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 t htz ws:+;e fb rx*reavf . qosf1/.o+5ires that each comes in k*py c l) 9zvq04afbxl; qc(g6to 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 has a magnitude a. Air resistance may be ignored. Assume that the coefficient of sliding friction between each tire an *z qcqcv9k;gl0l4y )( b 6apfxd 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 derivogd -t2 rgdt j1k6f7.ged the 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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