Question
An Atwood machine is constructed using two wheels (with the masses concentrated at the rims). The left wheel has a mass of 2.4 kg (m1) and a radius of 23.02 cm. The right wheel has a mass of 2.8 kg (m2) and a radius of 32.21 cm. The hanging mass on the left is 1.97 kg (m3) and on hanging mass on the right is 1.35 kg (m4). What is the acceleration of the hanging mass on the left? Use down as the positive direction. The acceleration of gravity is 9.8 m/s^2. Answer in units of m/s^2.
Answers
Answered by
GPT-4o mini
To find the acceleration of the hanging mass on the left (m3), we can apply the principles of Newton's second law and rotational dynamics, taking into account the moments of inertia of the wheels.
1. **Identify the forces acting on the masses:**
- For the left side (m3 = 1.97 kg):
- Weight force downwards: \( F_{g3} = m_3 \cdot g = 1.97 \cdot 9.8 \)
- For the right side (m4 = 1.35 kg):
- Weight force downwards: \( F_{g4} = m_4 \cdot g = 1.35 \cdot 9.8 \)
2. **Calculate the gravitational forces:**
\[
F_{g3} = 1.97 \times 9.8 = 19.366 \, \text{N}
\]
\[
F_{g4} = 1.35 \times 9.8 = 13.23 \, \text{N}
\]
3. **Set up the equations of motion:**
The net force acting on the left mass (m3) is:
\[
F_{\text{net}, 3} = F_{g3} - T
\]
The net force acting on the right mass (m4) is:
\[
F_{\text{net}, 4} = T - F_{g4}
\]
where \( T \) is the tension in the string connecting the two masses.
4. **Use Newton's second law** for each mass:
- For m3:
\[
m_3 a = F_{g3} - T \quad \text{(1)}
\]
- For m4:
\[
m_4 a = T - F_{g4} \quad \text{(2)}
\]
5. **Relate the tensions** and solve for \( a \). From (1):
\[
T = F_{g3} - m_3 a
\]
From (2):
\[
T = m_4 a + F_{g4}
\]
Setting the two expressions for \( T \) equal:
\[
F_{g3} - m_3 a = m_4 a + F_{g4}
\]
Rearranging gives:
\[
F_{g3} - F_{g4} = (m_3 + m_4)a
\]
6. **Substitute the gravitational force values:**
\[
19.366 - 13.23 = (1.97 + 1.35)a
\]
\[
6.136 = 3.32a
\]
7. **Solve for \( a \):**
\[
a = \frac{6.136}{3.32} \approx 1.847 \, \text{m/s}^2
\]
The acceleration of the hanging mass on the left is approximately:
\[
\boxed{1.85} \, \text{m/s}^2
\]
1. **Identify the forces acting on the masses:**
- For the left side (m3 = 1.97 kg):
- Weight force downwards: \( F_{g3} = m_3 \cdot g = 1.97 \cdot 9.8 \)
- For the right side (m4 = 1.35 kg):
- Weight force downwards: \( F_{g4} = m_4 \cdot g = 1.35 \cdot 9.8 \)
2. **Calculate the gravitational forces:**
\[
F_{g3} = 1.97 \times 9.8 = 19.366 \, \text{N}
\]
\[
F_{g4} = 1.35 \times 9.8 = 13.23 \, \text{N}
\]
3. **Set up the equations of motion:**
The net force acting on the left mass (m3) is:
\[
F_{\text{net}, 3} = F_{g3} - T
\]
The net force acting on the right mass (m4) is:
\[
F_{\text{net}, 4} = T - F_{g4}
\]
where \( T \) is the tension in the string connecting the two masses.
4. **Use Newton's second law** for each mass:
- For m3:
\[
m_3 a = F_{g3} - T \quad \text{(1)}
\]
- For m4:
\[
m_4 a = T - F_{g4} \quad \text{(2)}
\]
5. **Relate the tensions** and solve for \( a \). From (1):
\[
T = F_{g3} - m_3 a
\]
From (2):
\[
T = m_4 a + F_{g4}
\]
Setting the two expressions for \( T \) equal:
\[
F_{g3} - m_3 a = m_4 a + F_{g4}
\]
Rearranging gives:
\[
F_{g3} - F_{g4} = (m_3 + m_4)a
\]
6. **Substitute the gravitational force values:**
\[
19.366 - 13.23 = (1.97 + 1.35)a
\]
\[
6.136 = 3.32a
\]
7. **Solve for \( a \):**
\[
a = \frac{6.136}{3.32} \approx 1.847 \, \text{m/s}^2
\]
The acceleration of the hanging mass on the left is approximately:
\[
\boxed{1.85} \, \text{m/s}^2
\]