Question

1. child a weighs 225 n and sits 1.0 m from the pivot of an adjustable seesaw. if child b weighs 176 n, how far from the pivot should child b sit to balance the weight of child a?

Explanation:

Step1: Recall the principle of moments

For a seesaw to balance, the clockwise moment equals the counter - clockwise moment. The formula for moment is \(M = F\times d\), where \(F\) is the force (in this case, weight) and \(d\) is the distance from the pivot. Let \(F_A = 225\space N\), \(d_A=1.0\space m\), \(F_B = 176\space N\), and \(d_B\) be the distance of Child B from the pivot. So, \(F_A\times d_A=F_B\times d_B\).

Step2: Solve for \(d_B\)

We can re - arrange the formula \(F_A\times d_A=F_B\times d_B\) to get \(d_B=\frac{F_A\times d_A}{F_B}\). Substitute \(F_A = 225\space N\), \(d_A = 1.0\space m\), and \(F_B=176\space N\) into the formula: \(d_B=\frac{225\times1.0}{176}\approx1.28\space m\)

Answer:

Child B should sit approximately \(1.3\space m\) (or \(\frac{225}{176}\approx1.28\space m\)) from the pivot.