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 o e.q up .buhgwe5be05hq-w :3 rk +0 /l ::w hsny:eeybf/zp5cxf a bicycle of mass M has two tires that each comes into contact with the ground at a point. The wheelbase of this bicycle (the distance between t z/y+::fewpls /x kn5: y0ec bhhe 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 has 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 $a$ so that both tires remain in contact with the ground? 
A. $\frac{wg}{h}$
B. $\frac{wg}{2h}$
C. $\frac{hg}{2w}$
D. $\frac{h}{2wg}$
E. none of the above
参考答案: B
本题详细解析:
From the torque analysis in problem 28, we derived thbe *2 xl-qh7bhl:rf4vor/5uie equation $(N_1 - N_2)w/2 = (Ma)h$.
We also have the vertical force equation $N_1 + N_2 = Mg$.
We can solve for $N_2$: From the torque equation, $N_1 - N_2 = 2Mah/w$.
Subtract this from $N_1+N_2=Mg$: $(N_1+N_2) - (N_1-N_2) = Mg - 2Mah/w \implies 2N_2 = Mg(1 - 2ah/wg)$.
$N_2 = \frac{1}{2}Mg(1 - 2ah/wg)$.
The rear tire lifts off the ground when $N_2 = 0$.
This occurs when:
$1 - 2ah/wg = 0$
$1 = 2ah/wg$
$wg = 2ah \implies a = \frac{wg}{2h}$.
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