g subscript c

$g_c$ - ('g sub c' or 'g subscript c') is an often misunderstood factor in engineering. It is designed for inter-converting force and mass units in a number of equations.

Introduction

The value & units of $g_c$ are:

In FPS units      :  $32.2\frac{lb \cdot ft}{lb_f \cdot s^2}$

In SI units          : $9.8\frac{kg \cdot m}{kg_f \cdot s^2}$

The concept of $g_c$ arises from the fact that physical quantity “force” can be expressed in two types of units: mass units & force units.

Let’s understand in detail:

Force and its Units

We know the physical quantity force, the contribution of Newton.

Newton observed that force required to change the momentum of any body is directly proportional to its mass and acceleration.

That is, $F \propto m \cdot a$

Or $F = k \cdot m \cdot a$

At this point force was not defined mathematically and Newton's 2nd Law gave a chance to define it.

Scientists defined 1 unit of force as the force required to accelerate 1 kg mass with acceleration of $1 \frac{m}{s^2}$ . This 1 unit was named Newton and we observe the value of constant k comes out to be 1.

Okay, so let’s define force a bit differently.

Now we define a new unit of force as 1 Newerton as the force required to accelerate 1 kg mass with acceleration of $5 \frac{m}{s^2}$.

So what complications do we observe with this modification. Well, not much…

Just that our new 2nd Law of Newerton is F = $\frac{1}{5}$ ma, and our constant k is $\frac{1}{5}$. All the equations of motion would have to adjust force with the new constant $\frac{1}{5}$.

Okay, so which force do we know most commonly… Our Weight. Common man uses the word weight but doesn’t mention its units correctly. We still say “My weight is 65 kg” while kg is the unit of mass and not of weight, i.e. force. The person should have said $65 \times 9.8 = 637 N$.

What scientific world call mass, practical world calls it weight.

How to overcome this situation??… Let’s move back…

What if we define a new unit of force, and call it

$1 kg_f$ = weight of 1 kg mass = $m \cdot a = 1 \times 9.8 \frac{m}{s^2} = 9.8 N$.

So, $1 kg_f$ is the force that will produce an acceleration of $9.8 \frac{m}{s^2}$ in a body of 1 kg mass.

What does all this facilitate?? Now a person of 65 kg mass has a weight of $65 kg_f$. Easy!

So this is called force units of force and newton (N) was mass unit of force.

But the issue doesn’t end here… we need to refine mathematical form of 2nd law.

$F = kma$

and

$1 kg_f = k \times 1 kg \times 9.8 \frac{m}{s^2}$

Hence,   $k = \frac{1kg_f \cdot s^2}{9.8kg \cdot m}$   &   $\frac{1}{k} = \frac{9.8kg \cdot m}{1kg_f \cdot s^2}$

And what is this  $\frac{1}{k}$ …?  It’s our $g_c$ !!!

So, $g_c = 9.8\frac{kg \cdot m}{kg_f \cdot s^2}$

g sub c as a conversion factor

$g_c$ is a conversion factor to convert force units ( $kg_f$ ) to mass units ( N ) & vice versa.

As an example, to convert 637 N to force units we divide it by $g_c$:

637 N = $637 \frac{kg \cdot m}{s^2}$

Dividing this value with $g_c$ gives:

$\frac{637 N}{g_c}$ = $637 \frac{kg \cdot m}{s^2} \times \frac{1kg_f \cdot s^2}{9.8kg \cdot m}$ = $65 kg_f$

These are the units when we’re working in SI units. When working with FPS units, gc turns out to be:

$g_c = 32.2\frac{lb \cdot ft}{lb_f \cdot s^2}$

where mass m is in lb, acceleration due to gravity g in $\frac{ft}{s^2}$, and force F in $\frac{lb \cdot ft}{s^2}$ or $lb_f$

Remember Bernoulli’s equation? There are two common versions of it:

High School Form     : $p + \frac{1}{2} \rho v^2 + \rho gy = constant$

Engineering Form     : $\frac{p}{\rho} + \frac{v^2}{2g_c} + \frac{gy}{g_c} = constant$

Okay, so what is 'p' in the above equation… Pressure. And what’s pressure? It’s force per unit area. Now in engineering practice, force is generally expressed as $kg_f$ or $lb_f$ and hence pressure as

$\frac{kg_f}{cm^2}$ or $\frac{lb_f}{in^2}$.

So we introduce $g_c$ in Bernoulli’s equation to convert force units to mass units.

$p + \frac{1}{2} \rho v^2 + \rho gy = constant$

or ,   $pg_c + \frac{1}{2} \rho v^2 + \rho gy = constant$   (for force units of pressure)

or,     $\frac{p}{\rho} + \frac{v^2}{2g_c} + \frac{gy}{g_c} = constant$

If we are going to put Pressure directly in $\frac{N}{m^2}$, we need not consider the presence of $g_c$.

Whenever we need to work with force or any quantity derived from force, like torque, pressure, shear force, etc., we may need to introduce gc if we are putting the values of these quantities in force units.

Hope the article helps !!

About The Author

I'm Gaurav Chachra, chemical engineering graduate from Panjab University, Chandigarh, India. During my degree, I was often intimidated by the presence of g sub c in the equations until a day I challenged myself to decode it. And so this blog post came alive.