Posts Tagged ‘Earth’s gravity’

Calculating the Force of Friction

Wednesday, April 27th, 2016

    Last time we introduced the frictional force formula which is used to calculate the force of friction present when two surfaces move against one another, a situation which I as an engineering expert must sometimes negotiate.   Today we’ll plug numbers into that formula to calculate the frictional force present in our example scenario involving broken ceramic bits sliding across a concrete floor.

   Here again is the formula to calculate the force of friction,

FF = μ × m × g

where the frictional force is denoted as FF, the mass of a piece of ceramic sliding across the floor is m, and g is the gravitational acceleration constant, which is present due to Earth’s gravity.   The Greek letter μ, pronounced “mew,” represents the coefficient of friction, a numerical value predetermined by laboratory testing which represents the amount of friction at play between two surfaces making contact, in our case ceramic and concrete.

    To calculate the friction present between these two materials, let’s suppose the mass m of a given ceramic piece is 0.09 kilograms, μ is 0.4, and the gravitational acceleration constant, g, is as always equal to 9.8 meters per second squared.


Calculating the Force of Friction

Calculating the Force of Friction


    Using these numerical values we calculate the force of friction to be,

FF = μ × m × g

FF = (0.4) × (0.09 kilograms) × (9.8 meters/sec2)

FF = 0.35 kilogram meters/sec2

FF = 0.35 Newtons

    The Newton is shortcut notation for kilogram meters per second squared, a metric unit of force.   A frictional force of 0.35 Newtons amounts to 0.08 pounds of force, which is approximately equivalent to the combined stationary weight force of eight US quarters resting on a scale.

    Next time we’ll combine the frictional force formula with the Work-Energy Theorem formula to calculate how much kinetic energy is contained within a single piece of ceramic skidding across a concrete floor before it’s brought to a stop by friction.

Copyright 2016 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog



The Frictional Force Formula

Thursday, April 14th, 2016

    Last time we introduced the force of friction, another force in our ongoing discussion about changing forms of energy, and we learned that it’s often a counterproductive force which design engineers and engineering experts such as myself must work to minimize in order to optimize functionality of devices we’re designing.   Today we’ll introduce the frictional force formula, which computes the amount of friction present when two surfaces meet.

    To demonstrate frictional force, we’ve been working with the example of a shattered mug’s broken ceramic pieces and watching their progress as they slide across a concrete floor.   They eventually come to a stop not too far from the point where the mug shattered, because friction causes them to stop.   The mass of the ceramic pieces in combination with the downward pull of gravity causes the broken bits to “bear down” on the floor, thereby maximizing contact and creating friction.

    At first glance the floor and mugs’ surfaces may appear slippery smooth, but when viewed under magnification we see that both actually contain many peaks and valleys.   The peaks of one surface project into the valleys of the other and it’s fight, fight, fight for the ceramic pieces to continue their progress across the floor.   The strength of the frictional force acting upon the pieces is a factor of their individual weights coupled with the roughness of the two surfaces coming into contact — the shattered pieces and the floor.   If friction didn’t exist and no other impediments were in the way, the pieces might travel to the next state before stopping!

 Frictional Force Resists Motion

 Frictional Force Resists Motion


    Last time we introduced Charles-Augustin de Coulomb, a scientist whose work with friction led to the later development of a formula to calculate it.   It’s presented here, and frictional force is denoted as FF,

FF = μ × m × g

where, m is the mass of an object making contact with another surface and g is the gravitational acceleration constant, which is due to the force of Earth’s gravity.   The Greek letter μ, pronounced “mew,” represents the coefficient of friction, a number.   Numerical values for μ were determined by laboratory testing and are recorded in engineering books for many combinations of materials, including rubber on concrete, leather on steel, wood on aluminum, and our own example of ceramic on concrete.

    Next time we’ll plug the numbers that apply to our ceramic-on-concrete example into the friction formula and calculate the frictional force at play.

Copyright 2016 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog



Gravity and the Mass of the Sun

Friday, December 12th, 2014

      As a young school boy I found it hard to believe that scientists were able to compute the mass of our sun.  After all, a galactic-sized measuring device does not exist.  But where there’s a will, there’s a way, and by the 18th Century scientists had it all figured out, thanks to the work of others before them.  Newton’s two formulas concerning gravity were key to later scientific discoveries, and we’ll be working with them again today to derive a third formula, bringing us a step closer to determining our sun’s mass.

      Newton’s Second Law of Motion allows us to compute the force of gravity, Fg, acting upon the Earth, which has a mass of m. It is,

Fg = m × g                     (1)

      Newton’s Universal Law of Gravitation allows us to solve for g, the sun’s acceleration of gravity value,

g = (G × M) ÷  r2              (2)

where, M is the mass of the sun, r is the distance between the sun and Earth, and G is the universal gravitational constant.

Force of gravity, falling objects, engineering expert witness

      You will note that g is a common factor between the two equations, and we’ll use that fact to combine them.  We’ll do so by substituting the right side of equation (2) for the g in equation (1) to get,

Fg = m × [(G × M) ÷  r2]

then, using algebra to rearrange terms, we’ll set up the combined equation to solve for M, the sun’s mass:

M = (Fg × r2) ÷ (m × G)           (3)

      At this point in the process we know some values for factors in equation (3), but not others.  Thanks to Henry Cavendish’s work we know the value of m, the Earth’s mass, and G, the universal gravitational constant.  What we don’t yet know is Earth’s distance to the sun, r, and the gravitational attractive force, Fg, that exists between them.

      Next time we’ll introduce some key scientists whose work contributed to a method for computing the distance of our planet Earth to its sun.


The Force of Gravity

Thursday, November 20th, 2014

      Last time we saw how Henry Cavendish built upon the work of scientists before him to calculate Earth’s mass and its acceleration of gravity factor, as well as the universal gravitational constant.   These values, together with the force of gravity value, Fg, which we’ll introduce today, moved scientists one step further towards being able to discover the mass and gravity of any heavenly body in the universe.

      According to Newton’s Second Law of Motion, the force of gravity, Fg, acting upon any object is equal to the object’s mass, m, times the acceleration of gravity factor, g, or,

Fg = m × g

      So what is Fg?  It’s a force at play way up there, in the outer reaches of the galaxy, as well as back home.   It keeps the moon in orbit around the Earth and the Earth orbiting around the sun.   In the same way, Fg keeps us anchored to Earth, and if we were to calculate it, it would be calculated as the force of our body’s mass under the influence of Earth’s gravity.   It’s common to refer to this force as weight, but it’s not quite so simple.

      Using the metric system, the unit of measurement most often used for scientific analyses, weight is determined by multiplying our body’s mass in kilograms by the Earth’s acceleration of gravity factor of 9.8 meters per second per second, or 9.8 meters per second squared.

      For example, suppose your mass is 100 kilograms.   Your weight on Earth would be:

Weight = Fg = m × g = (100 kg) × (9.8 m/sec2) = 980 kg · m/sec2 = 980 Newtons

     Newtons?   That’s right.   It’s easier than saying kilogram · meter per second per second.   It’s also a way to pay homage to the man himself.

      In the English system of measurement things are perhaps even more confusing.   Your weight is found by multiplying the mass of your body measured in slugs by the Earth’s acceleration of gravity factor of 32 feet per second per second.   Slugs is British English speak for pounds · second squared per foot.   We normally refer to weight in units of  pounds, and in engineering circles it’s pounds force.

      For example, suppose your mass is 6 slugs, or 6 pounds · second squared per foot.   Your weight on Earth would be:

Weight = Fg = m × g = (6 Lbs · sec2/ft) × (32.2 ft/sec2)= 193.2 Lbs

      To avoid any confusion, you could just step on the bathroom scale.

Weight Force -  Engineering Expert Witness

      Next time we’ll see how the force of gravity is influenced by an inverse proportionality phenomenon.


How Long is a Pendulum’s Swing?

Wednesday, October 29th, 2014

      What would you do to pass the time if you were stuck on a ship of the middle ages for weeks at a time?   Dutch mathematician Christiaan Huygens used the time to study the movement of clock pendulums.   He watched them for endless hours, and he eventually came to realize that the pendulums’ swing was uneven due to the ship’s listing on the waves, a phenomenon which also affected the ship’s clocks’ accuracy.

      Eager to devise a solution to the problem of inaccurate time keeping, Huygens dedicated himself to finding a solution to the problem, and in so doing increase the navigational accuracy of ships as well.   His efforts eventually resulted in a formula that shared a common variable with Isaac Newton’s gravitational formula, namely, g, Earth’s acceleration of gravity factor, a value which Huygens posited was indeed a non varying constant.

engineering expert witness on falling objects

      Building upon Newton’s work, Huygens devised a formula which demonstrated the mathematical relationship between the motion of a clock’s pendulum and g.   That formula is,

T = 2 × × (L ÷ g)1/2

where, T is the period of time it takes a pendulum to make one complete swing, the Greek symbol pi, valued at 3.14, and L the length of the pendulum.

      Since devices capable of directly measuring the Earth’s gravity did not exist then, as they still don’t exist today, how in the world (pardon the pun) was Huygens able to arrive at this formula?   Thinking outside the box, he posited that if one knows the length of the pendulum L, and then accurately measures the time it takes for the pendulum to complete its swings, taking into account the varied times that resulted due to the ship’s listing, one can calculate g using his equation.   He eventually determined g‘s value to be equal to 32.2 feet per second per second, or 32.2 ft/sec2.

      Fast forwarding to Henry Cavendish’s time, Huygens’ work with pendulums and his determination of g was well known.   We’ll see what Cavendish did with this knowledge next time.