关于分布微元力(df)积分等效于集中外力矢量和的严谨证明咨询

Awesome question—trusting intuition is fine, but digging into the rigorous proof is where real understanding happens. Let’s walk through this step by step, using core Newtonian mechanics as our foundation.

1. Break Down the System and Forces #

Think of the object as a collection of tiny mass elements (we’ll call each one ( dm )). Every single one of these elements has two types of forces acting on it:

• Internal forces (( d\mathbf{f}_{\text{int}} )): These are the pushes and pulls between elements inside the object itself.
• External forces (( d\mathbf{f}_{\text{ext}} )): These are forces coming from outside the object—including the concentrated forces you mentioned (like a hammer strike or a rope pull) and any distributed forces (like gravity or pressure, if we’re including those).

2. Apply Newton’s Second Law to a Single Element #

For any individual mass element, Newton’s second law tells us:
[ dm \cdot \mathbf{a} = d\mathbf{f}{\text{ext}} + d\mathbf{f}{\text{int}} ]
Here, ( \mathbf{a} ) is the acceleration of that tiny element.

3. Integrate Over the Entire Object #

Now we’ll add up (integrate) this equation across every single mass element in the object:
[ \int dm \cdot \mathbf{a} = \int d\mathbf{f}{\text{ext}} + \int d\mathbf{f}{\text{int}} ]

Let’s tackle the internal forces first #

Newton’s third law is key here: for every force that element A exerts on element B, element B exerts an equal and opposite force back on element A. When we integrate all internal forces, every pair of these opposing forces cancels each other out completely. That means:
[ \int d\mathbf{f}_{\text{int}} = \mathbf{0} ]
No net internal force acts on the object as a whole—they only affect how elements push against each other inside.

Now the external forces #

The external force integral is just the total sum of all forces coming from outside the object. For concentrated forces (like ( \mathbf{F}_1, \mathbf{F}2, ..., \mathbf{F}n )), each one acts on a specific small part of the object, so it’s included in the integral as part of that element’s external force. If we only have concentrated forces (no distributed ones like gravity), this simplifies to:
[ \int d\mathbf{f}{\text{ext}} = \sum{i=1}^n \mathbf{F}_i ]

What about the acceleration side? #

The left-hand side of our integrated equation is the total rate of change of the object’s momentum (from the momentum theorem for particle systems). We can also link this to the object’s center of mass: the integral of ( dm \cdot \mathbf{a} ) equals the total mass ( M ) of the object multiplied by the acceleration of its center of mass ( \mathbf{a}{\text{cm}} ):
[ \int dm \cdot \mathbf{a} = M \cdot \mathbf{a}{\text{cm}} ]

4. Pull It All Together #

Substituting our results back into the integrated equation gives us:
[ M \cdot \mathbf{a}{\text{cm}} = \sum{i=1}^n \mathbf{F}i ]
And crucially, we’ve shown that:
[ \int d\mathbf{f}{\text{ext}} = \sum_{i=1}^n \mathbf{F}_i ]
This is the rigorous proof you were asking for: the integral of the distributed external force elements (caused by the concentrated external forces) equals the vector sum of those concentrated forces.

Quick Note on Intuition vs. Rigor #

It makes sense that people rely on intuition here, but the missing piece for many is the cancellation of internal forces. Without Newton’s third law guaranteeing those internal forces cancel out, this conclusion wouldn’t hold. For example, in a fluid or deformable material, internal forces play a bigger role, but for rigid bodies (or any particle system), the third law locks in that internal forces don’t contribute to the net force on the whole object.

内容的提问来源于stack exchange，提问作者Suhaib Abu Raidah