Does static friction only depend on the forces acting on the axis perpedicular to the plane where an object is located?

Question

Suppose I have a vehicle with 4 wheels all connected to motors which is hauling a set of wagons. Suppose also that:

1. The contact between wheels and ground plane is uniform.
2. The weight of each wehicle is uniformely distributed on the wheels.
3. The vehicle "m1" is pulling all the other vehicles which do not have any motor (they have 4 wheels each which rotate freely).
4. The connection between each vehicle is rigid.

My "claim" is that the maximum traction force that can be applied by vehicle "m1" without its wheels starting to slip, depends only on its mass (i.e. not on the masses m2, m3 and m4). Is this true?

I think the question boils down to as: "Does any of the attached masses m2, m3 and m4 have any effect at all on the static friction between the weels of m1 and the ground?". Therfore, I am asking, does static friction depend only on forces acting on the vertical axis (in my drawing) and, if yes, does it mean that the maximum traction force applicable by m1 does not depend on how many vehicles it is hauling?

In my understanding, the mass of the hauled vehicles only reduces the maximum acceleration achievable but does not have any effect on the maximum traction force that can be applied in regards to static friction.

Answer 1

Yes, static friction depends only on the Normal force acting on the object and the friction coefficient.

Take the example below, of a block on a surface (not necessarily a plane, that is not relevant):

Here the normal force $F_N$ would be calculated from the weight of the block $mg$ and the angle of the tangent.

The static friction force is then, with $\mu$ the friction coefficient:

$$F_f=\mu F_N$$

In the case the of the 'convoy' the angle is of course $0$ and the maximum friction force ('traction' if you prefer) is simply:

$$F_f=\mu m_1g$$

In order for the convoy to start moving there must be some acceleration $a$, so with Newton's second:

$$\mu m_1g>\Sigma_{i=1}^{4}m_i a$$

Or:

$$\frac{\mu m_1g}{\Sigma_{i=1}^{4}m_i}>1$$

So your reasoning is correct: it doesn't matter how much power $m_1$ can deliver to its wheels, if the above condition isn't met the convoy will go nowhere!