Condition of translational equilibrium proof without calculus

Question

The principle of inertia or Newton's first law of dynamics affirms that if the sum of the forces acting on a body is null or the body is at rest or moves in uniform motion, i.e.

$$\bbox[5px,border:2px solid #138D75]{\sum_i \overline F_i =\overline 0} \tag 1$$

Using the $(1)$ is it possible to prove mathematically without the use of the derivates or calculus that a rigid body is in translational equilibrium i.e., the rigid body does not shift?

With the images using the vectors it is very easy using the tail-tip method or the parallelogram rule, but I don't remember if there is an analytical proof. If any user can help me because I have not found it in any book. Thank you.

PS: My request it is for students of an high school.

Answer 1

If you agree that Newton's second law means

$$ \bbox[5px,border:1px solid #044412]{ \sum_i^n \vec{F}_i = m \,\vec{a}_{\rm CM} } \tag{1} $$

then $\sum_i^n \vec{F}_i = \vec{0}$ means that $\vec{a}_{\rm CM} = \vec{0}$.

Zero acceleration means zero or constant velocity (of the center of mass).

Answer 2

Consider an isolated rigid body. From its definition we have that considering two different infinitesimal elements of it, calling them $1$ and $2$, the relative distance between them cannot change. So, considering the segment joining them, it cannot varies its length. So the relative acceleration between the two elements in the direction of the joining segment must be $0$ m/s$^2$: $m_2\cdot\vec{a_{12}}-m_1\cdot\vec{a_{21}}=\vec{0} \to \vec{F_{12}}=- \vec{F_{21}}$, where with $\vec{F_{ij}}$ the force exerted by $i$ element on $j$ element is indicated. Repeating this argument foreach pair of infinitesimal elements that make up the rigid body, the following relation is obtained:

$\sum_{i,j=1}^N F_{ij}=0$, $i\neq j$.

Then an isolated rigid body, must be in translational equilibrium.