Answer :
The center of mass of the system is located approximately 70.00 cm from the left end of the meter stick.
Here's how to calculate the center of mass location:
1. **Treat the system as point masses:** Since the meter stick is thin and uniform, its mass can be considered concentrated at its center. This simplifies the calculation as we only have three point masses to consider.
2. **Define variables:**
- Let M_s (mass of the stick) = 67 g = 0.067 kg (convert grams to kilograms)
- Let M1 (mass 1) = 1.2 kg
- Let M2 (mass 2) = 2.8 kg
- Let x_s (x-coordinate of the stick's center) = We need to solve for this.
- Let x1 (x-coordinate of mass 1) = 0.00 cm (left end)
- Let x2 (x-coordinate of mass 2) = 100 cm
3. **Center of mass formula:** The center of mass (x_cm) for a system of point masses is calculated by the weighted average of their positions:
x_cm = (Σ M_i * x_i) / Σ M_i
where:
- Σ (sigma) represents the sum over all masses (i) in the system.
- M_i is the mass of the i-th object.
- x_i is the x-coordinate of the i-th object.
4. **Apply the formula:** In this case, we have three masses:
x_cm = [(M_s * x_s) + (M1 * x1) + (M2 * x2)] / (M_s + M1 + M2)
x_cm = [(0.067 kg * x_s) + (1.2 kg * 0.00 cm) + (2.8 kg * 100 cm)] / (0.067 kg + 1.2 kg + 2.8 kg)
5. **Solve for x_s:**
- We can simplify the equation since M1 * x1 = 0 (mass 1 is at the origin).
x_cm = [(0.067 kg * x_s) + (2.8 kg * 100 cm)] / (4.067 kg)
To isolate x_s, we can multiply both sides by the total mass (denominator) and subtract (2.8 kg * 100 cm) from both sides:
(4.067 kg) * x_cm - (2.8 kg * 100 cm) = (0.067 kg * x_s)
x_s = (2.8 kg * 100 cm) / (4.00 kg)
x_s ≈ 70.00 cm
Therefore, the center of mass of the system is located approximately 70.00 cm from the left end of the meter stick.
