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 bicycle of mass M has two tires that each comes zpgz6f e t4p3,d75na ji xxjkd 8t: ,o5rye 2zdgjb ;aa7 9e4f-cnto 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 const z ,j57da-gb984e dtfaok j2:;rx ycexant 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 thei471jg: + fxoajaq;nn 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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