https://file.aiquickdraw.com/mathbot/file/1730433087650w6v1adhi.jpg

Let's break down this problem based on the forces, torques, and distances given in the image.

### Problem Analysis:
The child is being lifted at a constant speed, meaning there is no acceleration and the system is in **rotational equilibrium**. This implies that the sum of the torques about any point (typically the pivot point) must be zero.

Given:
- \( r_1 = 2.5 \, \text{cm} \)
- \( r'_1 = 10.0 \, \text{cm} \)
- \( r''_1 = 40.0 \, \text{cm} \)
- Mass of the leg, \( m_{\text{leg}} = 5.0 \, \text{kg} \)
- Mass of the child, \( m_{\text{child}} = 20.0 \, \text{kg} \)

We need to find the **force \( F_Q \)** that the upper leg muscle should exert to lift the child.

### Converting Units:
1. Convert distances to meters for SI consistency:
   - \( r_1 = 0.025 \, \text{m} \)
   - \( r'_1 = 0.10 \, \text{m} \)
   - \( r''_1 = 0.40 \, \text{m} \)

2. Calculate the **weights** \( W_{\text{leg}} \) and \( W_{\text{child}} \):
   - \( W_{\text{leg}} = m_{\text{leg}} \times g = 5.0 \times 9.81 = 49.05 \, \text{N} \)
   - \( W_{\text{child}} = m_{\text{child}} \times g = 20.0 \times 9.81 = 196.2 \, \text{N} \)

### Setting Up the Torque Equation:
Assuming the pivot point is at the hip joint:
1. **Torque due to the leg**:
   - This torque is counterclockwise and given by \( \tau_{\text{leg}} = W_{\text{leg}} \times r'_1 = 49.05 \times 0.10 \, \text{m} \).

2. **Torque due to the child**:
   - This torque is also counterclockwise, \( \tau_{\text{child}} = W_{\text{child}} \times r''_1 = 196.2 \times 0.40 \, \text{m} \).

3. **Torque due to \( F_Q \)** (force exerted by the muscle):
   - This torque is clockwise, and we’ll set it as \( \tau_{F_Q} = F_Q \times r_1 \).

### Applying Rotational Equilibrium:
Since the system is in equilibrium, the sum of clockwise and counterclockwise torques must be equal:
\[
F_Q \times r_1 = W_{\text{leg}} \times r'_1 + W_{\text{child}} \times r''_1
\]

Substitute the known values:
\[
F_Q \times 0.025 = (49.05 \times 0.10) + (196.2 \times 0.40)
\]

### Solving for \( F_Q \):
\[
F_Q = \frac{(49.05 \times 0.10) + (196.2 \times 0.40)}{0.025}
\]

Let me calculate this for you:The force \( F_Q \) that the upper leg muscle should exert to lift the child at a constant speed is approximately **3335.4 N**.

Would you like a detailed explanation of the steps or have any other questions?

### Follow-up Questions:
1. How would the required force \( F_Q \) change if the child’s weight increased?
2. What would happen to \( F_Q \) if the distance \( r_1 \) decreased?
3. Why is it important that the problem specifies lifting at a constant speed?
4. How would you calculate the force if there was acceleration involved?
5. How does the weight distribution affect the muscle force needed?

### Tip:
When analyzing torque problems, always set up a clear pivot point and be consistent with clockwise and counterclockwise directions.

What force (in N) should the upper leg muscle exert to lift the child at a constant speed? (Use the following values: r1 = 2.5 cm, r'1 = 10.0 cm, r''1 = 40.0 cm, m_leg = 5.0 kg, and m_child = 20 kg. Enter the magnitude.)

Determine the force the upper leg muscle must exert to lift a child at a constant speed. Using given values for distances and masses, solve for the force by applying torque and rotational equilibrium principles.

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