Posts Tagged ‘Newtons’

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



When Kinetic Energy Meets With Opposing Force

Tuesday, March 1st, 2016

    Objects in motion inevitably meet with opposing forces, a theme which I frequently encounter in my work as an engineering expert.   Today we’ll calculate the opposing force our exemplar coffee mug meets when it falls into a pan of kitty litter, thus transforming its freefalling kinetic energy into the work required to move through clay litter.

    Let’s revisit the Work-Energy Theorem formula, whose terms were explained in last week’s blog,

F × d = – ½ × m × v12          (1)

    The left side of this equation represents the mug’s work to move through the litter, while the right side represents its kinetic energy, which it gained through freefall.   To solve for F, the amount of force acting in opposition to the mug’s mass m as it plows a depth d into the litter, we’ll isolate it on one side of the equation, as shown here,

F = [- ½ × m × v12 ] ÷ d          (2)

    So how do we solve for F when we don’t know the value of v1, the mug’s freefall velocity at impact?   We’ll use the fact that The Law of Conservation of Energy tells us that all energies are equal, and we’ll eliminate the part of Equation (2) that contains this unknown variable, that is, the right side of the equation which deals with kinetic energy.   In its place we’ll substitute terms which represent the mug’s potential energy, that is, the latent energy held within it as it sat upon the shelf prior to falling.  Equation (2) then becomes,

F = [- m × g × h] ÷ d           (3)

where g is the Earth’s acceleration of gravity factor, a constant equal to 9.8 meters/sec2 , and h is the height from which the mug fell.

Kinetic Energy Meets With Opposing Force

Kinetic Energy Meets With Opposing Force


    So if we know the mug’s mass, the distance fallen, and the depth of the crater it made in the litter, we can determine the stopping force acting upon it at the time of impact.   It’s time to plug numbers.

    Let’s say our mug has a mass of 0.25 kg, it falls from a height of 2 meters, and it makes a crater 0.05 meters deep.   Then the stopping force acting upon it is,

F = [- (0.25 kg) × (9.8 meters/sec2) × (2 meters)] ÷ (0.05 meters)

=  – 98 Newtons

    The mug was subjected to -98 Newtons, or about -22 pounds of opposing force when it fell into the litter, that resistance being presented by the litter itself.

    Next time we’ll see what happens when our mug strikes a hard surface that fails to cushion its impact.    Energy is released, but where does it go?

Copyright 2016 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog



Applying the Work-Energy Theorem to Braking Distance

Friday, January 29th, 2016

    I’m sometimes called upon to render an engineering expert opinion on auto accidents, and in our last blog we stretched this application to a scenario in which Santa’s sleigh collided with the opposing force of a strong wind.   At that time we used the Work-Energy Theorem to calculate the amount of food energy Rudolph and his team required to regain speed and get back on schedule.   Today we’ll use the Theorem to analyze the forces at play in another deer scenario and calculate the braking distance a car needs to avoid hitting one on the highway.

    The average sedan has a mass of about 1,500 pounds, or 680 kilograms.   In our example it’s driving down the highway at a speed, or velocity, of 30 miles per hour, which equates to it covering a distance of 13.3 meters, or just under 44 feet, per second.

    A deer jumps onto the highway, 60 meters in front of the car.   The alert driver slams on the brakes, which exert 1200 Newtons of stopping force on the car.   If you’ll recall from past blogs in this series, the Newton is the metric unit used to measure force.


 What Is Safe Braking Distance?What is Safe Braking Distance?


    Did Bambi survive?   Let’s use the Work-Energy Theorem to find out.   Here it is again,

F × d = ½ × m × [v22v12]

where, F is the braking force used to slow a car of mass m, from an initial velocity of v1 to a final velocity of v2 in a braking distance, d.   The car will eventually come to a complete stop as the driver attempts to avoid hitting the deer, so its final velocity, v2, will be zero.   The Work-Energy Theorem is most often stated in terms of metric units, the measuring unit of choice in the scientific community, and we’ll follow suit with our math.

    Inserting these values into the equation, we get,

[1200 Newtons] × d

 = ½ × [680 kilograms] × [(0)2 – (13.3 meters per second)2]

    Using algebra to solve for d, the braking distance, we arrive at,

d = ½ × [680 kilograms] ×  [(0)2 – (13.3 meters per second)2] ÷  [1200 Newtons]

d = 50.11 meters

    The car stopped 50.11 meters from the point when the driver slammed on his brakes, just about 10 meters short of hitting the deer.   Bambi lives to leap another day!

    Next time we’ll use the Work-Energy Theorem to assess the fate of the falling coffee mug we introduced in past blogs when we opened our discussion on the different forms of energy.


Copyright 2016 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog



The Work-Energy Theorem Applied to Santa’s Sleigh

Monday, January 11th, 2016

    Last time my engineering expertise was put to the test when it was discovered that Santa’s sleigh was being hampered by a strong gust of wind.   At that time we introduced the Work-Energy Theorem to determine how strong the wind’s opposing force was, and today we’ll work with the Theorem to compute just what Rudolph was up against.   Here again is the expanded, workable version of the Work-Energy Theorem as introduced last time,

F × d = ½ × m × [v22 – v12]

where F is a force acting upon a moving object of mass m over a distance d to slow it from an initial velocity of v1 to a final velocity of v2.

    Applying the Theorem to the dynamics at play in Santa’s situation, F is the opposing wind force, which acts over a distance, d, to slow his sleigh from an initial velocity of v1 to a final velocity, v2, thus presenting Rudolph and his buddies with a real delivery challenge.

Rudolph Struggles Against a Fierce WindRudolph Struggles Against a Fierce Wind 



    If we know that Santa, his sleigh and reindeer have a combined mass of 900 kilograms — which is pretty standard for a fully loaded sleigh and reindeer team — and their initial velocity was 90 meters per second, final velocity 40 meters per second, and the distance over which the slowing took place was 760 meters, then the formula to calculate the opposing wind force becomes,

F × d = ½ × m × [v22 – v12]

F = ½ × m × [v22 – v12÷ d

F = ½ × (900kg) × [(40 meters/sec)2(90 meters/sec)2] ÷ 760 meters

F = -3848.7 Newtons = -865.2 Pounds

    The minus sign signifies that the wind must exert an opposing force of 865.2 pounds in order to slow Santa’s sleigh down.

    In order for Santa to get back on his delivery schedule, Rudolph is going to have to make up for lost time by expending extra energy.   We’ll see how he does that next time.

Copyright 2015 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog



Work and Energy Share an Interesting Relationship

Thursday, December 10th, 2015

      My work as an engineering expert has often required that I perform calculations to quantify the energy consumed by electric motors and steam turbines, such as when they work together at power plants to generate electricity.   Today we’ll see how work and energy share an interesting relationship that is brought out by examining the units by which they are measured.

     Last time we used de Coriolis’ formula to compute work to calculate the amount of work performed while pushing a loaded wheelbarrow a distance of 3 meters.   We found that in order to move the wheelbarrow that distance, a gardener must exert a force equal to 534 Newton • meters of work.   That relationship is shown here,

Work = 178 Newtons × 3 meters = 534 Newton • meters           (1)


Work is force times distance

de Coriolis’ Formula to Compute Work


     The Newton, as discussed previously in this blog series, is shorthand notation for metric units of force, and we’ll use those units today to demonstrate the special relationship between work and energy.

We’ll start by supposing that you’re unfamiliar with the Newton as a unit of measurement.   In that case you’d have to employ longhand notation to quantify things, which means you’d be measuring units of force in terms of kilogram • meters per second2.

     Putting equation (1) in longhand notation terms, we arrive at,

Work = 178 kilogram • meters per second2 × 3 meters       (2)

Work = 534 kilogram • meters2 per second2                    (3)

     If you’ve been following along in this blog series, you’ll recognize that the unit of measurement used to compute work, namely, kilogram • meters2 per second2, is the same as was used previously to measure energy.  That unit is the Joule, which is considerably less wordy.

     Equations (2) and (3) bear out the interesting relationship between work and energy — they share the same unit of measure.   This relationship would not be apparent if we only considered the units for work presented in equation (1).

     So following standard engineering convention where work and energy are expressed in the same units, the work required to push the wheelbarrow is expressed as,

Work = 534 Joules

     Yes, work and energy are measured by the same unit, the Joule.   But, energy isn’t the same as work.   Energy is distinguished from work in that it’s the measure of the ability to perform work.    Stated another way, work cannot be performed unless there is energy available to do it, just as when you eat it provides more than mere pleasure, it provides your body with the energy required to perform the work of pushing a wheelbarrow through the garden.

     Next time we’ll see how work factors into the Work Energy Theorem, which mathematically relates work to energy.

Copyright 2015 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog



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.


Vibration Analysis in Mechanical Engineering, Part II

Sunday, March 14th, 2010

     Last week we began a new article on vibration analysis.  This week we’ll continue by looking at how to solve a vibration analysis problem.  

     Analysis of vibration in machine designs typically involves advanced knowledge of kinetics and higher level mathematics like calculus.  For this discussion, let’s consider a relatively simple balancing problem for a rotating system of masses.

     Suppose you have a ball with 10kg of mass on the end of a stick.  If you’ll recall from our discussion about kinetics, mass is the weight of an object divided by the acceleration of gravity.  For this problem, let’s say that only the ball has mass and not the stick.  Although this may seem nonsensical, it is necessary in order not to complicate our analysis well beyond the basic discussion we are having here.  Getting back to our ball and stick combo, the other end of the stick is attached to a rotating shaft, and the center of the ball is 0.5 meters (m) away from the center of the shaft.  This rotating system is shown in Figure 1.

Figure 1 – A Rotating System With One 10kg Ball


     When the shaft rotates at 60 revolutions per minute (RPM), the ball moves in a circle around the shaft via the stick as shown in Figure 2.  The centrifugal force created by the circular motion of the ball always points at a 90 degree angle to the ball’s path of movement.  The net result is an effect similar to that felt when you drive quickly around a sharp right-hand curve and you feel as though you are being pushed to one side.

Figure 2 – Side View of the Rotating System in Figure 1


     Getting back to Figure 2, since the centrifugal force is at a 90 degree angle to its movement, it keeps changing direction.  When the ball is at the top of its movement, the force points straight up as shown in (a).  When the ball is horizontal right, the force points right as shown in (b).  When the ball is at the bottom of its movement, the force points down as shown in (c). When the ball is horizontal left, the force points left as shown in (d).  This change in the direction of the centrifugal force pulls the rotating system up, to the right, down, and to the left as the shaft rotates, causing vibration in the whole system. The mechanics of this vibration are very similar to the unbalanced load in the washing machine during spin cycle that we talked about last week.

     To answer this question, let’s apply the formula for centrifugal force on a ball as it rotates in our system above:

F = (Mass) x ((RPM) x (1 min. / 60 sec.))2 x (4 π2) x

                                                        (Distance From Center of Rotation)

Plugging in values, the centrifugal force for the 10kg ball is calculated to be:

F10kg = (10 kg) x ((60 RPM) x (1 min. / 60 sec.))2 x (4 π2) x (0.5 m)

= 197.39 kg m/sec2 = 197.39 Newtons

In case you’re wondering, scientists got tired of talking about the metric units of force as “kg m/sec2,” so they decided to rename these units in honor of the great Sir Isaac Newton.

     In order for the vibration in our example to go away, the centrifugal force for the 10kg ball must equal the centrifugal force of the 6kg ball.  Knowing this, we can work backwards to calculate the distance from the center of rotation for the 6kg ball.  This distance will be key to balancing the system out because it acts to compensate for the unequal masses of the balls.

   F6kg = 197.39 kg m/sec2 = (6 kg) x ((60 RPM) x (1 min. / 60 sec.))2 x

                                             (4 π2) x (Distance From Center of Rotation6kg)

And the distance from the center of rotation becomes:

     Distance From Center of Rotation6kg =

                 197.39 kg m/sec2 /[(6 kg) x ((60 RPM) x (1 min. / 60 sec.))2 x (4 π2)]

                                                                        = 0.83 meters

     This tells us that we’d have to cut our stick so that the center of the 6kg ball is 0.83 meters from the center of the shaft in order to compensate for the vibration caused by the rotating 10kg ball.  See Figure 3.

Figure 3 – A Rotating System With One 10kg Ball and One 6kg Ball

     This concludes our basic look at vibration analysis.  Next week we’ll get into the last installment of our mechanical engineering series to see how everything we’ve learned throughout this series ties together in machine design.