Two rectangles ABCD and EFGH represent two rectangular areas at sea. The points D, A, H and E are collinear. The point C, B , G , F are also collinear. The breadth of rectangles are a and the separation between the two is b. There is a current with constant speed V flowing in the direction along AH.
(a) A vessel P wishes to pass with constant velocity u, relative to the water, through the gap between the danger areas. Find the least value of |u| if P is just aviod entering either area.
(b) A second vessel Q also wishes to pass with constant velocity, relative to the water between the danger areas, but does not wish to approach closer than a distance c to the nearest point of either. Show that the least speed with which Q must move, relative to the water, is
V{2bc+a(a^2 + b^2 - 4c^2)^0.5 } / (a^2 + b^2)
Applied Maths - Mechanics 21
2009-08-01 2:09 am
回答 (1)
2009-08-01 5:36 am
✔ 最佳答案
Refer to the diagramhttp://img66.imageshack.us/img66/6034/vessel.jpg
(a) P should enter point B and sails at an angle y to AB at speed u
resultant speed AB direction ux = ucosy
resultant speed BG direction uy = v – usiny
ux/uy = a/b
bucosy = a(v – usiny)
u = av/(bcosy + asiny)
u is minimum when bcosy + asiny is maximum
Let tan p = a/b, so sin p = a/√(a2 + b2); cos p = b/√(a2 + b2)
bcosy + asiny =√(a2 + b2)(cospcosy + sinpsiny) = √(a2 + b2)cos(p-y)
Since maximum is √(a2 + b2), so minimum u = av/√(a2 + b2)
(b) Some geometry first
KHcosX = 2c
AK = b – KH
tanX = AK/a
tanX = (b – KH)/a
tanX = (b – 2c/cosX)/a
asinX = bcosX – 2c
a2sin2X = b2cos2X – 4bccosX + 4c2
a2 – a2cos2X = b2cos2X – 4bccosX + 4c2
(a2 + b2)cos2X – 4bccosX + 4c2 – a2 = 0
Let's assume c is so small that 4c2 – a2 is negative
Solving quadratic equation,
cosX = {4bc +/- √[16b2c2 – 4(a2+b2)(4c2 – a2)]}/2(a2 + b2)
cosX = {4bc + √[16b2c2 – 16a2c2 + 4a4 -16b2c2 + 4a2b2]}/2(a2 + b2), negative sign removed since it will cause cosX becomes negative.
cosX = {2bc + √[– 4a2c2 + a4 + a2b2]}/(a2 + b2)
cosX = {2bc + a√(a2 + b2 – 4c2)}/(a2 + b2)
The speed analysis is similar to part (a),
resultant speed AB direction ux = ucosz
resultant speed BG direction uy = v – usinz
ux/uy = ucosz / (v – usinz) = cotx
ucoszsinx = (v – usinz)cosx
u = vcosx/(coszsinx + sinzcosx)
u = vcosx/sin(z+x)
minimum u = vcosx when sin(z+x) = 1
minimum u = V{2bc + a√(a2 + b2 – 4c2)}/(a2 + b2)
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