Applied Maths - Mechanics 13

2009-07-31 2:11 am
A uniform circular disc of radius a rests in a vertical plane on two rough pegs distant sqrt(2)*a apart in a horizontal line; the coeficient of friction between the disc and each peg is m. A gradually increasing force acting in a direction parallel to the line of the pegs is applied at the highest point of the disc; prove that when equilibrium is first broken the disc will commence to rotate with its centre fixed or begin to turn about one of the pegs according as m > or < sqrt(2) - 1 .

回答 (1)

2009-08-01 1:40 am
✔ 最佳答案
Consider the disc is at the point of rotate at the point A. Therefore there is no normal reaction at the other peg.
http://img383.imageshack.us/img383/7604/disc.jpg
Resolve the forces in direction parallel to F & N,
F = W/√2 – X√2 … (1)
N = W/√2 + X√2 … (2)
Take moment about A,
X(1+√2/2)a = W√2/2a
W = X(√2 + 1)
Sub into (1) and (2)
F = X(√2 + 1)/√2 – X/√2 = X
N = X(√2 + 1)/√2 + X/√2 = X(1 + √2)
F/N = 1/(1 + √2)) = √2 – 1
If this is bigger than m, F cannot achieve the required friction for the disc to rotate at A, but rather will slip. So this is the cri


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