✔ 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