정답: $2$

직선상에 물체가 총 $nn$개 있고, $ii$번째 물체의 질량을 $m_{i}m\_\{i\}$, 기준점(원점)으로부터 $ii$번째 물체의 질량 중심까지의 거리를 $r_{i}r\_\{i\}$라 할 때,
이 물체들로 이루어진 계의 질량 중심의 위치는 ${\displaystyle \frac{{\displaystyle \sum _{i=1}^n}{m}_i{r}_i}{{\displaystyle \sum _{i=1}^n}{m}_i}}\backslash dfrac\{\{\backslash displaystyle\backslash sum\_\{i=1\}^\{n\}\}m\_\{i\}r\_\{i\}\}\{\{\backslash displaystyle\backslash sum\_\{i=1\}^\{n\}\}m\_\{i\}\}$이다.

물체가 총 $33$개이므로 $n=3n\; =\; 3$이고, $\mathrm{A}\backslash mathrm\{A\}$, $\mathrm{B}\backslash mathrm\{B\}$, $\mathrm{C}\backslash mathrm\{C\}$를 각각 $11$, $22$, $33$번째 물체라 하면
$m_1=1\mathrm{k}\mathrm{g}m\_\{1\}\; =\; 1\backslash ,\backslash mathrm\{kg\}$, $m_2=7\mathrm{k}\mathrm{g}m\_\{2\}\; =\; 7\backslash ,\backslash mathrm\{kg\}$, $m_3=3\mathrm{k}\mathrm{g}m\_\{3\}\; =\; 3\backslash ,\backslash mathrm\{kg\}$이다.

따라서 ${\displaystyle \sum _{i=1}^n}{m}_i=\{\backslash displaystyle\backslash sum\_\{i=1\}^\{n\}\}m\_\{i\}\; =\{\}$${\displaystyle \sum _{i=1}^3}{m}_i=\{\backslash displaystyle\backslash sum\_\{i=1\}^\{3\}\}m\_\{i\}\; =\{\}$$m_1+m_2+m_3=m\_\{1\}\; +\; m\_\{2\}\; +\; m\_\{3\}\; =\{\}$$1\mathrm{k}\mathrm{g}+7\mathrm{k}\mathrm{g}+3\mathrm{k}\mathrm{g}=1\backslash ,\backslash mathrm\{kg\}\; +\; 7\backslash ,\backslash mathrm\{kg\}\; +\; 3\backslash ,\backslash mathrm\{kg\}\; =\{\}$$11\mathrm{k}\mathrm{g}11\backslash ,\backslash mathrm\{kg\}$이다.

$x=0x\; =\; 0$을 기준점이라 하면 $r_1=1\mathrm{m}r\_\{1\}\; =\; 1\backslash ,\backslash mathrm\{m\}$, $r_2=2\mathrm{m}r\_\{2\}\; =\; 2\backslash ,\backslash mathrm\{m\}$, $r_3=6\mathrm{m}r\_\{3\}\; =\; 6\backslash ,\backslash mathrm\{m\}$이다.

그러므로 ${\displaystyle \sum _{i=1}^n}{m}_i{r}_i={\displaystyle \sum _{i=1}^3}{m}_i{r}_i\{\backslash displaystyle\backslash sum\_\{i=1\}^\{n\}\}m\_\{i\}r\_\{i\}\; =\; \{\backslash displaystyle\backslash sum\_\{i=1\}^\{3\}\}m\_\{i\}r\_\{i\}$,
${\displaystyle \sum _{i=1}^n}{m}_i{r}_i=m_1{r}_1+m_2{r}_2+m_3{r}_3\{\backslash displaystyle\backslash sum\_\{i=1\}^\{n\}\}m\_\{i\}r\_\{i\}\; =\; m\_\{1\}r\_\{1\}\; +\; m\_\{2\}r\_\{2\}\; +\; m\_\{3\}r\_\{3\}$,
${\displaystyle \sum _{i=1}^n}{m}_i{r}_i=(1\mathrm{k}\mathrm{g}\times 1\mathrm{m})+(7\mathrm{k}\mathrm{g}\times 2\mathrm{m})+(3\mathrm{k}\mathrm{g}\times 6\mathrm{m})\{\backslash displaystyle\backslash sum\_\{i=1\}^\{n\}\}m\_\{i\}r\_\{i\}\; =\; (1\backslash ,\backslash mathrm\{kg\}\; \backslash times\; 1\backslash ,\backslash mathrm\{m\})\; +\; (7\backslash ,\backslash mathrm\{kg\}\; \backslash times\; 2\backslash ,\backslash mathrm\{m\})\; +\; (3\backslash ,\backslash mathrm\{kg\}\; \backslash times\; 6\backslash ,\backslash mathrm\{m\})$,
${\displaystyle \sum _{i=1}^n}{m}_i{r}_i=33\mathrm{k}\mathrm{g}\cdot \mathrm{m}\{\backslash displaystyle\backslash sum\_\{i=1\}^\{n\}\}m\_\{i\}r\_\{i\}\; =\; 33\backslash ,\backslash mathrm\{kg·m\}$이다.

${\displaystyle \frac{{\displaystyle \sum _{i=1}^n}{m}_i{r}_i}{{\displaystyle \sum _{i=1}^n}{m}_i}}={\displaystyle \frac{{\displaystyle \sum _{i=1}^3}{m}_i{r}_i}{{\displaystyle \sum _{i=1}^3}{m}_i}}={\displaystyle \frac{33\mathrm{k}\mathrm{g}\cdot \mathrm{m}}{11\mathrm{k}\mathrm{g}}}=3\mathrm{m}\backslash dfrac\{\{\backslash displaystyle\backslash sum\_\{i=1\}^\{n\}\}m\_\{i\}r\_\{i\}\}\{\{\backslash displaystyle\backslash sum\_\{i=1\}^\{n\}\}m\_\{i\}\}\; =\; \backslash dfrac\{\{\backslash displaystyle\backslash sum\_\{i=1\}^\{3\}\}m\_\{i\}r\_\{i\}\}\{\{\backslash displaystyle\backslash sum\_\{i=1\}^\{3\}\}m\_\{i\}\}\; =\; \backslash dfrac\{33\backslash ,\backslash mathrm\{kg·m\}\}\{11\backslash ,\backslash mathrm\{kg\}\}\; =\; 3\backslash ,\backslash mathrm\{m\}$이므로 계의 질량 중심의 위치는 $x=3\mathrm{m}x\; =\; 3\backslash ,\backslash mathrm\{m\}$이다.

따라서 정답은 $22$번이다.

정답: $4$

직선상에 물체가 총 $nn$개 있고, $ii$번째 물체의 질량을 $m_{i}m\_\{i\}$, 기준점(원점)으로부터 $ii$번째 물체의 질량 중심까지의 거리를 $r_{i}r\_\{i\}$라 할 때,
이 물체들로 이루어진 계의 질량 중심의 위치는 ${\displaystyle \frac{{\displaystyle \sum _{i=1}^n}{m}_i{r}_i}{{\displaystyle \sum _{i=1}^n}{m}_i}}\backslash dfrac\{\{\backslash displaystyle\backslash sum\_\{i=1\}^\{n\}\}m\_\{i\}r\_\{i\}\}\{\{\backslash displaystyle\backslash sum\_\{i=1\}^\{n\}\}m\_\{i\}\}$이다.

물체가 총 $33$개이므로 $n=3n\; =\; 3$이고, $\mathrm{A}\backslash mathrm\{A\}$, $\mathrm{B}\backslash mathrm\{B\}$, $\mathrm{C}\backslash mathrm\{C\}$를 각각 $11$, $22$, $33$번째 물체라 하면
$m_1=2\mathrm{k}\mathrm{g}m\_\{1\}\; =\; 2\backslash ,\backslash mathrm\{kg\}$, $m_2=3\mathrm{k}\mathrm{g}m\_\{2\}\; =\; 3\backslash ,\backslash mathrm\{kg\}$, $m_3=1\mathrm{k}\mathrm{g}m\_\{3\}\; =\; 1\backslash ,\backslash mathrm\{kg\}$이다.

따라서 ${\displaystyle \sum _{i=1}^n}{m}_i=\{\backslash displaystyle\backslash sum\_\{i=1\}^\{n\}\}m\_\{i\}\; =\{\}$${\displaystyle \sum _{i=1}^3}{m}_i=\{\backslash displaystyle\backslash sum\_\{i=1\}^\{3\}\}m\_\{i\}\; =\{\}$$m_1+m_2+m_3=m\_\{1\}\; +\; m\_\{2\}\; +\; m\_\{3\}\; =\{\}$$2\mathrm{k}\mathrm{g}+3\mathrm{k}\mathrm{g}+1\mathrm{k}\mathrm{g}=2\backslash ,\backslash mathrm\{kg\}\; +\; 3\backslash ,\backslash mathrm\{kg\}\; +\; 1\backslash ,\backslash mathrm\{kg\}\; =\{\}$$6\mathrm{k}\mathrm{g}6\backslash ,\backslash mathrm\{kg\}$이다.

$x=0x\; =\; 0$을 기준점이라 하면 밀도가 균일하고 좌우대칭인 물체의 질량 중심은 물체의 중앙에 위치하므로
$r_1=1\mathrm{m}r\_\{1\}\; =\; 1\backslash ,\backslash mathrm\{m\}$이고, $r_2=5\mathrm{m}r\_\{2\}\; =\; 5\backslash ,\backslash mathrm\{m\}$, $r_3=7\mathrm{m}r\_\{3\}\; =\; 7\backslash ,\backslash mathrm\{m\}$이다.

그러므로 ${\displaystyle \sum _{i=1}^n}{m}_i{r}_i={\displaystyle \sum _{i=1}^3}{m}_i{r}_i\{\backslash displaystyle\backslash sum\_\{i=1\}^\{n\}\}m\_\{i\}r\_\{i\}\; =\; \{\backslash displaystyle\backslash sum\_\{i=1\}^\{3\}\}m\_\{i\}r\_\{i\}$,
${\displaystyle \sum _{i=1}^n}{m}_i{r}_i=m_1{r}_1+m_2{r}_2+m_3{r}_3\{\backslash displaystyle\backslash sum\_\{i=1\}^\{n\}\}m\_\{i\}r\_\{i\}\; =\; m\_\{1\}r\_\{1\}\; +\; m\_\{2\}r\_\{2\}\; +\; m\_\{3\}r\_\{3\}$,
${\displaystyle \sum _{i=1}^n}{m}_i{r}_i=(2\mathrm{k}\mathrm{g}\times 1\mathrm{m})+(3\mathrm{k}\mathrm{g}\times 5\mathrm{m})+(1\mathrm{k}\mathrm{g}\times 7\mathrm{m})\{\backslash displaystyle\backslash sum\_\{i=1\}^\{n\}\}m\_\{i\}r\_\{i\}\; =\; (2\backslash ,\backslash mathrm\{kg\}\; \backslash times\; 1\backslash ,\backslash mathrm\{m\})\; +\; (3\backslash ,\backslash mathrm\{kg\}\; \backslash times\; 5\backslash ,\backslash mathrm\{m\})\; +\; (1\backslash ,\backslash mathrm\{kg\}\; \backslash times\; 7\backslash ,\backslash mathrm\{m\})$,
${\displaystyle \sum _{i=1}^n}{m}_i{r}_i=24\mathrm{k}\mathrm{g}\cdot \mathrm{m}\{\backslash displaystyle\backslash sum\_\{i=1\}^\{n\}\}m\_\{i\}r\_\{i\}\; =\; 24\backslash ,\backslash mathrm\{kg·m\}$이다.

${\displaystyle \frac{{\displaystyle \sum _{i=1}^n}{m}_i{r}_i}{{\displaystyle \sum _{i=1}^n}{m}_i}}={\displaystyle \frac{{\displaystyle \sum _{i=1}^3}{m}_i{r}_i}{{\displaystyle \sum _{i=1}^3}{m}_i}}={\displaystyle \frac{24\mathrm{k}\mathrm{g}\cdot \mathrm{m}}{6\mathrm{k}\mathrm{g}}}=4\mathrm{m}\backslash dfrac\{\{\backslash displaystyle\backslash sum\_\{i=1\}^\{n\}\}m\_\{i\}r\_\{i\}\}\{\{\backslash displaystyle\backslash sum\_\{i=1\}^\{n\}\}m\_\{i\}\}\; =\; \backslash dfrac\{\{\backslash displaystyle\backslash sum\_\{i=1\}^\{3\}\}m\_\{i\}r\_\{i\}\}\{\{\backslash displaystyle\backslash sum\_\{i=1\}^\{3\}\}m\_\{i\}\}\; =\; \backslash dfrac\{24\backslash ,\backslash mathrm\{kg·m\}\}\{6\backslash ,\backslash mathrm\{kg\}\}\; =\; 4\backslash ,\backslash mathrm\{m\}$이므로 계의 질량 중심의 위치는 $x=4\mathrm{m}x\; =\; 4\backslash ,\backslash mathrm\{m\}$이다.

따라서 정답은 $44$번이다.