How would I calculate a force on a spring? Or more specifically the weight of an object that spring is supporting

I am trying to create a simulation for car suspension and tyre physics and to do the tyre physics I need to know how much weight is over the friction point that is under the spring.

The force associated with springs is well described by Hooke's law. The spring force is given by, $$F = k x$$ where $F$ is the spring force, $x$ is the displacement of the spring from its original resting position, and $k$ is the spring constant in units of force per length.

• Are we sure car suspension follows Hooke law? Mar 9 '17 at 3:16
• The force associated with displacement of the suspension system can be modeled using Hooke's law and an equivalent spring stiffness constant. If you are concerned with the dynamics of the suspension system, you would be dealing with linear differential equations of the form, $m d^2y/dx^2 + c dy/dx + k_{eq} x = f(t)$, where $k_{eq}$ would be an equivalent spring stiffness of the system, and the product $k_{eq} u$ is the force associated with displacement of the system (i.e. Hooke's law).
– TRF
Mar 10 '17 at 5:21
• I mean that the relationship between spring deformation and spring force can be nonlinear. In other words the spring constant k can change with the deformation. Maybe there are reasons for car suspension to become progessively stiffer (not sure, it's not my field) with deformation. Mar 10 '17 at 12:38
• I have never heard of non-linear suspension systems. The inherent nature of a non-linear spring mass system is chaotic. It wouldn't make sense to have a non-linear spring constant unless you were in fact yielding the material in the spring, which would defeat the purpose of the spring in the first place.
– TRF
Mar 15 '17 at 9:06
• Nonlinearity can be geometrical, not only due to yielding. As an example, leaf springs, which are commonly used in vehicle suspension systems become progressively stiffer with deformation because their deformed shape becomes more effective to transfer the vertical forces (their stiffness increases). Mar 15 '17 at 23:59

The weight supported by each spring will depend on how the weight is distributed by the vehicle. Most vehicles are slightly heavier at the front so, a 60:40% or 55:45% distribution would be a good approximation for most cars. Sports cars get closer to an even split. Then, you would subtract the weight of the tyre and supporting axle, brake, etc. since that weight is not supported by the spring.