Question

Question: Unreasonable Results

Suppose two children are using a uniform seesaw that is 3.00m long andhas its center of mass over the pivot. The first child has a mass of and sits from the pivot. (a) Calculate where the second 18.0kg child must sit to balance the seesaw. (b) What is unreasonable about theresult? (c) Which premise is unreasonable, or which premises areinconsistent?

Short answer

a) The distance of the second child from the pivot is .

b) The outcome of part(a) is illogical because the supporting pivot is in the middle of the seesaw, resulting in a m-long side for the second child. The other child has fallen off the seesaw.

c) The unreasonable premise is the first kid's position relative to the pivot; hence, shortening the distance between the first child and the pivot should yield a reasonable result in part (a).

Step-by-step solution

Mass

The mass of a body is used to determine how much matter it contains. Kilogram is the SI unit of mass (kg). A person's mass does not alter over time.

Data given

The length of the see saw is $L=3.0\text{m}$.

The mass of the first child is $m_1=30.0\text{kg}$.

The distance of the first child from the pivot is $r_1=1.40\text{m}$.

The mass of the second child is $m_2=18.0\text{kg}$.

Calculation of the distance of the second child from the pivot

(a)

Let the distance of the second child from the pivot is${\text{r}}_{\text{2}}$. The equilibrium under torques gives,

$$\begin{gathered} m_1\times g\times r_1=m_2\times g\times r_2 \\ {r}_2=\frac{m_1\times r_1}{m_2} \\ {r}_2=\frac{30.0\times 1.40}{18.0} \\ {r}_2=2.33\text{m} \end{gathered}$$

Hence, the distance is 2.33m.

Detection of the unreasonable fact

(b)

The second child is more than the half of the total length of the see saw which is not possible. Hence, this is unreasonable.

Reason of the unreasonable fact

(c)

If the mass of the second child were greater, the distance from the pivot would be less.

Hence, the reason behind the little fact is that the mass of the second child is small.