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 $2\,\mathrm{kg}$ rock is suspended by a massless string from one end of a uniform $1\,\mathrm{meter}$ measuring stick. What is the mass of the measuring stick if it is balanced by a support force at the $0.20\,\mathrm{meter}$ mark?
A. $0.20\,\mathrm{kg}$
B. $1.00\,\mathrm{kg}$
C. $1.33\,\mathrm{kg}$
D. $2.00\,\mathrm{kg}$
E. $3.00\,\mathrm{kg}$
参考答案: C
本题详细解析:
Let the stick have mass2b ok( as0zb;z0 uvi+n $M$ and length $L=1.0\,\mathrm{m}$.
The rock ($m=2\,\mathrm{kg}$) is at the $0.0\,\mathrm{m}$ mark. The pivot is at $x_p = 0.20\,\mathrm{m}$.
The center of mass of the uniform stick is at its center, $x_{cm} = 0.50\,\mathrm{m}$.
For static equilibrium, the net torque about the pivot must be zero.
The rock creates a counter-clockwise (CCW) torque, and the stick's weight creates a clockwise (CW) torque.
$\tau_{CCW} = \tau_{CW}$
$m g (x_p - 0) = M g (x_{cm} - x_p)$
$m (0.20) = M (0.50 - 0.20)$
$(2\,\mathrm{kg})(0.20\,\mathrm{m}) = M (0.30\,\mathrm{m})$
$0.40 = 0.30 M$
$M = 0.40 / 0.30 = 4/3 \approx 1.33\,\mathrm{kg}$.
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