- First mass
- 2kg
- Position of the first mass
- 0m
- Second mass
- 3kg
- Position of the second mass
- 10m
6.000m
Open with these values6.000m
Result: 6.000 mThe mass-weighted average of the two positions. Equal masses balance exactly halfway between them — the quickest sanity check there is — and any imbalance pulls the point toward the heavier one. Measure both positions from the same origin on the same axis; negative positions are fine.
Held fixed: First mass 2.00 kg, Position of the first mass 0.00 m, Second mass 3.00 kg.
| Position of the second mass (m) | Result (m) |
|---|---|
| 0.00 | 0.000 |
| 2.50 | 1.500 |
| 5.00 | 3.000 |
| 7.50 | 4.500 |
| 10.00Your value | 6.000 |
| 12.50 | 7.500 |
| 15.00 | 9.000 |
| 17.50 | 10.500 |
| 20.00 | 12.000 |
6.000m
Open with these values5.000m
Open with these values3.875m
Open with these valuesx = (m₁ × x₁ + m₂ × x₂) ÷ (m₁ + m₂)
| m₁, x₁, m₂, x₂ | Note | Balance point (m) |
|---|---|---|
| 1, 0, 1, 10 | Equal masses land on the midpoint | 5 |
| 5, 2, 5, 8 | Equal again, origin shifted | 5 |
| 10, -5, 10, 5 | Symmetric about the origin | 0 |
| 2, 0, 3, 10 | The heavier mass pulls it past 5 | 6 |
| 2, -4, 3, 6 | Negative positions are allowed | 2 |
| 2.5, 1.25, 7.5, 4.75 | Three quarters of the mass on the right | 3.875 |
Take the mass-weighted average of the two positions: x = (m1 × x1 + m2 × x2) / (m1 + m2). For example, a 2 kg mass at 0 m and a 3 kg mass at 10 m give (2 × 0 + 3 × 10) / (2 + 3) = 30 / 5 = 6 m.
The center of mass is the single point at which the whole system would balance — the average position of all the mass. For two point masses on a line it lies somewhere between them, and the system behaves, for many purposes, as if all the mass were concentrated there.
Yes. The heavier the mass, the more it weights the average, so the center of mass sits nearer to it. Only when the two masses are exactly equal does the point land precisely halfway between them.
When the masses are equal, the center of mass is simply the midpoint between the two positions. With equal weights, neither side pulls the average more than the other, so the balance point is the plain average of the two positions: (x1 + x2) / 2.
Yes. Positions are measured along an axis from an origin you choose, so anything to the left of that origin is negative. The masses themselves must be positive, but the positions — and the resulting center of mass — can be negative, zero, or positive.
Use consistent units: kilograms for both masses and metres for both positions, which gives the center of mass in metres. The result depends only on the ratio of the masses and on the positions, so any consistent mass and length units work as long as you do not mix them.
Information, not professional advice.
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