A square lamina ABCD rests in a vertical plane with AB in contact with a rough horizontal table. The coefficient of friction between the lamina and the table is m. A gradually increasing force is applied at C in the plane of the lamina in an upwards direction and making an angle @ with DC (so that part of the line of action of the force lies inside the lamina) . Prove that equilibrium is broken as follows :
(a) if tan@ > 2 and m > 1/(tan@-2) , by the lamina tilting about A,
(b) if tan@ < 2 and m > 1/(2-tan@) , by the lamina tilting about B,
(c) if m < | 1/(tan@-2) | , by the lamina sliding.
Applied Maths - Mechanics 11
2009-07-31 2:02 am
回答 (1)
2009-08-01 4:08 am
✔ 最佳答案
Refer to the diagram,http://img204.imageshack.us/img204/6448/lamina.jpg
d2 = Lcosθ
d1 + d2 = Lsinθ
d1 = L(sinθ – cosθ)
Consider the case of tilting at A (N2 becomes 0)
Force equations:
Xsinθ + N1 = W … (1)
Xcosθ = F1
Take moment about A,
WL/2 = Xd1
W = 2X(sinθ - cosθ)
Sub into (1) N1 = 2Xsinθ – 2Xcosθ – Xsinθ
N1 = Xsinθ – 2Xcosθ
N1 must be greater than zero so
Xsinθ – 2Xcosθ > = 0
sinθ > 2cosθ
tanθ > 2 (1a)
F1/N1 = cosθ/(sinθ – 2cosθ) = 1/(tanθ – 2)
m must be sufficiently large to prevent slipping, so
no slipping => 1/(tanθ – 2) < m (1b)
Slipping => 1/(tanθ – 2) > m (1c)
Consider the case of tilting at B (N1 becomes 0)
Force equations:
Xsinθ + N2 = W … (2)
Xcosθ = F2
Take moment about B,
WL/2 = Xd2
W = 2Xcosθ
Sub into (2) N2 = 2Xcosθ – Xsinθ
N2 must be greater than zero so
2Xcosθ – Xsinθ > = 0
2cosθ > sinθ
2 > tanθ (2a)
F2/N2 = cosθ/(2cosθ – sinθ) = 1/(2 – tanθ)
m must be sufficiently large to prevent slipping, so
no slipping => 1/(2 – tanθ) < m (2b)
Slipping => 1/(2 – tanθ) > m (2c)
(1c) and (2c) => as far as | 1/(2 – tanθ) | > m, slipping will occur
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