Particle Equilibrium and the Interpretation of Accelerations I have a simple question regarding the interpretation of acceleration and force in the context of a particle in equilibrium.
Given that the necessary and sufficient definition for particle equilibrium can be stated as;
$$
\sum\vec{F}=0
$$
The following substitution can be made;
$$
\sum(m\cdot\vec{a})=0
$$
Assuming equilibrium, the particle must be non-accelerating, and can only have a single mass. Yet intuitively the forces still pull or push. So given equilibrium, how can the forces pull with an acceleration vector of zero? I understand that it is the vector sum of the accelerations which must have a magnitude of zero, but how can any be non-zero anyway, if there is no observable acceleration? I think my interpretation is somewhat tangled. Can anybody clarify? Appreciate the time.
 A: You can't meaningfully make that substitution. Newton's second law (for constant mass) is actually:
$$ \sum F = ma $$
In other words, the acceleration is proportional to the total force.
A: Your substitution is almost correct. You can have multiple forces each supplying their own acceleration vector, but it doesn't make a lot of sense to include the mass in the sum because it would mean you have to distribute the mass among all the constituent forces just to add them again.
$$\sum _{i}{\vec F_i}=\vec F_{total}$$
$$\vec F_{total}=m\vec a_{total}=m\sum_{i}{\vec a_i}$$
The simplest example of forces cancelling each other out is of two equal forces in the opposite direction. Imagine an object of 1kg with two ropes attached to it. Now two teams (A and B) are having a tug of war and each team supplies 500N in the x-direction, resulting in a net force of zero. The total acceleration thus equals
$$a_{total}=a_A+a_B=\left(\begin{array}{c}{-500+500\\0+0}\end{array}\right)=\left(\begin{array}{c}{0\\0}\end{array}\right)$$
Here is another example with three forces/accelerations. Again you can imagine an object wich is pulled by three ropes.
$$\quad\color{green}{\left(\begin{array}{c}{0\\15}\end{array}\right)}+
  \color{red}{\left(\begin{array}{c}{10\\0}\end{array}\right)}+
  \color{blue}{\left(\begin{array}{c}{-10\\-15}\end{array}\right)}=
\left(\begin{array}{c}{0\\0}\end{array}\right)
$$

A: You cannot substitute the force $\vec F$ with $m\vec a$. This is second law of Newton. The law talks about equivalency not sameness. Indeed we have $\Sigma \vec F\equiv m\vec a$.
For substitution of the force, you should consider to the first law of Newton that defines the force.
A: Say $\Sigma F=F_1+F_2+...=0$. Then your substitution, which is mathematically correct, would physically imply the following: if $F_1$ alone had acted on the body then it would accelerate to $a_1$, if $F_2$ alone had acted on the body then it would accelerate to $a_2$, and so on. So $F_1+F_2+...=0$ implies $a_1+a_2+...=0$, while individual components themselves are not necessarily zero. Each $a_i$ can be non-zero because each of them refers to a different physical condition, which is when the body is being acted upon by $F_i$ alone.
