Engineering Mechanics I Mansoura University 2011-2012 PDF

Mansoura University Engineering Mechanics I Faculty of Engineering First term (2011-2012) CIE Program. Final Exam Time: Tw...

Mansoura University Engineering Mechanics I Faculty of Engineering First term (2011-2012) CIE Program. Final Exam Time: Two hours Instructor: Prof. Dr. Bishri Abdel-Mo'emen Answer the Following Questions (Total mark 50 points) y F=1000 N o 60 Question 1: ( 5 Points ) o 30 Determine the x and y components of the 1000 N force shown in Figure. x v F=1000 N Fu= -1000 cos 30o = -866 N Fv = 1000 sin 30o o Fu 30 Fv o = 500 N 30 u Question 2: ( 5 Points ) Determine the magnitude of FA and its direction θ so that the resultant force is directed along the positive x axis and has a magnitude of 1250 N. 1250 N 30o α FA FB=800 N √( ) ( ) ( )( ) FA = 686 N ( ) 35.7 54.3o Another Solution: ∑ 1250 = FA sin θ +800 cos 30o FA sin θ = 557.2 (a) ∑ 0 = FA cos θ -800 sin 30o FA cos θ = 400 (b) Solving (a) and (b), gives : FA= 686 N and θ = 54.3o Question 3: ( 8 Points ) Determine the tension developed in each rope used to support the 40-kg chandelier. Equilibrium of Point D: ∑ ------ (1) ∑ ( ) ------ (2) Solving (1) and (2) : Thus, and FDB Equilibrium of Point B: FDC D ∑ 45 o 30 o ------ (3) 40(9.8) N ∑ ------ (4) Solving (3) and (4): FBA o B FBC 30 o 45 FDB Question 4: ( 8 Points ) The articulated crane boom has a weight of 400 N 1.5 m A and a center of gravity at G. If it supports load of 0.4m G 1200 N, determine the force acting at the pin A and the force in the hydraulic cylinder BC when the B 3m boom is in the position shown. 40o 0.4 m C Ay 400 N 1.5 m Ax A 0.4m G B 3m FB 40o 0.4 m C 1200 N Question 4: Solution ∑⬚ 𝑀𝐴 𝐹𝐵 ∗( ) 𝐹𝐵 ∗( ) ∗( ) ∗( ) 𝐹𝐵 𝑁𝑚 𝐴𝑛𝑠 ∑⬚ 𝐹𝑥 𝐴𝑥 𝐹𝐵 → 𝐴𝑥 𝑁 𝐴𝑛𝑠 ∑⬚ 𝐹𝑦 𝐴𝑦 𝐹𝐵 → 𝐴𝑦 𝑁 𝐴𝑛𝑠 Question 5: ( 5 Points ) Two forces F and F1 act on a bracket as shown in Figure. If the resultant force of F and F1 is directed along the positive y axis, determine the magnitude of the resultant force and the coordinate direction angles of F so that β < 90o.  B A Question 6: ( 5 Points ) A boy exerts a force F= ( 20 i – 200 k ) N on a z swing as shown. The force acts at the boy's center of mass G. Determine the moment created by F about G O the axis AB. The coordinates of bearings at A is y F (1,0.5, 2.5) and at bearing B is (1, 1.5, 2.5). x 0.5 m 1m Given: A: (1,0.5, 2.5) ; B: (1, 1.5, 2.5); F= ( 20 i – 200 k ) N G: (1.5,1, 0.5) , rAG=0.5i+0.5j-2k MA = | | = -100i – (-100+40)j+(-10 k) = (-100i + 60j -10 k) N.m The axis AB is parallel to y-axis. So, the moment about the axis is: 60 N.m. OR MAB= MA.UAB = (-100i + 60j -10 k).(j) = 60 N.m Question 7: ( 8 Points ) Two forces act on the post as shown in Figure. Replace the two forces by an equivalent force-couple system at point O. Question 8: ( 8 Points ) z 400 N Replace the loading shown by an equivalent single 3m resultant force and specify the x and y coordinates 200 N A y of its line of action. 2m 600 N 3m C B x A: ( 0, 3, 0) ; B: ( 3, 3, 0); C: (2, 0, 0) The Equivalent system at the Origin: 1. Summing the moment about the origin: M= 3j ^ (-400 k) + (3i+3j ^ (-600 k) + 2i ^ (200 k) = -1200 i + ( 1800 j – 1800 i) – 400 j = -3000 i + 1400 j 2. The resultant : R = -400 k -600 k + 200 k = -800 k The Single Resultant is : R = -800 k and acts at point (x, y, 0), where: (x i + y j )^ (R) = M (x i + y j )^ (-800 k) = -3000 i + 1400 j 800 x j – 800 y i= -3000 i + 1400 j x = (1400 /800 ) = 1.75 m and y = (3000 /800 ) = 3.75 m Good Luck and Happy New Year Dr. Bishri Abdel-Mo'emen