In a previous question-and-answer sequence, the spring constant ( k ) for a car’s shock absorbers was determined. Interestingly enough, when the net force exerted by all four shock absorbers was determined, an entirely different spring constant of ( k’ ) was derived. Why would the fraction of the system’s net force ( ¼ Fs ) maintained by one shock absorber give rise to one spring constant while the ( Fs ) value for the entire system ( four springs ) give rise to some other??? The differences observed are somewhat analogous to the force multiplication observed when dealing with systems that consist of enclosed fluids:
The force imparted upon the cylinder to the left creates an increase in pressure ( ΔP ) throughout the system. Since the area of the cylinder to the right is 50 times greater than that on the left, fifty times as much force is exerted in the upward direction in the rightmost cylinder. The key point is that new forces introduced into the system are transmitted equally throughout, so the concept of force within this system, by itself, cannot give an appropriate context of what has occurred within the system. What’s most pertinent is the force experienced throughout the fluid system per unit area. The relationship between force and area are summed up by Pascal’s principle:
P = ( F / A )
The units of pressure are Newtons per square meter ( N / m2 ).
It is somewhat easy to conceptually accept the fact that the energetics of a liquid are changed throughout by forces exerted upon it; however, the same concept applies to solid systems. In regard to the aforementioned car, it could have been more useful ( depending upon circumstances ) to evaluate the system in a way that severs reliance upon the number of springs involved. When dealing with solids, this task is accomplished via usage of an equation for Stress:
σ = Stress = ( F / A )
In the previous car system, the different ( k ) values for 1 vs. 4 springs simply showed that greater forces ( Fs ) were being evaluated:
Fs = ks
4Fs = ( 4 k )( s )
4 k = k’
Nonetheless, the material nature of each spring is one and the same. The stress equation severs our reliance upon forces that may vary, as was the case leading to the derivation of ( k ) and ( k’ ). Instead, stress is a measure of the force distribution within a given material.
Q: A punch card is used to poke holes in a stack of papers. Its cylinder has a cross-sectional diameter of 1.0 cm that is driven downward with a force of 100 kN. What is the value of the shear stress within the 2.0-mm-thick stack of papers within it?
A: With shear stress, force pairs act parallel to the area of a surface. In regard to the punch card, something akin to a pinch occurs between the paper’s surface and the card itself. Fortunately, the same equation is used to evaluate tensile stress ( σt ), normal compressive stress ( σc ), and shear stress as well ( σs ):
σs = ( F / A )
The vertical cylindrical surface of the card puncher contains our area of interest:
A = ( 2𝝅r )( d )
A = ( 2𝝅 )( D / 2 )( d )
A = ( 𝝅 )( 1.0 x 10-2 m )( 2.0 x 10-3 m )
A = 6.283 x 10-5 m2
σs = ( 10 x 104 N / 6.283 x 10-5 m2 )
σs = 1.6 GPa
RENAISSANCE PHYSICS 