PHYSICS 170 MECHANICS I Final Exam: 12 noon to 3:00 pm, Monday, April 20, 2020 SOLUTION TO QUESTION 1 Determine the components of reaction at the ball-and-socket joint A and the tension in each cable necessary for equilibrium of the rod. x B SOLUTION Force And Position Vectors. The coordinates of points A, B, C, D and E are A(0, 0, 0), B(6, 0, 0), C(0, -2, 3) m, D(0, 2, 3) m and E(3, 0, 0) m respectively. FBC = FBC = FBC TBC. FBC (0 - 6)i + (-2 - 0)j + (3 - 0)k V(0 - 6)2 + (-2 - 0)2 + (3 - 0)2 i = – Find - ? Faci + Fack IBD FBD = FBD = FBD IBD (0 - 6)i + (2 - 0)j + (3 - 0)k + N/A m/A V(0 - 6)2 + (2 - 0)2 + (3 - 0)2 := – Fedi + ? Fooj+ Fak FA = A,i + Ayj + A.k F = {-600k) N TAB = {6i} m TAE = [3] m Equations of Equilibrium. Referring to the FBD of the rod shown in Fig. a, the force equation of equilibrium gives ?? = 0; FBC + FBD + FA + F = 0 (?FBC - ? FBD + Ai)+ (? FHD -? FBC + A3)?+( FAC +3 FBD+A2-600 k=0 * 2m TAB Ax 3m FBC FOD Any y 3m Az MÃE 3m x F=600N (a) page 1 Z 2 m Ce 2 m D 0 A 3 m E -3 m -3 m V y 600 N
PHYSICS 170 MECHANICS I Final Exam: 12 noon to 3:00 pm, Monday, April 20, 2020 SOLUTION TO QUESTION 1 Equating i, j and k components, 9FBC - 6 FBD + A = 0 (1) N/A ? FBD -? FBC +A1 =0 (2) N/A FBC + FBD + A2 - 600 = 0 (3) The moment equation of equilibrium gives ŽMA = 0; TAE X F + TAB X (FBC + FBD) = 0 ?i j k 3 0 0 + 0 0 -600 i 6 6 (FBC + FBD) j 0 (FBD - FBC) k 0 =0 3 7(FBC + FBD) 1800 - 7 (FBC + FBD) j + 7 (FBD - FBC)k = 0 12 Equating j and k components, 1800 - (FBC + FBD) = 0 (4) (FBD - FBC) = 0 (5) Solving Eqs. (1) to (5), FBD = FBC = 350 N Ans. Ar = 600 N Ans. Ay = 0 Ans. Az = 300 N Ans. page 2 Ans: FBD= FBC = 350 N Ar = 600 N Ay = 0 A2 = 300 N
PHYSICS 170 MECHANICS I Final Exam: 12 noon to 3:00 pm, Monday, April 20, 2020 SOLUTION TO QUESTION 2 page 3 Block A has a weight of 8 lb and block B has a weight of 6 lb. They rest on a surface for which the coefficient of kinetic friction is ps = 0.2. If the spring has a