By Steven Holzner
Any physicist knows that if an object applies a force to a spring, then the spring applies an equal and opposite force to the object. Hooke’s law gives the force a spring exerts on an object attached to it with the following equation:
F = –kx
the minus sign shows that this force is in the opposite direction of the force that’s stretching or compressing the spring. The variables of the equation are: F which represents force, k which is called the spring constant and measures how stiff and strong the spring is, and x is the distance the spring is stretched or compressed away from its equilibrium or rest position.
The force exerted by a spring is called a restoring force; it always acts to restore the spring toward equilibrium. In Hooke’s law, the negative sign on the spring’s force means that the force exerted by the spring opposes the spring’s displacement.
Understanding springs and their direction of force
The direction of force exerted by a spring.
The preceding figure shows a ball attached to a spring. You can see that if the spring isn’t stretched or compressed, it exerts no force on the ball. If you push the spring, however, it pushes back, and if you pull the spring, it pulls back.
Hooke’s law is valid as long as the elastic material you’re dealing with stays elastic — that is, it stays within its elastic limit. If you pull a spring too far, it loses its stretchy ability. As long as a spring stays within its elastic limit, you can say that F = –kx. When a spring stays within its elastic limit and obeys Hooke’s law, the spring is called an ideal spring.
How to find the spring constant (example problem)
Suppose that a group of car designers knocks on your door and asks whether you can help design a suspension system. “Sure,” you say. They inform you that the car will have a mass of 1,000 kilograms, and you have four shock absorbers, each 0.5 meters long, to work with. How strong do the springs have to be? Assuming these shock absorbers use springs, each one has to support a mass of at least 250 kilograms, which weighs the following:
F = mg = (250 kg)(9.8 m/s2) = 2,450 N
where F equals force, m equals the mass of the object, and g equals the acceleration due to gravity, 9.8 meters per second2. The spring in the shock absorber will, at a minimum, have to give you 2,450 newtons of force at the maximum compression of 0.5 meters. What does this mean the spring constant should be? In order to figure out how to calculate the spring constant, we must remember what Hooke’s law says:
F = –kx
Now, we need to rework the equation so that we are calculating for the missing metric which is the spring constant, or k. Looking only at the magnitudes and therefore omitting the negative sign, you get
Time to plug in the numbers:
The springs used in the shock absorbers must have spring constants of at least 4,900 newtons per meter. The car designers rush out, ecstatic, but you call after them, “Don’t forget, you need to at least double that if you actually want your car to be able to handle potholes.”
stretch in meters. Hooke’s law is easily applied to springs, but also applies to any solid object which can have its shape deformed by a given force. Although Hooke’s law is in a linear (y=mx+b) form, if too large a force, or too large a change in “x” is applied to any object, the deformation of an object can become permanent. This is similar to the flexing of a thin metal rod, or the flex of a healthy small tree. These objects are considerably flexible, and will return to an indistinguishable original form unless a threshold is exceeded causing the object to not return to its original form. In everyday life most object stretch or deform to some degree , all of which causes a deterioration to the given material. Although this happens, it is most commonly occurs to such a miniscule degree there will not be a visible, or noticeable difference between before and after the force to the object was applied. In this laboratory report, two methods will be used to calculate the “k” constant of the spring. The first one will be a static method where weight is hung from the investigated spring. This Results in a force of gravity on the spring; F = Mg Using this equation of mass multiplied by the gravitational constant, the force hanging on the spring will be able to be calculated, but the force sensor we will be using finds an identical value in a similar fashion. A distance detection device will also be used, it determines the change in displacement on the vertical-axis as a force is applied to the spring. Both devices will be zeroed with the spring hanging, and no weight placed on it, so all the following trials can be compatible with Hooke’s law. The data for every mass interval will be analyzed using logger pro software to generate a slope, and visual representation of Hooke’s law.