A parent of mass m₂ = 60 kg sits on a uniform beam seesaw of mass M = 40 kg, at a distance r₁ = 1.0 m to the right of the pivot P (as pictured below). Each end of the beam is equidistant from P. a) What distance r, to the left should a child of mass m, = 30 kg sit in order to balance the seesaw (hint: the center of mass of the beam is right at its center, since it is taken to be uniform)? Show any equations you use, or justify your intuition. m₂ = 30 kg m, = 60 kg M = 40 kg P √₂ = ? r₁=1.0 m b) What is the torque exerted by the weight of the parent, the torque from the child's weight, and the torque from the seesaw weight (magnitude and direction for each case, with direction indicated as into or out of the paper, and state the proper units)?
A parent of mass m₂ = 60 kg sits on a uniform beam seesaw of mass M = 40 kg, at a distance r₁ = 1.0 m to the right of the pivot P (as pictured below). Each end of the beam is equidistant from P. a) What distance r, to the left should a child of mass m, = 30 kg sit in order to balance the seesaw (hint: the center of mass of the beam is right at its center, since it is taken to be uniform)? Show any equations you use, or justify your intuition. m₂ = 30 kg m, = 60 kg M = 40 kg P √₂ = ? r₁=1.0 m b) What is the torque exerted by the weight of the parent, the torque from the child's weight, and the torque from the seesaw weight (magnitude and direction for each case, with direction indicated as into or out of the paper, and state the proper units)?
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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![**Text and Diagram Explanation for Educational Website**
**Problem Statement:**
A parent of mass \( m_1 = 60 \, \text{kg} \) sits on a uniform beam seesaw of mass \( M = 40 \, \text{kg} \), at a distance \( r_1 = 1.0 \, \text{m} \) to the right of the pivot \( P \) (as pictured below). Each end of the beam is equidistant from \( P \).
**a) What distance \( r_2 \), to the left should a child of mass \( m_2 = 30 \, \text{kg} \) sit in order to balance the seesaw (hint: the center of mass of the beam is right at its center, since it is taken to be uniform)? Show any equations you use, or justify your intuition.**
**Diagram:**
The diagram shows a seesaw balanced on a central pivot \( P \). To the left, a child of mass \( m_2 = 30 \, \text{kg} \) is sitting at a distance \( r_2 \), which needs to be determined. To the right, the parent of mass \( m_1 = 60 \, \text{kg} \) is sitting at \( r_1 = 1.0 \, \text{m} \).
**b) What is the torque exerted by the weight of the parent, the torque from the child's weight, and the torque from the seesaw weight (magnitude and direction for each case, with direction indicated as into or out of the paper, and state the proper units)?**
**Calculations:**
For part (a), to balance the seesaw:
\[ m_1 \cdot r_1 = m_2 \cdot r_2 \]
\[ 60 \, \text{kg} \cdot 1.0 \, \text{m} = 30 \, \text{kg} \cdot r_2 \]
\[ r_2 = 2.0 \, \text{m} \]
For part (b), torque is calculated as \( \tau = r \cdot F \cdot \sin(\theta) \), where \( F \) is the force, \( r \) is the distance from the pivot, and \( \theta \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5fafa0a8-e885-4f45-9f8c-34c617df8f38%2F3100beee-5a1e-4bef-bc00-29e6c0f56c86%2Fxcgha6_processed.png&w=3840&q=75)
Transcribed Image Text:**Text and Diagram Explanation for Educational Website**
**Problem Statement:**
A parent of mass \( m_1 = 60 \, \text{kg} \) sits on a uniform beam seesaw of mass \( M = 40 \, \text{kg} \), at a distance \( r_1 = 1.0 \, \text{m} \) to the right of the pivot \( P \) (as pictured below). Each end of the beam is equidistant from \( P \).
**a) What distance \( r_2 \), to the left should a child of mass \( m_2 = 30 \, \text{kg} \) sit in order to balance the seesaw (hint: the center of mass of the beam is right at its center, since it is taken to be uniform)? Show any equations you use, or justify your intuition.**
**Diagram:**
The diagram shows a seesaw balanced on a central pivot \( P \). To the left, a child of mass \( m_2 = 30 \, \text{kg} \) is sitting at a distance \( r_2 \), which needs to be determined. To the right, the parent of mass \( m_1 = 60 \, \text{kg} \) is sitting at \( r_1 = 1.0 \, \text{m} \).
**b) What is the torque exerted by the weight of the parent, the torque from the child's weight, and the torque from the seesaw weight (magnitude and direction for each case, with direction indicated as into or out of the paper, and state the proper units)?**
**Calculations:**
For part (a), to balance the seesaw:
\[ m_1 \cdot r_1 = m_2 \cdot r_2 \]
\[ 60 \, \text{kg} \cdot 1.0 \, \text{m} = 30 \, \text{kg} \cdot r_2 \]
\[ r_2 = 2.0 \, \text{m} \]
For part (b), torque is calculated as \( \tau = r \cdot F \cdot \sin(\theta) \), where \( F \) is the force, \( r \) is the distance from the pivot, and \( \theta \
![**Description of the Problem:**
Consider the same seesaw (M = 40 kg and length L = 5.0 m), but now the pivot P is at the far end with a child of \( m_2 = 30 \) kg sitting a distance \( r_2 = 4.0 \) m to the left of the pivot. To balance the net torque on the beam, a perpendicular rope that can exert a torque via upwards tension force is attached to the beam at a distance \( r = 1.0 \) m from the pivot. **Calculate** the upward force \( F \) (tension in the rope) in terms of \( M, m_2, r_2, r, L, \) and \( g \), and then plug in the values above to find the force \( F \) that is needed for equilibrium.
**Diagram Explanation:**
The diagram shows a seesaw with the following components and dimensions:
- A beam of mass \( M \) and length \( L \).
- A pivot point, labeled \( P \), located at the far right end of the beam.
- A child of mass \( m_2 \) sitting on the beam at a distance \( r_2 \) from the pivot.
- An upward force \( F \) exerted by a rope at a distance \( r \) from the pivot.
- The distances \( r \) and \( r_2 \) are marked along the beam.
### Steps to Solve:
1. **Calculate the Torque Due to the Child:**
\[
\text{Torque}_{m_2} = m_2 \cdot g \cdot r_2
\]
2. **Calculate the Torque Due to the Upward Force \( F \):**
\[
\text{Torque}_F = F \cdot r
\]
3. **Set the Torques Equal for Equilibrium:**
\[
m_2 \cdot g \cdot r_2 = F \cdot r
\]
4. **Solve for the Upward Force \( F \):**
\[
F = \frac{m_2 \cdot g \cdot r_2}{r}
\]
5. **Insert Known Values:**
- \( m_2 = 30 \) kg
- \( r_2 =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5fafa0a8-e885-4f45-9f8c-34c617df8f38%2F3100beee-5a1e-4bef-bc00-29e6c0f56c86%2Fk1h7jl_processed.png&w=3840&q=75)
Transcribed Image Text:**Description of the Problem:**
Consider the same seesaw (M = 40 kg and length L = 5.0 m), but now the pivot P is at the far end with a child of \( m_2 = 30 \) kg sitting a distance \( r_2 = 4.0 \) m to the left of the pivot. To balance the net torque on the beam, a perpendicular rope that can exert a torque via upwards tension force is attached to the beam at a distance \( r = 1.0 \) m from the pivot. **Calculate** the upward force \( F \) (tension in the rope) in terms of \( M, m_2, r_2, r, L, \) and \( g \), and then plug in the values above to find the force \( F \) that is needed for equilibrium.
**Diagram Explanation:**
The diagram shows a seesaw with the following components and dimensions:
- A beam of mass \( M \) and length \( L \).
- A pivot point, labeled \( P \), located at the far right end of the beam.
- A child of mass \( m_2 \) sitting on the beam at a distance \( r_2 \) from the pivot.
- An upward force \( F \) exerted by a rope at a distance \( r \) from the pivot.
- The distances \( r \) and \( r_2 \) are marked along the beam.
### Steps to Solve:
1. **Calculate the Torque Due to the Child:**
\[
\text{Torque}_{m_2} = m_2 \cdot g \cdot r_2
\]
2. **Calculate the Torque Due to the Upward Force \( F \):**
\[
\text{Torque}_F = F \cdot r
\]
3. **Set the Torques Equal for Equilibrium:**
\[
m_2 \cdot g \cdot r_2 = F \cdot r
\]
4. **Solve for the Upward Force \( F \):**
\[
F = \frac{m_2 \cdot g \cdot r_2}{r}
\]
5. **Insert Known Values:**
- \( m_2 = 30 \) kg
- \( r_2 =
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