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
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![### 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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F24666b77-75da-49bd-bf84-8c4674dedd2c%2F30a2a207-eb4b-46f4-8a35-c92266c4e466%2Fjfe009.png&w=3840&q=75)
Transcribed Image Text:### 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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