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lecture10 - 10 Equilibrium of Rigid Bodies R=in ij lek=0...

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Unformatted text preview: 10 Equilibrium of Rigid Bodies R=in+ij+lek=0 C=Mxi+Myj+Mzk=0 Free -Body Diagrams Step 1. Decide which body or combination of bodies is to be isolated or cut free from its surroundings. Step 2. Prepare a drawing or sketch of the outline of this isolated or free body. Step 3. Carefully trace around the boundary of the free body and identify and label all the forces and moments exerted by contacting or interacting bodies that were removed during the isolation process. Step 4. Choose the set of coordinate axes to be used in solving the problem and indicate these directions on the free-body diagram. A Supports l . Supports used in two-Dimensional Applications R lactions One Collincur Force A One Force. Normal to Lhe. Supporiing Surface 7...).1 V - ~ A: T ‘ A "l‘wo Force, Components Equivalents A One. Force Normal to the Supporting Surface 1" ' One Normal Force Two Force Components and One, Couple Supports used in Three-Dimensional Applications Roller Support One Normal Force Supports used in Three-Dimensional Applications ‘7.“ A ,-" - It’d-JV-v-J -_x -l a x V Hinge 3' (Thezaxis is par-(11101 3 to the hinge axis.) «5-- . ‘ )‘ Bearing (The : axis is paraliel to the axis of the supported shaft.) [MIA v Cb ' I '61-?“ x J A A , Mm- ‘ Three Force Components. Two Couple Components y A I q A _/ A “ ,\' 7 / (When no couples are exerted) ----)> ~—— A: A .1" I / 4. (When no couples and no axiai force are exerted) Built-in (Fixed) Support Three Force Components. Three Couple Components L . 691 Draw a freeebod§hdiagramr for the cantilever beam shown in Fig. P671 which has a weight w} SOLUTION MECHANICAL ENGINEERING The action of the fiXEd support at the left end of the beam is represented by force components fix and Ky and a moment HA. The weight 9 of the beam acts through the center df gravity G of the beam and is directed-toward the center of the earth. 6—2 Draw a free—body diagram for the beam shown in Fig. P6—2 which has a mass m. SOLUTION The action of the pin at support A TOPS PRINTING College Station ,TX is represented by force components. Ax and Ky. Force 3 acts normal to the supporting surface at B. The weight W = m5 of the beam acts through the center of gravity G of the beam and is directed toward the center of the earth. 2500 lb Beam weighs 250 lbs What are the reactions at A and B? Frictionless Draw Free Body Diagram (a) Draw the free-body diagram of the 60—1b drill press: assuming that the surfaces at A and B are smooth. (b) Determine..‘twhe’re—actionsatannd B. A Axfi—x—“axfkwemgm c2 ‘1 Ain'BS-i-«kfI OIL COuL-‘B U$tT @EMBW ) Abbrmwtormmw AatSSLJs fi Draw Free Body Diagram 3’ 1:2 -....,.... 1 _______ 7 .. ‘1 3 : 4' r 1 h, ‘ .... ,M I; __,_ .k i 1 . in é" 2 in 5:4A— 2 n] _ l'? 1/” (20(30):;93 1; N Dn-m-‘éng the. free—body diagram. Find the reaction forces at A and Ben the strut A-B t ‘ é g; 't ' r “”1. w" . ‘1‘ -M...W . ‘n‘ - ;, ‘1 , I) a)??? .-’ “W" (@155) C" (3 £1“; 5 A (Q (m) " C) \ Y... W” m“ "'”’ ] Draw Free Body Diagram of Fish B,Fish D & Fish C, Bottom Bar & Top Bar The mobile is in equilibrium. The fish B weighs 27 oz. Determine the weights of the fish A, C, and D. (The weights of the crossbars are negligible.) amflwm «- Fatal m3: 0 Fig 3"“ 3 (3&3, “ F: +34. g; "1.7532 - o) ‘ \ n! . z f , , .1.ng A , v, . ‘ V _ . H _ .1 a, _ . . ,, . l n u .. ‘ m a n w :3 a A . . m a M 1 w, n a . M u 5 2 H _ ., J ,, , w (”y-‘4 '1 Beam weighs 80 lbs 10 deg @EM ___‘_’ '0 4' +M A,“ 31:33:41» Lb‘fi'éfi'ffiimfi { ea:- M - IO GLECW 20$} 357%”? M MA __,, 37/1,0€; :1 O ‘ MAS: 37/9,,92 4~T~w3w§33 {3 57"“ ~‘. I 'U ...
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