As John Rennie explained, in the American Engineering System, force is expressed in lb$_f$, mass in lb$_m$, and acceleration in ft/sec$^2$. This system is not coherent. Hence, a conversion factor other than one must be used in the equation for force; that is, $F=\frac{ma}{g_c}$, where $g_c=32.174 \frac{lbm \cdot ft}{sec^2 \cdot lbf}$ is a constant, known as the gravitational conversion constant.
Derivation of $g_c$
The principle of conservation of units is used to derive $g_c$. We wish to convert $F=ma$ from SI to AES.
First, find the dimension of the "hidden constant" in the given equation. In other words, $1=\frac{F}{ma}$ has dimension force divided by the product of mass and acceleration. Next, consider the units of the hidden constant -- the given units are $\frac{N \cdot s^2}{kg \cdot m}$ and the required units are $\frac{lbf \cdot sec^2}{lbm \cdot ft}$.
Now, convert the given units of the hidden constant to the required units. Thus, $$\left(\frac{1 N \cdot s^2}{kg \cdot m}\right) \left(\frac{0.45359 kg}{lbm}\right) \left(\frac{0.3048 m}{ft}\right)\left(\frac{lbf}{4.4482 N}\right)=\frac{1}{32.174}\frac{lbf \cdot sec^2}{lbm \cdot ft}$$
The hydrostatic pressure is calculated as follows:
$$p=D\frac{(\rho g_c)}{144g} \ ft \frac{ft^2 \ lbm \ ft \ sec^2 \ lbf}{in^2 \ ft^3 \ lbm \ ft \ sec^2}$$
where $p$ is hydrostatic pressure in psig, $D$ is depth (or height) of the fluid in ft, $\rho$ is the fluid density (lbm/ft^3). Note that the density of fresh water is 62.4 lbm/ft^3, $g_c$ is the gravitational constant of acceleration (32.2 ft/sec^2), and $g$ is units conversion to lbf (32.2 $\frac{lbm \ ft/sec^2}{lbf}$)
In AES the fluid density is often express in lbm/gal. There is 7.48 gal/ft^3, therefore $p=0.05194D\rho$.
It is often convenient to express the hydrostatic pressure as a fluid pressure gradient or hydrostatic pressure developed per unit height of fluid.
$\nabla p =0.052\rho$, where $\nabla p$ is the hydrostatic pressure gradient (psig/ft). For fresh water the pressure gradient is: $$\nabla p=0.052(8.33)=0.433\ \text{psi/ft}$$
The intensity depends on the average of the pressure squared so it is that average which must be used. Taking the squared bit out of the log bracket leaves you, as you have noted, with the rms pressure which is not the same as the mean of the magnitude of the pressure. The same is true in ac theory where the rms value is used because the power depends on the current squared.
In the end it boils down to the measurement of energy transfer per second and because that measure is proportional to the pressure squared it is that quantity that is averaged giving greater weight in the averaging process to the larger pressure variations.
To get the loudness you must factor in the sensitivity of the ear (response to different frequencies) to your $L_p$ which is a measure of the intensity of a sound wave relative to some chosen standard intensity. The process of factoring in the sensitivity of the ear is a complex process which is usually present as graphs of the type shown here. This is not of course the response of your ears but a "typical" ear for a person of a certain age.
So if you increase the intensity of a sound that is within the audible range of frequencies then the sound will become louder but that will not be so for frequencies outside that range.
Best Answer
"Bite force" is just one of many anachronism that pop up in science when you look for them. Physics and the other sciences haven't always been such sticklers for accurately naming phenomena, both because they didn't care about rigor as much and because they didn't know better. In this case, the problem isn't bad units, it's a bad name. It really should be called "bite pressure" since it is the pressure that the teeth exert on whatever it is the animal is biting.
Strong jaw muscles and well shaped jaw bones will increase the amount of force that can be exerted, while sharp teeth decrease the biting surface, creating large pressures. When trying to pierce, cut, or grind, pressure is the important factor, not force.