Question: 2. When one boy is sitting 1.2m from the center a see-saw, another boy must to sit on the other side 1.5m from the center to maintain an even balance. However, when the first boy carries an additional weight of 14kg and sit 1.8m from the center, the second boy must move to 3m from the center to balance. Neglecting the weight of the see-saw, find the weight (in kg) of the heavier boy. Answer: __________
1. **State the problem:** We have a see-saw balanced by two boys sitting at different distances from the center. We want to find the weight of the heavier boy.
2. **Define variables:** Let $W_1$ be the weight of the first boy and $W_2$ be the weight of the second boy.
3. **Use the principle of moments:** For balance, the clockwise moment equals the anticlockwise moment.
4. **First balance condition:** When the first boy sits $1.2m$ from the center and the second boy sits $1.5m$ from the center,
$$W_1 \times 1.2 = W_2 \times 1.5$$
5. **Express $W_2$ in terms of $W_1$:**
$$W_2 = \frac{1.2}{1.5} W_1 = \frac{4}{5} W_1$$
6. **Second balance condition:** When the first boy carries an additional $14kg$ and sits $1.8m$ from the center, the second boy moves to $3m$ from the center,
$$ (W_1 + 14) \times 1.8 = W_2 \times 3 $$
7. **Substitute $W_2$ from step 5 into step 6:**
$$ (W_1 + 14) \times 1.8 = \frac{4}{5} W_1 \times 3 $$
8. **Simplify the right side:**
$$ (W_1 + 14) \times 1.8 = \frac{12}{5} W_1 $$
9. **Expand the left side:**
$$ 1.8 W_1 + 25.2 = \frac{12}{5} W_1 $$
10. **Rewrite $\frac{12}{5}$ as $2.4$ for easier calculation:**
$$ 1.8 W_1 + 25.2 = 2.4 W_1 $$
11. **Bring terms involving $W_1$ to one side:**
$$ 25.2 = 2.4 W_1 - 1.8 W_1 = 0.6 W_1 $$
12. **Solve for $W_1$:**
$$ W_1 = \frac{25.2}{0.6} $$
13. **Simplify the fraction:**
$$ W_1 = 42 $$
14. **Find $W_2$ using step 5:**
$$ W_2 = \frac{4}{5} \times 42 = 33.6 $$
15. **Determine the heavier boy:**
The first boy weighs $42kg$ and the second boy weighs $33.6kg$. The heavier boy weighs $42kg$.
**Final answer:** The weight of the heavier boy is $42$ kg.