Question

1. two children are playing on a seesaw trying to balance. if the child on the left sits 1.5 meters away from the fulcrum and weighs 25 kg, and the child on the right weighs 35 kg, how far away from the fulcrum will the child on the right have to sit? 0.75 m 1.5 m 1.75 m 1.07 m

Explanation:

Step1: Apply the principle of moments

For a balanced seesaw, the principle of moments states that the product of the weight and the distance from the fulcrum on one - side is equal to the product of the weight and the distance from the fulcrum on the other side. Let \(d_1\) be the distance of the child on the left from the fulcrum, \(m_1\) be the mass of the child on the left, \(d_2\) be the distance of the child on the right from the fulcrum, and \(m_2\) be the mass of the child on the right. Then \(m_1\times d_1=m_2\times d_2\).

Step2: Substitute the given values

We are given that \(m_1 = 25\space kg\), \(d_1=1.5\space m\), and \(m_2 = 35\space kg\). We need to find \(d_2\). Rearranging the formula \(d_2=\frac{m_1\times d_1}{m_2}\).

Step3: Calculate the distance

Substitute the values into the formula: \(d_2=\frac{25\times1.5}{35}=\frac{37.5}{35}\approx1.07\space m\)

Answer:

\(1.07\space m\)