PHYSICS 170 MECHANICS 1: March Exam: March, 2024 SOLUTION TO QUESTION 1 (15 MARKS) page 1 a) Free-Body Diagram Bz 1 X 4 B(2,0,0) 30? C ( 3, 02 -2 ) 600 C Coordinates: A= (0,0,0) m B = (2,0,0) m C= (3,0, -2) m Mx Ax (4 marks) Z 1 Mz A (0,0, 0) J At My
PHYSICS 170 MECHANICS 1: March Exam: March, 2024 SOLUTION TO QUESTION 1 (15 MARKS) page 2 b) Cartesian component force equations of equilibrium (4 marks) Forces: À=A?+A,j ? = B_k C = 800(cos 60° cos 30°[ + cos 60° sin 30° ] - sin 60°k) EFT =0 : A + 800cos 60° cos30° = 0 (1) >Fy =0 : A, +800cos 60° sin 30° = 0 (2) EF2 =0 : B2 - 800 sin 60° = 0 (3) c) Vector moment equation of equilibrium at point C (3 marks) FCA =FA -Fc =- 30 +2k FCB =¥B-Ï ¿== { +2k Couple Moment: MA=M? +M j +Mk ( M ) = ? M + ? ( r × F = ) R C x 1. + y + A + r × B = 1 .~ jki z + × jk M?+M J+Mk+-3 Ax A y 0 2+-1 0 2 =0 O 0 0 B. N
PHYSICS 170 MECHANICS 1: March Exam: March, 2024 SOLUTION TO QUESTION 1 (15 MARKS) page 3 d) Cartesian component moment equations of equilibrium: (2 marks) Mx-2A, =0 (4) My+2Ax+B=0 (5) M2-3Ay=0 (6) e) Solution to the equations of equilibrium: (2 marks) AR =- 346 N A =- 200 N B2 = 693 N M x = - 400 Nm My =0.00 Nm M = - 600 Nm The negative signs indicate that the reaction components are directed along the negative coordinate axes.