Question
What is the minimum distance from the pivot point required for a 32 kg person to achieve rotational equilibrium on a 7 m long, 41 kg beam when a 20 kg person is sitting 3.5 m away from the pivot point?
Answers
GPT-4o mini
To achieve rotational equilibrium, the torques acting on the beam around the pivot point must be balanced.
Let’s denote:
- \( m_1 = 20 \, \text{kg} \) (mass of the first person)
- \( d_1 = 3.5 \, \text{m} \) (distance of the first person from the pivot)
- \( m_2 = 32 \, \text{kg} \) (mass of the second person, trying to achieve equilibrium)
- \( d_2 \) (distance of the second person from the pivot, which we need to find)
- \( m_{\text{beam}} = 41 \, \text{kg} \) (mass of the beam, which we will consider as acting at its center of mass)
- The length of the beam \( L = 7 \, \text{m} \), so the center of mass is at \( 3.5 \, \text{m} \) from the pivot.
1. **Calculate the torque due to the first person located at \( 3.5 \, \text{m} \)**:
\[
\tau_1 = m_1 \cdot g \cdot d_1 = 20 \, \text{kg} \cdot g \cdot 3.5 \, \text{m}
\]
2. **Calculate the torque due to the beam, acting at its center of mass \( 3.5 \, \text{m} \)**:
\[
\tau_{\text{beam}} = m_{\text{beam}} \cdot g \cdot \frac{L}{2} = 41 \, \text{kg} \cdot g \cdot 3.5 \, \text{m}
\]
3. **Calculate the torque due to the second person**:
\[
\tau_2 = m_2 \cdot g \cdot d_2 = 32 \, \text{kg} \cdot g \cdot d_2
\]
Since we are looking for equilibrium, the sum of torques around the pivot should be zero:
\[
\tau_1 + \tau_{\text{beam}} = \tau_2
\]
Substituting the expressions for torque:
\[
(20 \, \text{kg} \cdot g \cdot 3.5 \, \text{m}) + (41 \, \text{kg} \cdot g \cdot 3.5 \, \text{m}) = 32 \, \text{kg} \cdot g \cdot d_2
\]
4. **Factoring out \( g \)** from the equation:
\[
g \left( 20 \cdot 3.5 + 41 \cdot 3.5 \right) = 32 \cdot g \cdot d_2
\]
We can simplify this as follows:
\[
(20 + 41) \cdot 3.5 = 32 \cdot d_2
\]
Calculating \( 20 + 41 \):
\[
61 \cdot 3.5 = 32 \cdot d_2
\]
Calculating \( 61 \cdot 3.5 \):
\[
213.5 = 32 \cdot d_2
\]
Now, to find \( d_2 \):
\[
d_2 = \frac{213.5}{32}
\]
Calculating \( d_2 \):
\[
d_2 \approx 6.671875 \, \text{m}
\]
Since the beam is only 7 m long and the maximum distance from the pivot is indeed less than the length of the beam, we can state:
\[
d_2 \approx 6.67 \, \text{m}
\]
Thus, the minimum distance required for the 32 kg person to achieve rotational equilibrium is approximately:
\[
\boxed{6.67 \, \text{m}}
\] from the pivot point.
Let’s denote:
- \( m_1 = 20 \, \text{kg} \) (mass of the first person)
- \( d_1 = 3.5 \, \text{m} \) (distance of the first person from the pivot)
- \( m_2 = 32 \, \text{kg} \) (mass of the second person, trying to achieve equilibrium)
- \( d_2 \) (distance of the second person from the pivot, which we need to find)
- \( m_{\text{beam}} = 41 \, \text{kg} \) (mass of the beam, which we will consider as acting at its center of mass)
- The length of the beam \( L = 7 \, \text{m} \), so the center of mass is at \( 3.5 \, \text{m} \) from the pivot.
1. **Calculate the torque due to the first person located at \( 3.5 \, \text{m} \)**:
\[
\tau_1 = m_1 \cdot g \cdot d_1 = 20 \, \text{kg} \cdot g \cdot 3.5 \, \text{m}
\]
2. **Calculate the torque due to the beam, acting at its center of mass \( 3.5 \, \text{m} \)**:
\[
\tau_{\text{beam}} = m_{\text{beam}} \cdot g \cdot \frac{L}{2} = 41 \, \text{kg} \cdot g \cdot 3.5 \, \text{m}
\]
3. **Calculate the torque due to the second person**:
\[
\tau_2 = m_2 \cdot g \cdot d_2 = 32 \, \text{kg} \cdot g \cdot d_2
\]
Since we are looking for equilibrium, the sum of torques around the pivot should be zero:
\[
\tau_1 + \tau_{\text{beam}} = \tau_2
\]
Substituting the expressions for torque:
\[
(20 \, \text{kg} \cdot g \cdot 3.5 \, \text{m}) + (41 \, \text{kg} \cdot g \cdot 3.5 \, \text{m}) = 32 \, \text{kg} \cdot g \cdot d_2
\]
4. **Factoring out \( g \)** from the equation:
\[
g \left( 20 \cdot 3.5 + 41 \cdot 3.5 \right) = 32 \cdot g \cdot d_2
\]
We can simplify this as follows:
\[
(20 + 41) \cdot 3.5 = 32 \cdot d_2
\]
Calculating \( 20 + 41 \):
\[
61 \cdot 3.5 = 32 \cdot d_2
\]
Calculating \( 61 \cdot 3.5 \):
\[
213.5 = 32 \cdot d_2
\]
Now, to find \( d_2 \):
\[
d_2 = \frac{213.5}{32}
\]
Calculating \( d_2 \):
\[
d_2 \approx 6.671875 \, \text{m}
\]
Since the beam is only 7 m long and the maximum distance from the pivot is indeed less than the length of the beam, we can state:
\[
d_2 \approx 6.67 \, \text{m}
\]
Thus, the minimum distance required for the 32 kg person to achieve rotational equilibrium is approximately:
\[
\boxed{6.67 \, \text{m}}
\] from the pivot point.