By A.G. Buckley, J-.L. Goffin

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**Additional info for Algorithms for Constrained Minimization of Smooth Nonlinear Functions**

**Sample text**

Collide if/when AB = 0, then t = 2 sop = 2. When t = I, OA = 2i- 3j and OB = 3i- 2j. After further time t 1 , OA = 2i - 3j + t 1 (2i + 3j), OB = 3i- 2j + t 1 (i- 4j), AB=(l-t 1 )i+(l-7t 1 )j. This is never zero, so collision is avoided. 8 m s- 1 , rp = (9t + c 1 ) i + (6t + c 2 )j. 4ms- 1 ,rQ =(5t+c 3 )i+(4t+c 4 )j. Conditions when t = 4 give c 1 = 60, c 2 = 20, c 3 = 80, c 4 = 80. (a) When t = 0, rp = 60i + 20j, rQ = 80i + 80j. (b) PQ = rQ - rp = (20- 4t) i + (60- 2t)j. (c) PQ = y(400- I60t + 16t 2 + 3600- 240t + 4t 2 ) = y[20 (t- 10) 2 + 2000].

A). (a) Show that the weight will slide down the plane. ) . sm(a- + {3) Obtain the least value of P as {3 varies and the corresponding value of {3. < ~,what is the direction and magnitude of the smallest force which will just move the weight up the plane? (SUJB) • (a) Let F be the force required for equilibrium. Resolving normal to the plane, R = W cos 0'. Resolving parallel to the plane, F = W sin a. F =tan a R w F=R tana>R tan A. ). R. R. Thus the weight will slide down the plane. (b) When the weight is just prevented from sliding down the plane the maximum possible friction F acts up the plane The resultant of R and F is R 1 acting at an angle A.

Resultant is I 0 N along 4y = 3x, x increasing. 2 --+, ~, R Rx = 1 cos60° + 3 + 5 cos60°- 6 + 2y3 cos30° = 3 N. Ry =-I cos30° + 5 cos30°- 2y3 sin30° = y3 N. e. 3ff to AB on opposite side to D and C. \ Taking moments about A, A 6 3 (a ~3) + 5 (ay3) = RyAX. Therefore AX= 1; a. 3 Taking moments about P, T= 30yl3 ~52. T (21) cos 60° :; 60 (I) sin 60°. 4 Moments about A: P(3a) = S (a)+ W (2a) cos()+ 2W ( 4a) cos(). Resolving vertically, 3 W + S cos() = P cos(). 2W Solving, P= W (10cos 2 S= w00cos 2 0- 9)' 2 cos () ()- 2 cos () 3).