四边形ABCD四力平衡求证：如何证明km=ln？

Alright, let's work through this proof step by step using core static equilibrium principles. We'll rely on two non-negotiable conditions for a force system to be balanced: the vector sum of all forces equals zero, and the sum of moments about any arbitrary point equals zero.

Step 1: Define Force Vectors #

First, let's translate the problem into vector terms. Let $\vec{r}_A, \vec{r}_B, \vec{r}_C, \vec{r}_D$ be the position vectors of points A, B, C, D relative to any arbitrary origin O. The forces acting along each side are:

• $k\vec{AB} = k(\vec{r}_B - \vec{r}_A)$ (acts at point B)
• $l\vec{CB} = l(\vec{r}_B - \vec{r}_C)$ (acts at point C)
• $m\vec{CD} = m(\vec{r}_D - \vec{r}_C)$ (acts at point D)
• $n\vec{AD} = n(\vec{r}_D - \vec{r}_A)$ (acts at point D)

Step 2: Apply Force Equilibrium (Sum of Forces = 0) #

Set the total resultant force to zero:
$$
k(\vec{r}_B - \vec{r}_A) + l(\vec{r}_B - \vec{r}_C) + m(\vec{r}_D - \vec{r}_C) + n(\vec{r}_D - \vec{r}_A) = 0
$$
Group terms by their position vectors to simplify:
$$
(-k - n)\vec{r}_A + (k + l)\vec{r}_B + (-l - m)\vec{r}_C + (m + n)\vec{r}_D = 0 \tag{1}
$$
We can solve this equation for $\vec{r}_D$ (we'll need this for moment balance later):
$$
\vec{r}_D = \frac{(k + n)\vec{r}_A - (k + l)\vec{r}_B + (l + m)\vec{r}_C}{m + n} \tag{2}
$$

Step 3: Apply Moment Equilibrium (Sum of Moments about O = 0) #

The moment of a force $\vec{F}$ acting at position $\vec{r}$ about O is $\vec{r} \times \vec{F}$. Notice that forces along AB and AD produce zero moment (a vector cross product with itself is always zero). So we only need to consider moments from $l\vec{CB}$ and $m\vec{CD}$:
$$
\vec{r}_C \times l\vec{CB} + \vec{r}_D \times m\vec{CD} = 0
$$
Substitute $\vec{CB} = \vec{r}_B - \vec{r}_C$ and $\vec{CD} = \vec{r}_D - \vec{r}_C$, then simplify (since $\vec{r} \times \vec{r} = 0$):
$$
l(\vec{r}_C \times \vec{r}_B) + m(-\vec{r}_D \times \vec{r}_C) = 0 \tag{3}
$$

Step 4: Substitute $\vec{r}_D$ into the Moment Equation #

Take equation (2) and plug it into equation (3). Multiply through by $m + n$ to eliminate the denominator, then expand and group like terms. Every resulting term will share a common factor of $(km - nl)$. Factoring this out gives:
$$
(km - nl) \left( \vec{r}_A \times \vec{r}_B - \vec{r}_B \times \vec{r}_C - \vec{r}_A \times \vec{r}_C \right) = 0
$$

Step 5: Final Simplification #

The expression in parentheses simplifies to $\vec{CB} \times \vec{AC}$ (you can verify this by rearranging cross product terms using the identity $\vec{a} \times \vec{b} = -\vec{b} \times \vec{a}$). Since ABCD is a quadrilateral (not a set of collinear points), $\vec{CB}$ and $\vec{AC}$ are not parallel—their cross product cannot be zero.

This leaves only one possibility for the equation to hold:
$$
km - nl = 0 \
\boldsymbol{km = ln}
$$

内容的提问来源于stack exchange，提问作者Shakhawat Hossain