The center of mass of a person may be determined by an arrangement such as the one shown in the figure below. A light plank rests on two scales separated by a distance of 
d = 1.80 m
 and reading 
Fg1 = 470 N
 and 
Fg2 = 350 N.
 Determine the distance of the girl's center of mass from her feet.

To determine the distance of the girl's center of mass from her feet, we can set up a torque balance about one end of the plank.

### Given Data:
- Distance between the scales (total length of the plank), \( d = 1.80 \, \text{m} \)
- Force on scale 1 (near the feet), \( F_{g1} = 470 \, \text{N} \)
- Force on scale 2 (near the head), \( F_{g2} = 350 \, \text{N} \)

### Step-by-Step Solution:

1. **Total Weight of the Girl (and Plank, if any):**
   The total downward force due to the girl is the sum of the two readings on the scales:
   \[
   F_{\text{total}} = F_{g1} + F_{g2} = 470 \, \text{N} + 350 \, \text{N} = 820 \, \text{N}
   \]

2. **Choosing the Pivot Point:**
   Let's choose the pivot point at the position of scale 1 (near her feet). This allows us to ignore the torque due to \( F_{g1} \) because its distance to the pivot is zero.

3. **Setting Up the Torque Equation:**
   To keep the plank in equilibrium, the clockwise torques must equal the counterclockwise torques around the pivot.
   
   Let \( x \) be the distance of the girl's center of mass from her feet (pivot point).

   - The counterclockwise torque is due to the center of mass, located at distance \( x \) from the pivot.
     - Torque due to girl's weight at her center of mass: \( 820 \times x \)
   
   - The clockwise torque is due to \( F_{g2} \), located \( d = 1.80 \, \text{m} \) from the pivot.
     - Torque due to \( F_{g2} \): \( F_{g2} \times d = 350 \times 1.80 \)

   Setting the torques equal gives:
   \[
   820 \, x = 350 \times 1.80
   \]

4. **Solving for \( x \):**
   \[
   x = \frac{350 \times 1.80}{820}
   \]
   \[
   x = \frac{630}{820} \approx 0.768 \, \text{m}
   \]

### Answer:
The distance of the girl's center of mass from her feet is approximately \( 0.768 \, \text{m} \).

Would you like a more detailed breakdown of any step?

---

Here are five related questions for further exploration:
1. How would the center of mass location change if the girl's weight were unevenly distributed along the plank?
2. If the length of the plank changed, how would it affect the readings on the scales?
3. What if the plank itself had a significant weight? How would this impact the calculation?
4. How could this setup be adjusted to measure the center of mass of an irregularly shaped object?
5. What other methods are there for finding a person’s center of mass?

**Tip:** When dealing with torque and center of mass problems, it helps to set up the pivot at a point where unknown forces (like weights of the plank) can be ignored.

The center of mass of a person may be determined by an arrangement such as the one shown in the figure below. A light plank rests on two scales separated by a distance of d = 1.80 m and reading Fg1 = 470 N and Fg2 = 350 N. Determine the distance of the girl's center of mass from her feet.

Find the distance of a girl's center of mass from her feet using a torque-based method. This problem involves key concepts of equilibrium and center of mass, with forces of 470 N and 350 N applied across a 1.80 m plank. Suitable for Grades 9-12 students, it provides a step-by-step solution with practical applications of torque equations.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation