A uniform rod AB of weight W and length l leans over a smooth vertical wall of height h at the point C and on a smooth horizontal floor at the end B. A horizontal force P acts on the rod at its lower end B. In equilibrium, the rod makes an angle θ with the horizontal floor where sin-1(h/l) < θ < π/2. Let R be the normal reaction at the end B.
Find R and P in terms of θ, h, l and W.
If (√3)l/9 < h < l, show that R > 0 for sin-1(h/l) < θ < π/2.
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Applied maths: Statics
2009-08-24 9:59 pm
回答 (1)
2009-08-25 5:33 am
✔ 最佳答案
Let the reaction at C be F, perpendicular to the length of the rod.Taking moment about B,
Fh/sinθ = Wlcosθ/2 … (1)
Vertical force,
Fcosθ + R = W … (2)
Horizontal force,
Fsinθ = P … (3)
(1) => F = Wlsinθcosθ/2h
(3) => P = Wlsin2θcosθ/2h
(2) => R = W – Fcosθ
R = W – cosθWlsinθcosθ/2h
R = W(1 – sinθcos2θl/2h)
Firstly l > h in order for the rod to take support at C.
R = W(1 – sinθcos2θl/2h)
dR/dθ = -(Wl/2h)[cosθcos2θ + sinθ(2cosθ)(-sinθ)]
dR/dθ = -(Wl/2h)(cosθcos2θ – 2sin2θcosθ)
dR/dθ = -(Wl/2h)(cosθ – 3sin2θcosθ)
d2R/dθ2 = -(Wl/2h)[-sinθ – 3sin2θ(-sinθ) – 6sinθcosθcosθ]
d2R/dθ2 = -(W/2h)(-sinθ + 3sin3θ – 6sinθ + 6sin3θ)
d2R/dθ2 = -(Wl/2h)(9sin3θ – 7sinθ)
dR/dθ = 0 => cosθ = 0 or sin2θ = 1/3 or θ = π/2 or θ = sin-1(1/√3)
d2R/dθ2 at θ = π/2 is -Wl/h < 0 => maximum point
d2R/dθ2 at θ = sin-1(1/√3) is 2Wl/(√3h) > 0 => minimum point
Minimum R is W[1 – (1/√3)(2/3)l/2h] = W(1 – √3l/9h)
When h>√3/9l, minimum R = W(1 – √3l/9h) > W(1 – 1) = 0
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