Posts Tagged ‘distance’
Sunday, November 6th, 2016
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In our last blog we saw how adding extra pulleys resulted in mechanical advantage being doubled, which translates to a 50% decreased lifting effort over a previous scenario. Pulleys are engineering marvels that make our lives easier. Theoretically, the more pulleys you add to a compound pulley arrangement, the greater the mechanical advantage — up to a point. Eventually you’d encounter undesirable tradeoffs. We’ll examine those tradeoffs, but before we do we’ll need to revisit the engineering principle of work and see how it applies to compound pulleys as a work input-output device.
Pulleys as a Work Input-Outut Device
The compound pulley arrangement shown includes distance notations, d1 and d2. Their inclusion allows us to see it as a work input-output device. Work is input by Mr. Toga, we’ll call that WI, when he pulls his end of the rope using his bicep force, F. In response to his efforts, work is output by the compound pulley when the urn’s weight, W, is lifted off the ground against the pull of gravity. We’ll call that work output WO.
In a previous blog we defined work as a factor of force multiplied by distance. Using that notation, when Mr. Toga exerts a force F to pull the rope a distance d2 , his work input is expressed as,
WI = F × d2
When the compound pulley lifts the urn a distance d1 above the ground against gravity, its work output is expressed as,
WO = W × d1
Next time we’ll compare our pulley’s work input to output to develop a relationship between d1 and d2. This relationship will illustrate the first undesirable tradeoff of adding too many pulleys.
Copyright 2016 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: compound pulley, distance, engineering, engineering principle, force, mechanical advantage, pulley, weight, work, work input-output device, work of lifting
Posted in Engineering and Science, Expert Witness, Forensic Engineering, Innovation and Intellectual Property, Personal Injury, Product Liability | Comments Off on Pulleys as a Work Input-Outut Device
Monday, June 6th, 2016
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As an engineering expert, I often use the fact that formulas share a single common factor in order to set them equal to each other, which enables me to solve for a variable contained within one of them. Using this approach we’ll calculate the velocity, or speed, at which the broken bit of ceramic from the coffee mug we’ve been following slides across the floor until it’s finally brought to a stop by friction between it and the floor. We’ll do so by combining two equations which each solve for kinetic energy in their own way.
Last time we used this formula to calculate the kinetic energy, KE, contained within the piece,
KE = FF × d (1)
and we found that it stopped its movement across the floor when it had traveled a distance, d, of 2 meters.
We also solved for the frictional force, FF, which hampered its free travel, and found that quantity to be 0.35 kilogram-meters/second2. Thus the kinetic energy contained within that piece was calculated to be 0.70 kilogram-meters2/second2.
Now we’ll put a second equation into play. It, too, provides a way to solve for kinetic energy, but using different variables. It’s the version of the formula that contains the variable we seek to calculate, v, for velocity. If you’ll recall from a previous blog, that equation is,
KE = ½ × m × v2 (2)
Of the variables present in this formula, we know the mass, m, of the piece is equal to 0.09 kilograms. Knowing this quantity and the value derived for KE from formula (1), we’ll substitute known values into formula (2) and solve for v, the velocity, or traveling speed, of the piece at the beginning of its slide.
Combining Kinetic Energy Formulas to Calculate Velocity
The ceramic piece’s velocity is thus calculated to be,
KE = ½ × m × v2
0.70 kilogram-meters2/second2= ½ × (0.09 kilograms) × v2
now we’ll use algebra to rearrange things and isolate v to solve for it,
v2 = 2 × (0.70 kilogram-meters2/second2) ÷ (0.09 kilograms)
v = 3.94 meters/second =12.92 feet/second = 8.81 miles per hour
Our mug piece therefore began its slide across the floor at about the speed of an experienced jogger.
This ends our series on the interrelationship of energy and work. Next time we’ll begin a new topic, namely, how pulleys make the work of lifting objects and driving machines easier.
Copyright 2016 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: distance, energy, engineering expert, friction, frictional force, kinetic energy, mass, velocity, work
Posted in Engineering and Science, Expert Witness, Forensic Engineering, Innovation and Intellectual Property, Personal Injury, Product Liability | Comments Off on Combining Kinetic Energy Formulas to Calculate Velocity
Wednesday, May 25th, 2016
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My activities as an engineering expert often involve creative problem solving of the sort we did in last week’s blog when we explored the interplay between work and kinetic energy. We used the Work-Energy Theorem to mathematically relate the kinetic energy in a piece of ceramic to the work performed by the friction that’s produced when it skids across a concrete floor. A new formula was derived which enables us to calculate the kinetic energy contained within the piece at the start of its slide by means of the work of friction. We’ll crunch numbers today to determine that quantity.
The formula we derived last time and that we’ll be working with today is,
 Calculating Kinetic Energy By Means of the Work of Friction
where, KE is the ceramic piece’s kinetic energy, FF is the frictional force opposing its movement across the floor, and d is the distance it travels before friction between it and the less than glass-smooth floor brings it to a stop.
The numbers we’ll need to work the equation have been derived in previous blogs. We calculated the frictional force, FF, acting against a ceramic piece weighing 0.09 kilograms to be 0.35 kilogram-meters/second2 and the measured distance, d, it travels across the floor to be equal to 2 meters. Plugging in these values, we derive the following working equation,
KE = 0.35 kilogram-meters/second2 × 2 meters
KE = 0.70 kilogram-meters2/second2
The kinetic energy contained within that broken bit of ceramic is just about what it takes to light a 1 watt flashlight bulb for almost one second!
Now that we’ve determined this quantity, other energy quantities can also be calculated, like the velocity of the ceramic piece when it began its slide. We’ll do that next time.
Copyright 2016 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: distance, electrical energy, energy, engineering expert, frictional force, kinetic energy, mass, velocity, Watt, work, work of friction, work-energy theorem
Posted in Engineering and Science, Expert Witness, Forensic Engineering, Innovation and Intellectual Property, Personal Injury, Product Liability | Comments Off on Calculating Kinetic Energy By Means of the Work of Friction
Thursday, May 12th, 2016
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We’ve been discussing the different forms energy takes, delving deeply into de Coriolis’ claim that energy doesn’t ever die or disappear, it simply changes forms depending on the tasks it’s performing. Today we’ll combine mathematical formulas to derive an equation specific to our needs, an activity my work as an engineering expert frequently requires of me. Our task today is to find a means to calculate the amount of kinetic energy contained within a piece of ceramic skidding across a concrete floor. To do so we’ll combine the frictional force and Work-Energy Theorem formulas to observe the interplay between work and kinetic energy.
As we learned studying the math behind the Work-Energy Theorem, it takes work to slow a moving object. Work is present in our example due to the friction that’s created when the broken piece moves across the floor. The formula to calculate the amount of work being performed in this situation is written as,
W = FF ×d (1)
where, d is the distance the piece travels before it stops, and FF is the frictional force that stops it.
We established last time that our ceramic piece has a mass of 0.09 kilograms and the friction created between it and the floor was calculated to be 0.35 kilogram-meters/second2. We’ll use this information to calculate the amount of kinetic energy it contains. Here again is the kinetic energy formula, as presented previously,
KE = ½ × m × v2 (2)
where m represents the broken piece’s mass and v its velocity when it first begins to move across the floor.
The Interplay of Work and Kinetic Energy
The Work-Energy Theorem states that the work, W, required to stop the piece’s travel is equal to its kinetic energy, KE, while in motion. This relationship is expressed as,
KE = W (3)
Substituting terms from equation (1) into equation (3), we derive a formula that allows us to calculate the kinetic energy of our broken piece if we know the frictional force, FF, acting upon it which causes it to stop within a distance, d,
KE = FF × d
Next time we’ll use this newly derived formula, and the value we found for FF in our previous article, to crunch numbers and calculate the exact amount of kinetic energy contained with our ceramic piece.
Copyright 2016 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: distance, energy, engineering expert, frictional force, kinetic energy, mass, velocity, work, work-energy theorem
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Tuesday, March 1st, 2016
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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
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
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Tags: distance, engineering expert, falling objects, force, kinetic energy, law of conservation of energy, mass, Newtons, potential energy, velocity, work-energy theorem
Posted in Engineering and Science, Expert Witness, Forensic Engineering, Innovation and Intellectual Property, Personal Injury, Product Liability | Comments Off on When Kinetic Energy Meets With Opposing Force
Monday, January 11th, 2016
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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 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
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Tags: distance, engineering expert, force, mass, Newtons, pounds, velocity, Work-EnergyTheorem
Posted in Engineering and Science, Expert Witness, Innovation and Intellectual Property, Personal Injury | Comments Off on The Work-Energy Theorem Applied to Santa’s Sleigh
Friday, January 1st, 2016
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As an engineering expert I’ve applied the Work-Energy 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 v1 to a final velocity, v2, over a distance, d. Today we’ll begin using the Work-Energy 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 Work-Energy Theorem, we must first revisit the formula it’s predicated upon, de Coriolis’ formula for kinetic energy,
KE = ½ × m × v2 (1)
where, KE is kinetic energy, m is the moving object’s mass, and v its velocity.
The equation behind the Work-Energy Theorem is,
W = KE2 – KE1 (2)
where W is the work performed, KE1 is the moving object’s initial kinetic energy and KE2 its final kinetic energy after it has slowed or stopped. In cases where the object has come to a complete stop KE2 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 × v22 ] – [½ × m × v12] (3)
Factoring out like terms, equation (3) is simplified to,
W = ½ × m × [v22 – v12] (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 Work-Energy Theorem becomes,
F × d = ½ × m × [v22 – v12] (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, work-energy theorem
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Friday, December 25th, 2015
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I hope all the good little boys and girls got Santa’s presents, because his sleigh was spotted yesterday on radar moving at an alarmingly slow pace.
As the radar screen here shows, the progress of Santa and his reindeer had slowed from an initial velocity of v1 to a much slower velocity of v2 over the distance, d, he traveled from his workshop.
Santa’s Sleigh Tracked On Radar
Upon closer examination, the radar operator noted that the North Wind was especially powerful this year, which caused Rudolph and the gang to double their efforts to overcome its negative force.
Santa and His Reindeer Fighting The Wind
Did Santa make it in time to deliver Christmas presents? Find out next time, when we put the developing drama into the form of mathematical equations.
Copyright 2015 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: distance, force, velocity, work-energy theorem
Posted in Engineering and Science | Comments Off on The Work-Energy Theorem Affects Santa’s Sleigh!
Sunday, November 29th, 2015
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Although I’m an engineering expert in the 21st Century, I often have to employ engineering principles that are centuries old. A case in point is Gaspard Gustave de Coriolis‘ formula to compute work, as set out in his Principle of Work. We’ll work with his formula today, and we’ll introduce a unit of measurement used to quantify work known as the Newton.
de Coriolis’ formula to compute work is used to determine the amount of work, that is, the amount of dynamic energy available to influence the movement of an object, and is calculated by the formula,
Work = Force × Distance
where F represents the force acting upon an object that travels a distance of D. Force is most often expressed in metric units as kilogram • meter per second2, a wordy expression which is more conveniently referred to as the Newton.
In the image below, F is the force of 178 Newtons exerted by the gardener to push his filled wheelbarrow a distance of 3 meters. The quantity 178 Newtons was obtained by way of direct personal experience working in my own garden. I’ve found that it takes approximately 40 pounds of force to push a wheelbarrow loaded with dirt across level ground. Because one pound of force is equal to 4.45 Newtons, the amount of force I exerted is expressed as,
[40 pounds of force] × [4.45 Newtons per pound force] = 178 Newtons
Work = Force × Distance
If 178 Newtons of force is required to push the wheelbarrow a distance of 3 meters, then the work performed is expressed as,
Work = 178 Newtons × 3 meters
= 534 Newton • meters
Next time we’ll explore the special relationship between work and energy and introduce another unit used to quantify work.
Copyright 2015 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: de Coriolis' formula to compute work, distance, energy, engineering expert, force, Gaspard Gustave de Coriolis' principle of work, Newton, work formula
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Wednesday, April 2nd, 2014
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Last time we introduced the mathematical formula for torque, which is most simply defined as a measure of how much a force acting upon an object causes that object to rotate around a pivot point. When manipulated, torque can produce a mechanical advantage in gear trains and tools, which we’ll see later. The formula is:
Torque = Distance × Force × sin(ϴ)
We learned that the factors Distance and Force are vectors, and sin(ϴ) is a trigonometric function of the angle ϴ which is formed between their two vectors. Let’s return to our wrench example and see how the torque formula works.
Vectors have both a magnitude, that is, a size or extent, and a direction, and they are typically represented in physics and engineering problems by straight arrows. In our illustration the vector for distance is represented by an orange arrow, while the vector for force is represented by a red arrow. The orange distance vector has a magnitude of 6 inches, while the red force vector has a magnitude of 10 pounds, which is being supplied by the user’s arm muscle manipulating the nut. That muscle force follows a path from the arm to the pivot point located at the center of the nut, a distance of 6 inches.
Vector arrows point in a specific direction, a direction which is indicative of the way in which the vectors’ magnitudes — in our case inches of distance vs. pounds of force — are oriented with respect to one another. In our illustration the orange distance vector points away from the pivot point. This is according to engineering and physics convention, which dictates that, when a force vector is acting upon an object to produce a torque, the distance vector always points from the object’s pivot point to the line of force associated with the force vector. The angle, ϴ, that is formed between the two vectors in our example is 90 degrees, as measured by any common, ordinary protractor.
Next we must determine the trigonometric value for sin(ϴ). This is easily accomplished by simply entering “90” into our calculator, then pressing the sin button. An interesting fact is that when the angle ϴ ranges anywhere between 0 and 90 degrees, the values for sin(ϴ) always range between 0 and 1. To see this in action enter any number between 0 and 90 into a scientific calculator, then press the sin button.
For our angle of 90 degrees we find that,
sin(90) = 1
Thus the formula for torque in our example, because the sin(ϴ) is equal to 1, simply becomes the product of the magnitudes of the Distance and Force vectors:
Torque = Distance × Force × sin(90)
Torque = Distance × Force × 1
Torque = Distance × Force
Next time we’ll insert numerical values into the equation and see how easily torque can be manipulated.
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Tags: angle, direction, distance, engineering, expert witness, force, forensic engineer, magnitude, nut, physics, torque, torque formula, trigonometry, vector, wrench, wrench and nut
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