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 (torque balance) for a seesaw. The principle states that for the seesaw to be balanced, the clockwise moment must equal the counter - clockwise moment. The moment \( M \) is given by the formula \( M = F\times d \), where \( F \) is the force (weight in this case) and \( d \) is the distance from the pivot. Let the weight of Child A be \( F_A = 225\ N \), the distance of Child A from the pivot be \( d_A=1.0\ m \), the weight of Child B be \( F_B = 176\ N \), and the distance of Child B from the pivot be \( d_B \) (which we need to find). For balance, \( F_A\times d_A=F_B\times d_B \).

Step2: Rearrange the formula to solve for \( d_B \). We get \( d_B=\frac{F_A\times d_A}{F_B} \).

Step3: Substitute the given values into the formula. \( F_A = 225\ N \), \( d_A = 1.0\ m \), and \( F_B=176\ N \). So \( d_B=\frac{225\ N\times1.0\ m}{176\ N} \).

Step4: Calculate the value. \( \frac{225\times1.0}{176}=\frac{225}{176}\approx1.28\ m \).

Answer:

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