Distribution of force over springs in parallel I have a problem, I am not sure about.
Assume there is a system of 4 springs with a mass resting/ supported by the springs.

*

*the springs all have the same length, minimum deflection/ same distance of compression,

*same k value

*and are the same distance from each other.

what is the force value on each spring?
What I reason is that the force is distributed evenly across the springs and as such to find the force on, say spring 1, one just divides the total force by the number of springs?
I think this is wrong and I am forgetting something about basic Newtonian mechanics
 A: The tension in each spring will depend on their positions relative to the object’s centre of mass.
To simplify things, suppose we have just two springs supporting a rod. If the springs are attached to the ends of the rod, at equal distances from its centre of mass, then each spring supports half of the weight of the rod.
But if one spring is attached to the middle of the rod, at its centre of mass, and the other spring is attached to one end, then the spring in the middle must support the whole weight of the rod. This is because if there was any tension in the spring at the end of the rod then there would be a torque or turning moment on the rod, making the rod tip up until the tension in the end spring became zero.
A: Since all have the same initial length, and the same deflection $x$, then the load on each spring is simply $F = k x$.
The total load applied would then be $4 F = (4 k) x$
This leads to the rule of parallel springs
$$ k = \sum_i^n k_i $$
On the other hand, springs in series share the load $F$, and each has deflection $x_i = F/k_i$, and the total deflection $x = x_1 + x_2 + \ldots$ which leads to the rule for series springs
$$ x=\frac{F}{k} = \frac{F}{x_1} + \frac{F}{x_2}  + \ldots = \sum_i^n \frac{F}{x_i}  $$
or
$$ \frac{1}{k} = \sum_i^n \frac{1}{x_i}  $$
