Question

1. child a weighs 225 n and sits 1.0 m from the pivot of an adjustable seesaw. if child b weighs 175 n, how far from the pivot should child b sit to balance the weight of child a?

Explanation:

Step1: Recall the lever principle

For a lever to be balanced, the product of the weight (force) and its distance from the fulcrum (effort arm or resistance arm) should be equal on both sides. So, \( W_A \times d_A = W_B \times d_B \), where \( W_A \) is the weight of Child A, \( d_A \) is the distance of Child A from the fulcrum, \( W_B \) is the weight of Child B, and \( d_B \) is the distance of Child B from the fulcrum.

Step2: Identify the given values

We know that \( W_A = 225\space N \), \( d_A = 0.8\space m \), and \( W_B = 175\space N \). We need to find \( d_B \).

Step3: Rearrange the formula to solve for \( d_B \)

From \( W_A \times d_A = W_B \times d_B \), we can rearrange it to \( d_B=\frac{W_A \times d_A}{W_B} \).

Step4: Substitute the values into the formula

Substitute \( W_A = 225\space N \), \( d_A = 0.8\space m \), and \( W_B = 175\space N \) into the formula:
\( d_B=\frac{225 \times 0.8}{175} \)
First, calculate the numerator: \( 225\times0.8 = 180 \)
Then, divide by the denominator: \( d_B=\frac{180}{175}=\frac{36}{35}\approx1.03\space m \) (rounded to two decimal places)

Answer:

Child B should sit approximately \( \frac{36}{35}\space m \) (or about \( 1.03\space m \)) from the point (fulcrum) to balance the weight of Child A.