \u2234 =A BM d F where ( ) ( )2 26.5 ft 4.4 ft 3.1 ft 6.6287 ftd = + \u2212 = ( ) min 282.56 lb ft 42.627 lb 6.6287 ft \u22c5\u2234 = = =AB MF d and 1 6.5 fttan 78.690 4.4 ft 3.1 ft \u3b8 \u2212 \uf8eb \uf8f6= = °\uf8ec \uf8f7\u2212\uf8ed \uf8f8 or ( ) min 42.6 lb=BF 78.7°W PROBLEM 3.9 The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts a 125-N force directed along its center line on the ball and socket at B, determine the moment of the force about A. SOLUTION First note ( ) ( )2 2240 mm 46.6 mmCBd = + 244.48 mm= Then 240 mmcos 244.48 mm \u3b8 = 46.6 mm sin 244.48 mm \u3b8 = and cos sinCB CB CBF F\u3b8 \u3b8= \u2212F i j ( ) ( )125 N 240 mm 46.6 mm 244.48 mm \uf8ee \uf8f9= \u2212\uf8f0 \uf8fbi j Now /A B A CB=M r F× where ( ) ( )/ 306 mm 240 mm 46.6 mmB A = \u2212 +r i j ( ) ( )306 mm 286.6 mm= \u2212i j Then ( ) ( ) ( )125 N306 mm 286.6 mm 240 46.6 244.48A \uf8ee \uf8f9= \u2212 \u2212\uf8f0 \uf8fbM i j i j× ( ) ( )27878 N mm 27.878 N m= \u22c5 = \u22c5k k or 27.9 N mA = \u22c5M W PROBLEM 3.10 The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts a 125-N force directed along its center line on the ball and socket at B, determine the moment of the force about A. SOLUTION First note ( ) ( )2 2344 mm 152.4 mm 376.25 mmCBd = + = Then 344 mmcos 376.25 mm \u3b8 = 152.4 mmsin 376.25 mm \u3b8 = and ( ) ( )cos sinCB CB CBF F\u3b8 \u3b8= \u2212F i j ( ) ( )125 N 344 mm 152.4 mm 376.25 mm \uf8ee \uf8f9= +\uf8f0 \uf8fbi j Now /A B A CB=M r F× where ( ) ( )/ 410 mm 87.6 mmB A = \u2212r i j Then ( ) ( ) ( )125 N410 mm 87.6 mm 344 152.4 376.25A \uf8ee \uf8f9= \u2212 \u2212\uf8f0 \uf8fbM i j i j× ( )30770 N mm= \u22c5 k ( )30.770 N m= \u22c5 k or 30.8 N mA = \u22c5M W