### Balancing a Seesaw: A Physics ProblemThree children are trying to balance on a seesaw, which includes a fulcrum rock acting as a pivot at the center (see figure). The board is 5.6 m long and weighs 330 N. Two children sit at either end: Child A has a weight of 400 N and Child B has a weight of 320 N. Where should Child C, who has a weight of 140 N, sit so that the seesaw is balanced? Take x = 0 at the pivot.#### Illustration of the Problem:The illustration shows a seesaw with a fulcrum rock at the center and the following setup:- Child A on the left end with a weight of 400 N.- Child B on the right end with a weight of 320 N.- Child C is standing below the seesaw, waiting to position themselves.#### Balancing Principle:To balance the seesaw, the clockwise and counterclockwise moments around the pivot must be equal. Using the principle of moments:\[ \text{Clockwise Moment} = \text{Counterclockwise Moment} \]The moments are calculated as:\[ \text{Moment} = \text{Force} \times \text{Distance from pivot} \]#### Given Data:- Length of seesaw (L) = 5.6 m (split into 2.8 m on either side of the pivot)- Weight of Child A (W_A) = 400 N at a distance of 2.8 m- Weight of Child B (W_B) = 320 N at a distance of 2.8 m- Weight of Child C (W_C) = 140 N#### Calculation:\[ (W_A \times \text{distance of A}) + (W_{\text{board} \, \text{left of pivot}} \times \text{their distances}) = (W_B \times \text{distance of B}) + (W_{\text{board right}} \times \text{distance}) + (W_C \times x) \]Given that:\[ 400 \times 2.8 = 320 \times 2.8 + 140 \times x \]Solving for 'x' gives the position where Child C should sit.#### Multiple Choice:- \[-1.6 m\]- \[-0.

Three children are trying to balance on a seesaw, which includes a fulcrum rock acting as a pivot at the center (see figure). The board is 5.6 m long and weighs 330 N. Two children sit at either end: Child A has a weight of 400 N'and Child B has a weight of 320 N. Where should Child C, who has a weight of 140 N, sit so that the seesaw is balanced? Take x = 0 at the pivot. 400 N 320N 140 N -1.6 m -0.57 m 0.57 m 1.6 m

Three children are trying to balance on a seesaw, which includes a fulcrum rock acting as a pivot at the center (see figure). The board is 5.6 m long and weighs 330 N. Two children sit at either end: Child A has a weight of 400 N'and Child B has a weight of 320 N. Where should Child C, who has a weight of 140 N, sit so that the seesaw is balanced? Take x = 0 at the pivot. 400 N 320N 140 N -1.6 m -0.57 m 0.57 m 1.6 m

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Rotational Equilibrium And Rotational Dynamics

In physics, the state of balance between the forces and the dynamics of motion is called the equilibrium state. The balance between various forces acting on a system in a rotational motion is called rotational equilibrium or rotational dynamics.

Equilibrium of Forces

The tension created on one body during push or pull is known as force.

Question

### Balancing a Seesaw: A Physics Problem

Three children are trying to balance on a seesaw, which includes a fulcrum rock acting as a pivot at the center (see figure). The board is 5.6 m long and weighs 330 N. Two children sit at either end: Child A has a weight of 400 N and Child B has a weight of 320 N. Where should Child C, who has a weight of 140 N, sit so that the seesaw is balanced? Take x = 0 at the pivot.

#### Illustration of the Problem:
The illustration shows a seesaw with a fulcrum rock at the center and the following setup:
- Child A on the left end with a weight of 400 N.
- Child B on the right end with a weight of 320 N.
- Child C is standing below the seesaw, waiting to position themselves.

#### Balancing Principle:
To balance the seesaw, the clockwise and counterclockwise moments around the pivot must be equal. Using the principle of moments:

\[ \text{Clockwise Moment} = \text{Counterclockwise Moment} \]

The moments are calculated as:

\[ \text{Moment} = \text{Force} \times \text{Distance from pivot} \]

#### Given Data:
- Length of seesaw (L) = 5.6 m (split into 2.8 m on either side of the pivot)
- Weight of Child A (W_A) = 400 N at a distance of 2.8 m
- Weight of Child B (W_B) = 320 N at a distance of 2.8 m
- Weight of Child C (W_C) = 140 N

#### Calculation:
\[ (W_A \times \text{distance of A}) + (W_{\text{board} \, \text{left of pivot}} \times \text{their distances}) = (W_B \times \text{distance of B}) + (W_{\text{board right}} \times \text{distance}) + (W_C \times x) \]

Given that:
\[ 400 \times 2.8 = 320 \times 2.8 + 140 \times x \]

Solving for 'x' gives the position where Child C should sit.

#### Multiple Choice:
- \[-1.6 m\]
- \[-0.

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