本题目来源于试卷: 2007美国US F=MA物理竞赛，类别为 美国F=MA物理竞赛
[单选题]
A simplified model of a bicy 80ya0 whsn; bqcle of mass M has two qvu2 2,g1t trd1s oy(r*3pgox tires that each comes into contact with the ground at a point. The wheelbase of this bicycle (the distance between the points of contact with the ground) is w, and the center of mass of the bicycle is located midway between the tires and a height h above the ground. The bicycle is moving to the right, but slowing down at a constant rate. The acceleration o2ug ,y3o1*dvr2xg1 (pstrqthas a magnitude a. Air resistance may be ignored. Assume that the coefficient of sliding friction between each tire and the ground is $\mu$, and that both tires are skidding: sliding without rotating. What is the maximum value of $\mu$ so that both tires remain in contact with the ground? 
A. $\frac{w}{2h}$
B. $\frac{h}{2w}$
C. $\frac{2h}{w}$
D. $\frac{w}{h}$
E. none of the above
参考答案: A
本题详细解析:
The bicycle is "slowing down", which mean2wz2 7si jgwuzx;a.7b s the net friction force $f = f_1+f_2$ causes a deceleration $a$.
This braking torque tends to lift the rear tire ($N_2$).
We find the maximum $a$ (and $\mu$) just as $N_2$ becomes 0.
Vertical forces: $N_1 + N_2 = Mg$.
When $N_2=0$, $N_1=Mg$.
Horizontal forces: $f = f_1 + f_2 = Ma$.
Since $f_1 = \mu N_1$ and $f_2 = \mu N_2$, we have $f = \mu(N_1+N_2) = \mu Mg$.
So, $Ma = \mu Mg \implies a = \mu g$.
Torques about the center of mass (CM): The normal forces $N_1$ (at $w/2$ left) and $N_2$ (at $w/2$ right) create torques, as do the friction forces $f_1$ and $f_2$ (at height $h$ below CM).
$\tau_{net} = N_1(w/2) - N_2(w/2) - (f_1+f_2)h = 0$ (for no rotational acceleration).
$(N_1 - N_2)w/2 = (f_1+f_2)h = (Ma)h$.
$(N_1 - N_2)w/2 = M(\mu g)h$.
At the limit where the rear tire lifts, $N_2=0$ and $N_1=Mg$:
$(Mg - 0)w/2 = M\mu g h$.
$Mg w / 2 = M\mu g h$.
$w/2 = \mu h \implies \mu = \frac{w}{2h}$.
(The solution PDF solves for acceleration, which lifts the front wheel, but the problem's symmetry yields the same algebraic result.)
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