forcesfree-body-diagramnewtonian-mechanicsspring
In a normal system of parallel springs,
We say that their extension of the each spring is same, so the equivalent spring constant k is the sum of all k of spring.
But how can this make sense if one spring in a system of two springs, has a very large k1, and one has a very small k2? Wouldn't the system, like a load,when force is applied, "turns over" and "slanted" ？
The displacement from equilibrium is not L-x but would be (L-x)-L That is the final position which is L-x (assuming your origin is on the left wall) minus the initial position which is L, the equilibrium position. Hence your net force is (k1+k2)x.
A simple model for a coil spring would be that, when the spring is subjected to a force, the entire coil is subjected to a torsion $\tau$. This torque causes the coil the twist by an angle, which can be approximated with,
$$ \theta = \frac{l\, \tau}{I\, G}, $$
where $\theta$ is the angle of twist in radians, $l$ the length of the coil (not to be confused with the length of the spring), $I$ the second moment of inertia of the cross-section of the coil and $G$ the shear modulus of the material the coil is made of.
Assuming that the coil is circular rod, then $I$ would be equal to,
$$ I = \frac{\pi}{2} r^4 = \frac{\pi}{32} d^4, $$
where $r$ and $d$ are the radius and diameter of the rod respectively.
However the spring constant of such a spring does not relate $\tau$ and $\theta$, but the elongation and (linear) force. The entire spring has free length $L$ and consists of $N$ turns with mean diameter $D$. The coil angle, $\alpha$, is defined as the the angle the coil makes with the plane normal to the length axis of the spring. Also see the Figure below.
If you unroll the spring onto a flat plain, the rod will be the diagonal of a rectangle with height $L$ and width $\pi\, N\, D$. Combining this with the fact that $\alpha$ is the angle between the diagonal and horizontal of this rectangle and therefore will be equal to,
$$ \alpha= \tan^{-1}\left(\frac{L}{\pi\, N\, D}\right). $$
The relationship between $\theta$ and the elongation, $\Delta L$, can be found by looking at the change in $\alpha$ due to the elongation,
$$ \Delta\alpha = \tan^{-1}\left(\frac{L+\Delta L}{\pi\, N\, D}\right) - \tan^{-1}\left(\frac{L}{\pi\, N\, D}\right) = \frac{\pi\, N\, D}{L^2 + \pi^2 N^2 D^2}\Delta L + O(\Delta L^2). $$
Namely the change in $\alpha$ is equal to the twist angle of a quarter of a turn of the spring, thus,
$$ \theta = 4\, N\, \Delta\alpha \approx \frac{4\, \pi N^2 D}{L^2 + \pi^2 N^2 D^2} \Delta L. $$
The relationship between $\tau$ and $F$ can be found by looking at the lever of this torque in the spring,
$$ F = \frac{2 \cos(\alpha)}{D} \tau = \frac{2\, \pi\, N\, \tau}{\sqrt{L^2 + \pi^2 N^2 D^2}}. $$
By using Pythagorean theorem it can be shown that the length of the coil is equal to,
$$ l = \sqrt{L^2 + \pi^2 N^2 D^2}. $$
By substituting $F$ and $\Delta L$ from these equations, the spring constant can be approximated by,
$$ k = \frac{F}{\Delta L} \approx \frac{\pi^3 N^3 d^4 D\, G}{4 \left(L^2 + \pi^2 N^2 D^2\right)^2} = \frac{\pi^3 N^3 d^4 D\, G}{4\,l^4}. $$
To test this we can try to calculate the spring constant of a spring from a ballpoint pen.
One end of the spring has four tightly packed windings, which are measured to have a height of 1.5 mm, thus the diameter of the the rod to be equal to 0.375 mm. The mean diameter of the spring is measured to be about 4 mm. The spring counts 8.5 windings. The length of the spring is measured to be 2 cm. The shear modulus is harder to determine, because I am not certain from which metal it is made of, but because it is attracted to a magnet I will assume it is iron, which has a shear modulus of about 64 GPa. The estimation for the spring constant would then be equal to roughly 173 N/m. Since the spring is relatively small I did not had an accurate way of measuring the elongation of the spring while applying a force. I used my mobile phone, which according to the the online specs weighs 162 g, which causes a compression of roughly 0.7 cm, which would yield a spring constant of roughly 227 N/m. So the predicted spring constant is of by 23.8%, which is quite a large relative error, but at least the same order of magnitude. Sources for this error might be: the fact that the compression was quite large, so the linear approximation in $\Delta L$ might not hold; the measured dimensions of the spring might not be totally accurate, especially the values for $d$ and $l$, which are raised to the fourth power could contribute a lot if the of my a little.
Best Answer
Yes; a resultant torque causes rotation. If you don’t see this addressed in a problem, it’s because the block is being assumed to translate only, not rotate (e.g., perhaps it’s riding on frictionless tracks).
Other likely assumptions for this problem are that the wall and block are rigid, the springs are ideal, the system doesn’t sag from gravity, and that no out-of-plane motion occurs, for instance. Sometimes the problem will state these assumptions, and sometimes you have to deduce and apply them through experience.