Posts Tagged ‘mass’
Thursday, February 18th, 2016
It’s not uncommon in my work as an engineering expert to encounter a situation in which I’m missing information. At that point I’ve got to find a creative solution to working the problem. We’ll get creative today when we combine the Law of Conservation of Energy and the WorkEnergy Theorem to get around the fact that we’re missing a key quantity to calculate forces exerted upon the falling coffee mug we’ve been following in this blog series.
Last time we applied the WorkEnergy Theorem to our mug as it came to rest in a pan of kitty litter. Today we’ll set up the Theorem formula to calculate the force acting upon it when it meets the litter. Here’s where we left off,
F × d = –½ × m × v_{1}^{2}
where, F is the force acting to slow the progress of the mug with mass m inside the litter pan. The mug eventually stops and comes to rest in a crater with a depth, d. The left side of the equation represents the mug’s work expenditure, as it plows through the litter, which acts as a force acting in opposition to the mug’s travel.
Kinetic Energy Meets Up With Displacement
The right side of the equation represents the mug’s kinetic energy, which it gained in freefall, at its point of impact with the litter. The right side is in negative terms because the mug loses energy when it meets up with this opposing force.
Let’s say we know the values for variables d and m, quantities which are easily measured. But the kinetic energy side of the equation also features a variable of unknown value, v_{1}, the mug’s velocity upon impact. This quantity is difficult to measure without sophisticated electronic equipment, something along the lines of a radar speed detector used by traffic cops. For the purpose of our discussion we’ll say that we don’t have a cop standing nearby to measure the mug’s falling speed.
If you’ll recall from past blog discussions, the Law of Conservation of Energy states that an object’s — in this case our mug’s — kinetic energy is equal to its potential energy. It’s this equivalency relationship which will enable us to solve the equation and work around the fact that we don’t have a value for v_{1}.
We’ll do the math and plug in the numbers next time.
Copyright 2016 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: displacement, engineering expert, falling objects, impact force, kinetic energy, law of conservation of energy, mass, opposing force, potential energy, velocity, workenergy theorem
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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 WorkEnergy 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?
Did Bambi survive? Let’s use the WorkEnergy Theorem to find out. Here it is again,
F × d = ½ × m × [v_{2}^{2} – v_{1}^{2}]
where, F is the braking force used to slow a car of mass m, from an initial velocity of v_{1} to a final velocity of v_{2 }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, v_{2, }will be zero. The WorkEnergy 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 WorkEnergy 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
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Tags: automobile, braking distance, braking force, engineering expert, mass, Newtons, velocity, workenergy theorem
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Wednesday, January 20th, 2016
In my work as an engineering expert I’ve never had to convert Joules of workenergy into calories, but that’s exactly what we’ll be doing together today. Last time we applied the WorkEnergy Theorem to the progress of Santa’s sleigh and found that an opposing wind force of 3848.7 Newtons –or 865.2 pounds for those of us who are American– slowed his team from an initial velocity of 90 meters per second to a final velocity of 40 meters per second and that it happened over a distance of 760 meters. Today we’ll see how many calories the reindeer need to expend to get them back up to full delivery speed.
Prancer Loves Oats
Now we know that Santa successfully made all his deliveries on time this past Christmas, so that means that at some point his reindeer team was able to get back up to full sleighflying speed. They did it by expending extra energy. We’ll use the WorkEnergy Theorem to find out how much energy that equates to. Here’s the Theorem again,
W = ½ × m × [v_{2}^{2} – v_{1}^{2}]
where W is the work/energy required to speed up the sleigh team’s mass, m, from an initial velocity v_{1} to a final velocity v_{2}. For a refresher on the special relationship between work and energy, see our past blog on the subject.
If Santa’s sleigh has a mass of 900 kilograms and its speed must increase from 40 to 90 meters per second, then the work required to do so is calculated as,
W = ½ × (900 kilograms) × [(90 meters/second)^{2} – (40 meters/second)^{2}]
W = ½ × (900 kilograms) × (6,500 meters^{2}/second^{2})
W = 2,925,000 kilogram^{2} · meters^{2} per second^{2} = 2,925,000 Joules
So Rudolph and his buddies had to expend 2,925,000 Joules of energy to perform 2,925,000 Joules of work. To understand where Rudolph and his team got that energy, we must state things in terms of nutritional value, that is, units of calories.
Did you know that 1 calorie is equal to 4,184.43 Joules? Applying that equivalency to our situation we get,
Nutritional Energy Required = (2,925,000 Joules) × (4,184.43 Joules/calorie)
= 699.02 calories
The net result is Santa’s team expended a total of 699.02 calories for all the reindeer to regain their full speed of 90 meters per second. That’s the nutritional energy found in slightly more than one cup of oats. Now everybody knows that Santa takes good care of his reindeer, so undoubtedly they were fed plenty of oats and hay before takeoff. This was stored in their body fat for future, ondemand use.
Sadly, Christmas is over, and it’s time to get back to the more mundane aspects of life. Next time we’ll apply the principles behind the WorkEnergy Theorem to calculate the braking force required to stop a car in motion.
Copyright 2015 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: calories, energy, engineering expert, Joules, mass, speed, velocity, work, workenergy theorem
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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 WorkEnergy 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 WorkEnergy Theorem as introduced last time,
F × d = ½ × m × [v_{2}^{2} – v_{1}^{2}]
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 v_{1} to a final velocity of v_{2}.
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 v_{1 }to a final velocity, v_{2}, thus presenting Rudolph and his buddies with a real delivery challenge.
Rudolph 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 × [v_{2}^{2} – v_{1}^{2}]
F = ½ × m × [v_{2}^{2} – v_{1}^{2}] ÷ 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
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Tags: distance, engineering expert, force, mass, Newtons, pounds, velocity, WorkEnergyTheorem
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Friday, January 1st, 2016
As an engineering expert I’ve applied the WorkEnergy Theorem to diverse situations, but none as unique as its most recent application, the progress of Santa’s sleigh. Last week we saw how Santa and his reindeer team encountered a wind gust which generated enough force to slow them from an initial velocity of v_{1} to a final velocity, v_{2,} over a distance, d. Today we’ll begin using the WorkEnergy Theorem to see if Santa was able to keep to his Christmas delivery schedule and get all the good boys and girls their gifts in time.
Before we can work with the WorkEnergy Theorem, we must first revisit the formula it’s predicated upon, de Coriolis’ formula for kinetic energy,
KE = ½ × m × v^{2} (1)
where, KE is kinetic energy, m is the moving object’s mass, and v its velocity.
The equation behind the WorkEnergy Theorem is,
W = KE_{2 }– KE_{1} (2)
where W is the work performed, KE_{1} is the moving object’s initial kinetic energy and KE_{2} its final kinetic energy after it has slowed or stopped. In cases where the object has come to a complete stop KE_{2} is equal to zero, since the velocity of a stationary object is zero.
In order to work with equation (2) we must first expand it into a more useful format that quantifies an object’s mass and initial and final velocities. We’ll do that by substituting equation (1) into equation (2). The result of that term substitution is,
W = [½ × m × v_{2}^{2} ] – [½ × m × v_{1}^{2}] (3)
Factoring out like terms, equation (3) is simplified to,
W = ½ × m × [v_{2}^{2} – v_{1}^{2}] (4)
Now according to de Coriolis, work is equal to force, F, times distance, d. So substituting these terms for W in equation (4), the expanded version of the WorkEnergy Theorem becomes,
F × d = ½ × m × [v_{2}^{2} – v_{1}^{2}] (5)
Next time we’ll apply equation (5) to Santa’s delivery flight to calculate the strength of that gust of wind slowing him down.
Copyright 2015 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: de Coriolis, distance, energy, engineering expert, force, kinetic energy, mass, velocity, wind force, work, workenergy theorem
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Friday, September 18th, 2015
Last time we introduced kinetic energy as the energy of movement. Today we’ll see how to calculate it, using French mathematician GaspardGustave de Coriolis’ formula as set out in his textbook, Calculation of the Effect of Machines. We’ll then apply his formula to our example of a coffee mug falling from its shelf.
GaspardGustave de Coriolis’ book presented physics concepts, specifically the study of mechanics, in an accessible manner, without a lot of highbrow theory and complicated mathematics. His insights made complicated subjects easy to understand, and they were immediately put to use by engineers of his time, who were busily designing mechanical devices like steam engines during the early stages of the Industrial Revolution.
Within its pages the mathematics of kinetic energy was presented in the scientific form that persists to present day. That formula is,
KE = ½ × m × v^{2}
where m is the moving object’s mass and v its velocity.
In the case of our coffee mug, its kinetic energy will be zero so long as it remains motionless on the shelf. A human arm had lifted it to its perch against the force of gravity, thereby investing it with gravitational potential energy. If the mug was sent freefalling to the ground by the mischievous kitty, its latent potential energy would be realized and converted into the kinetic energy of motion.
To illustrate, let’s say a mug with a mass equal to 0.25 kg rests on a shelf 2 meters above the floor. Its potential energy would then be equal to 4.9 kg • meter^{2}/second^{2}, as was computed in our previous blog, Computing Potential Energy.
Once kitty nudges the mug from its perch and it begins to fall, its latent gravitational potential energy begins a conversion process from potential to kinetic energy. It will continue to convert into an amount of kinetic energy that’s precisely equal to the mug’s potential energy while at rest on the shelf, that is, 4.9 kg • meter^{2}/second^{2}. Upon impact with the floor, all the mug’s gravitational potential energy will have been converted into kinetic energy.
Next time we’ll apply the Law of Conservation of Energy to the potential and kinetic energy formulas to calculate the mug’s velocity as it freefalls to the floor.
Copyright 2015 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: Calculation of the Effect of Machines, engineering expert witness, engines, falling objects, GaspardGustave de Coriolis, kinetic energy, machinery, mass, mechanical energy, mechanical engineer, mechanics, physics, potential energy, velocity
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Friday, September 11th, 2015
Last time we introduced The Law of Conservation of Energy, which holds that energy can neither be created nor destroyed. We then applied the concept to a mug resting on a shelf, brimming with latent gravitational potential energy. Today we’ll continue our discussion with a focus on kinetic energy and how Willem Gravesande’s experimentation contributed to our understanding of the subject.
The concept of kinetic energy was first posited by mathematicians Gottfried Leibniz and Johann Bernoulli in the early 18^{th} Century when they theorized that the energy of a moving object is a factor of its mass and speed. Their theory was later proven by Willem Gravesande, a Dutch lawyer, philosopher, and scientist.
Gravesande conducted experiments in which he dropped identical brass balls into a soft clay block. See Figure 1.
Figure 1
Figure 1 shows the results obtained when balls of the same mass m are dropped from various heights, resulting in different velocities as they fall and different clay penetrations. The ball on the left falls at velocity v and penetrates to a depth d. The center ball falls at twice the left ball’s velocity, or 2v, and penetrates four times as deep, or 4d. The right ball falls at three times the left ball’s velocity, 3v, and it penetrates nine times deeper, 9d. The results indicate an exponential increase in clay penetration, dependent on the balls’ speed of travel.
In fact, all the kinetic energy that the balls exhibited during freefall was converted into mechanical energy from the instant they impacted the clay until their movement within it stopped. This change in forms of energy from kinetic to mechanical demonstrates what Julius Robert von Mayer had in mind when he derived his Law of Conservation of Energy. For a refresher on the subject, see last week’s blog, The Law of Conservation of Energy.
As a result of his experimentation, Gravesande was able to conclude that the kinetic energy of all falling objects is a factor of their mass multiplied by their velocity squared, or m × v^{2}.
We’ll see next time how Gravesande’s work paved the way for later scientists to devise the actual formula used to calculate kinetic energy and then we’ll apply it all to our coffee mug falling from the shelf.
Copyright 2015 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: falling objects, forensic engineer, Gottfried Leibniz, impact, Johann Bernoulli, Julius Robert von Mayer, kinetic energy, mass, mechanical energy, mechanical engineering expert witness, penetration, potential energy, velocity, Willem Gravesande
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Tuesday, August 25th, 2015
Last week we discovered that objects acquire potential energy as it relates to gravity based on the height those objects are elevated above the ground. We also introduced an equation to calculate the potential energy of a coffee mug perched on a shelf. We’ll work with that equation today and compute the latent energy that’s hidden within that mug.
Here again is the equation to determine potential energy, put in terms relating to gravity,
PE_{gravitational} = m × g × h
where m is the mass of the mug, h is the height it’s been elevated above the floor, and g is the Earth’s acceleration of gravity factor, as explained in my previous blog entitled, Sir Isaac Newton and the Acceleration of Gravity.
The equation above can be solved using either English or metric units. In the US it’s generally standard practice to perform calculations using English units, such as feet and pounds. But when measuring mass a less familiar English unit, the slug, comes into play. If you’re interested in learning more about this unit, go to a previous blog article entitled, The Force of Gravity.
The kilogram is the metric equivalent of a slug. Since it’s the unit of mass most commonly used throughout the world, we’ll use it to perform our calculation.
Let’s say our mug has a mass of 0.25 kilograms, the shelf it’s resting on is 2 meters above the floor, and g is 9.8 meters/second^{2}. The mug’s gravitational potential energy would then be expressed as,
PE_{gravitational} = (0.25 kg) × (9.8 meters/second^{2}) × (2 meters)
PE_{gravitational} = 4.9 kg • meter^{2}/second^{2}
Next time we’ll expand on our discussion of potential energy and discuss the Law behind the phenomenon and the fact that energy can only be converted from one form to another.
Copyright 2015 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: acceleration of gravity, engineering expert witness, falling objects, forensic engineer, joule, kilogram, mass, potential energy mechanical engineering expert witness, slug
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Thursday, July 9th, 2015
Last time we saw how centripetal force is responsible for keeping a child’s ball moving in a circular path when twirled above his head. Today we’ll see how that same force is responsible for keeping Earth in its rotation around the sun. We’ll use Isaac Newton’s Second Law of Motion as it relates to centripetal force to bear that out today.
Our example from last week,
Figure 1
According to Newton’s Second Law of Motion, F_{c}, the centripetal force acting upon the ball, is calculated by the equation,
F_{c} = [m_{ball} × v^{2}] ÷ r_{string}
where, m_{ball} is the ball’s mass, v the ball’s velocity, and r_{string} is the length of the string.
The mechanics are obviously different with regard to Earth and its orbit around the sun, where there’s no obvious physical link between the objects in rotation. In this case the sun’s gravitational force, F_{g}, acts to tether Earth to it. And just like the string and ball, the sun’s gravitational tether creates a centripetal force, F_{c, } which prevents Earth from leaving its orbital path. See Figure 2.
Figure 2
Newton concluded that for the Earth to remain on its permanent, fixed path of orbit around the sun, F_{c} must be equal to F_{g}, and that the forces in play between the sun and Earth were like a gigantic tug of war where neither side is able to pull any harder than the other. This stalemate is responsible for keeping Earth stably in place in its orbit.
According to Newton’s Second Law, the sun’s gravitational force acting upon the Earth is calculated by,
F_{g} = F_{c} = [m × v^{2}] ÷ r
where, m is Earth’s mass, v its orbital velocity, and r the distance between Earth and the sun. Thanks to early scientists like Edmund Halley and Henry Cavendish, the values for m and r had already been determined. That left v, Earth’s orbital velocity, the only variable remaining to be solved.
We’ll see how 19^{th} Century scientists accomplished that next time.
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Tags: Cavendish, centripetal force, engineering expert witness, forensic engineer, gravitational force, Halley, Issac Newton's Second Law of Motion, mass, velocity
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Friday, December 5th, 2014
In this blog series we’ve been working our way towards determining mass values for heavenly bodies in our universe. Today we’ll explore the inverse relationship phenomenon that’s present when examining the acceleration of gravity between two heavenly bodies, for our purposes Earth and its moon.
To demonstrate the inverse relationship phenomenon, we’ll use Newton’s Law of Gravitation formula:
g = (G × M) ÷ r^{2} (1)
In this formula g represents Earth’s acceleration of gravity that’s acting upon the moon. The center of the moon is located at a distance r from Earth’s center. Earth’s mass is represented by M, and G is the universal gravitational constant, which never varies from its value of 3.49 × 10^{8} cubic feet per slug per second squared.
We’ll solve for Earth’s g factor relative to the moon, which we’ll position at varying distances from Earth’s core, thereby demonstrating how the attracting force of Earth’s gravity becomes weaker as the moon’s distance from Earth’s core increases — hence the inverse relationship.
To begin, we know that the present distance from Earth’s center to the center of the moon is about 238,900 miles, or 1,261,392,000 feet. We also know that Earth’s mass, M, is equal to 4.09 × 10^{23} slugs.
Plugging these values into Newton’s Law of Gravitation equation, we calculate the acceleration of gravity exerted upon the moon in this, its normal orbit, to be:
g = ((3.49 × 10^{8} ft^{3}/slug/sec^{2}) × (4.09 × 10^{23} slugs)) ÷ (1,261,392,000 ft)^{2}
g = 0.0089 ft/sec^{2}
Now let’s suppose that the moon’s orbit was caused to increase so that it became situated 400,000 miles, or 2,112,000,000 feet from Earth. Earth’s acceleration of gravity exerted upon the moon at this distance would be calculated as:
g = ((3.49 × 10^{8} ft^{3}/slug/sec^{2}) × (4.09 × 10^{23} slugs)) ÷ (2,112,000,000 ft)^{2}
g = 0.0032 ft/sec^{2}
From these two examples we can see that the further the moon is positioned from Earth, the weaker Earth’s gravitational pull upon it is. This gravitational pull is the force of gravity, F_{g}, introduced in our last blog, a term which originates in Newton’s Second Law of Motion, as given by the equation:
F_{g} = m × g (2)
where, in this case, m represents the mass of the moon.
Next time we’ll combine equations (1) and (2) and derive a third formula which will allow us to calculate the mass of the sun.
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Tags: acceleration of gravity, engineering expert witness, falling objects, forensic engineer, gravitational force, gravity, mass, Newton's Second Law of Motion, Newton's Universal Theory of Gravitation, pull of gravity
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