Question

A rectangle is formed by attaching two rods of length 28 cm and two rods of length 11 cm as shown below. А. P B The ends of the rods are attached to one another by hinges so that the rods can move. The blue hinges are anchored to the ground and a weight is attached to the top rod at point P, which causes @ to decrease. Suppose that when 0 = the vertical distance between the top and bottom rods is decreasing at a rate of 0.7 cm per second. At what rate is the distance between A and B changing when e cm per second (Round to the nearest 0.001 cm per second.)

Answer 1

A 11cm 180-9 28cm "B From Parallelogram law of vector addition, AB² = DB² + Bc² + 2(DB) (BC) Cos (180-0) AB2 = (28)2 + cw2 + 2(28) (11) Cos (180-0) with Ae? - (28)2 + (12 – 2(28) (1) cose Differ entrating on (ABZ) = [(28*+ C902- 2(28)(w) C050] + 2 (28 (117) sine do 2 A0 (demo) Oto Sino do when sino / 308 dt AB dt AB=√(28)2 + (1112 -2 (28) (11) Cos (7/6) = 19.2751 de @ ? dt between AC & BD =h AD Sino - dh AD COSo do dt at dh = -0.7 cm/s ; dt AB= 11 0.7 (2) -104 Il 3 TV3 Vertical distance h= Coso = 3 Given, det

d(AB) dt Sino. 308 do dt AB 1.4 Sin (I). - 308 19.2751 = -0.5871 cmls Distance between A & B is decreasing at the rate of 0.587 cm/s. Hope This Helps ..