t IFBE-[(-=- "Z- tj ..e:::: /1 0s6 (I :::-13N -p11 /(1 196 /If: t3 (Z3':::+ </6;+9 2 J)~ ltt 6 ~= f(3~-+-2bl.,ls2~)iij :.llrj

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3 Na me So..., 270 Fall 2011 Problem #2 (20 points) A uniform rod AB is 6 meters long, weighs 52 N and is in static equilibrium The 52N weight, W, is depicted as acting Yo way along the length of the rod at point C. At point A, a ball and socket is attached to the rod. At point B, two cables (B and BD) are attached to the rod. Please use the answer box for final answers. You must show A work for parts boe to get full credit. / / c..f~ '::) a. Draw the free-body diagram of the rod AB (4 points) on the figure provided in the answer box. b. Determine the position vector from A to B (R AB ) (2 points) c. Determine the position vector from A to C (RAcl (2 points) d. Determine the magnitude of the tensions in cables B and BD (6 points) e. Determine the reactions at A. (6 points, Answers: a. z 1- 'T i T 0O i ~ C (. A ~._-_._._...:.._.,. x/ A t 1 a. i /? " "'- b.(r AB )= z...;:.1- 'f.:~..,l Vk,..., '3 - / '"'-- z...:.... c. (RAe> = ~ <f~ ~ z.k.,..",., d(i). agnitude of the tension in Cable B= z::w d(ii). agnitude of the tension in Cable BD= * y w i- e. Reactions at point A: (~V>~ +- i Wi 4- W 1:)~ Z,.:O- c> ;. R; xw -+-2.:/ to -4- ~D) e: =- o -. (i(;-+- 7d-: ~ ":. _1= - 1;1~ <,/ k:..,. -7 -},;'-0$(,.) -7 -!3D tfo &(; ~ C:: Z-<. ~ Y {;.d':::: -+ {-fttld';' c.))i +- ( - Z."d... '~) t. ~..r c..- -., o.-.ll ~ w+ '-' 2.Q:o ut. "l Sf ~ 7' T: a ~ ::.i z.. W 6 = C = >'78.o? ~: f;c h ~ z. 78 :-~D vd..cit:..s 4"" = ~ / ~ - 1;6=0 '5t: li j,4~=

4 t FB-[(-=- "Z- t ltt 6 -p11 :.llr ~ ~ ~ -- Z <:.. 'i~- </{/, Ai /1 0s6 /( ~ Z{, A! BD ( :::-13N ~= f(3~-+-2bl.,s2~)..e:::: /(1 196 / ~ stb ~ /f: t3 D/1 ::: 3,< (Z3':::+ </6;+9 2 )~

5 Problem 3 (20 points) The truss is in static equilibrium. t is pinned at A and has a roller at C. The overall length is 20m and the overall height is 6 m (i.e. each grid line is 1 m). A load is applied at point. A B Solution version A K C D F a) Draw an overall free body diagram and find the reactions at the supports (7 pts) b) dentify A zero force members (3 point) c) Find the forces in members and F. ndicate if they are in tension or compression (7 pts) d) Find the forces in member AB. ndicate if it is in tension or compression (3 pts) Σ A =0=4Cy-20(40) K Cy = 200 kn =0= = 0 kn =0=Ay+Cy-40 = Ay Ay = -160 kn Cy D F b) B, C,, are zero force c) section cut. Σ F =0=6F -4(40) = arctan(6/4)=56.3 =0= F F sin 40 =0= -F F F F F cos 0 = F F 48(0.554) F = 26.6 kn (tension) F F = 48.0 kn (tension) F d) oint A =0= -F AB sin +Ay 0 = -F AB (0.832) -160 F AB = kn (comp)

6 Problem 3 (20 points) The truss is in static equilibrium. t is pinned at A and has a roller at C. The overall length is 20m and the overall height is 6 m (i.e. each grid line is 1 m). A load is applied at point. A B Solution version B K C D F a) Draw an overall free body diagram and find the reactions at the supports (7 pts) b) dentify A zero force members (3 point) c) Find the forces in members K and K. ndicate if they are in tension or compression (7 pts) d) Find the forces in member AB. ndicate if it is in tension or compression (3 pts) Σ A =0=4Cy-20(40) K Cy = 200 kn =0= = 0 kn =0=Ay+Cy-40 = Ay Ay = -160 kn Cy D F b) B, C,, are zero force c) section cut. Σ =0=6F K -8(40) = arctan(6/4)=56.3 =0= F K sin 40 =0= -F K F D F K cos 0 = F F 48(0.554) F K = 53.3 kn (tension) F K = 48.0 kn (tension) F d) oint A =0= -F AB sin +Ay 0 = -F AB (0.832) -160 F AB = kn (comp)

7 Problem 3 (20 points) The truss is in static equilibrium. t is pinned at A and has a roller at C. The overall length is 20m and the overall height is 6 m (i.e. each grid line is 1 m). A load is applied at point. A B Solution version C K C D F a) Draw an overall free body diagram and find the reactions at the supports (7 pts) b) dentify A zero force members (3 point) c) Find the forces in members and F. ndicate if they are in tension or compression (7 pts) d) Find the forces in member AB. ndicate if it is in tension or compression (3 pts) Σ A =0=4Cy-20(30) K Cy = 150 kn =0= = 0 kn =0=Ay+Cy-30 = Ay Ay = -120 kn Cy D F b) B, C,, are zero force c) section cut. Σ F =0=6F -4(30) = arctan(6/4)=56.3 =0= F F sin 30 =0= -F F F F F cos 0 = -20 F F 36.1(0.554) F = 20.0 kn (tension) F F = 36.1 kn (tension) F d) oint A =0= -F AB sin +Ay 0 = -F AB (0.832) -120 F AB = kn (comp)

Problem 3 (20 points) The truss is in static equilibrium. t is pinned at A and has a roller at C. The overall length is 20m and the overall height is 6 m (i.e. each grid line is 1 m).

8 Problem 3 (20 points) The truss is in static equilibrium. t is pinned at A and has a roller at C. The overall length is 20m and the overall height is 6 m (i.e. each grid line is 1 m). A load is applied at point. A B Solution version D K C D F a) Draw an overall free body diagram and find the reactions at the supports (7 pts) b) dentify A zero force members (3 point) c) Find the forces in members K and K. ndicate if they are in tension or compression (7 pts) d) Find the forces in member AB. ndicate if it is in tension or compression (3 pts) Σ A =0=4Cy-20(30) K Cy = 150 kn =0= = 0 kn =0=Ay+Cy-30 = Ay Ay = -120 kn Cy D F b) B, C,, are zero force c) section cut. Σ =0=6F K -8(30) = arctan(6/4)=56.3 =0= F K sin 30 =0= -F K F D F K cos 0 = -40 F F 36(0.554) F K = (tension) F K = 36.0 kn (tension) d) oint A =0= -F AB sin +Ay F AB = -144 kn (comp) 0 = -F AB (0.832) -120 F

Engineering Mechanics: Statics in SI Units, 12e

Engineering Mechanics: Statics in SI Units, 12e 5 Equilibrium of a Rigid Body Chapter Objectives Develop the equations of equilibrium for a rigid body Concept of the free-body diagram for a rigid body