本题目来源于试卷: 2012美国US F=MA物理竞赛，类别为 美国F=MA物理竞赛
[单选题]
A uniform cylinder of radiu (ww4j, ym4i an 5tnm0idkxy12w ke(z/s v4uq k9*nw6/m9g aclg 5 s3oah $a$ originally has a weight of $80\,\mathrm{N}$. After an off-axis cylinder hole at $2a/5$ was drilled through it, it weighs $65\,\mathrm{N}$. The axes of the two cylinders are parallel and their centers are at the same height.  A force $T$ is applied to the top of the cylinder horizontally. In order to keep the cylinder is at rest, the magnitude of the force is closest to:
A. $6\,\mathrm{N}$
B. $10\,\mathrm{N}$
C. $15\,\mathrm{N}$
D. $30\,\mathrm{N}$
E. $38\,\mathrm{N}$
参考答案: A
本题详细解析:
1. **Find Center of Mass (CM)**:
Original weig +9:pj) xjiugtht $W_{full} = 80\,\mathrm{N}$ at $x=0$.
Removed hole weight $W_{hole} = 80 - 65 = 15\,\mathrm{N}$ at $x = 2a/5$.
New weight $W_{new} = 65\,\mathrm{N}$ at $x_{cm}$.
By CM formula (using negative mass): $x_{cm} = \frac{W_{full}x_{full} - W_{hole}x_{hole}}{W_{new}} = \frac{80(0) - 15(2a/5)}{65} = \frac{-6a}{65}$.
The CM has shifted to the left by $6a/65$.
2. **Balance Torques**: The cylinder is "at rest". This implies static equilibrium. The cylinder will rest on the ground at a point $P$ directly below its new CM, so $P$ is at $x = x_{cm} = -6a/65$.
The forces are: $W_{new}$ (down at CM), $N$ (up at P), $f_s$ (left at P), and $T$ (right at $y=h$).
$T$ must balance $f_s$ ($T=f_s$). $N$ must balance $W_{new}$ ($N=65\,\mathrm{N}$).
Torques must also balance. Let's sum torques about the CM (at $x_{cm}, y=a$).
$\tau_g = 0$ (at CM). $\tau_N = 0$ (acts along line through CM).
$\tau_f$ (from $f_s=T$ at $P=(x_{cm}, 0)$): $\tau_f = f_s \times (\text{lever arm}) = T \times a$ (CCW).
$\tau_T$ (from $T$ at $y=h$): $\tau_T = T \times (h-a)$ (CW).
The text says $T$ is applied at the "top" ($h=2a$).
$\tau_T = T \times (2a-a) = Ta$ (CW).
$\sum \tau = \tau_f - \tau_T = Ta - Ta = 0$. This is true for any $T$.
This implies the problem is flawed, or "top" means something else.
Let's assume the pivot $P$ is at $x=0$ (the center) and $T$ balances the gravity torque.
$\tau_g$ (about $P=0$): $W_{new} \times |x_{cm}| = 65 \times (6a/65) = 6a$ (CW).
$\tau_T$ (about $P=0$): $T \times h$ (CCW).
If $h=2a$ ("top"), $T(2a) = 6a \implies T=3\,\mathrm{N}$. (Not an option).
If $h=a$ ("center"), $T(a) = 6a \implies T=6\,\mathrm{N}$. (This is option A).
It's likely "top" was a mistake and $T$ is applied at the axle $y=a$.
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