Archive for the ‘Professional Malpractice’ Category
Saturday, January 7th, 2017
Last time we saw how the presence of friction reduces mechanical advantage in an engineering scenario utilizing a compound pulley. We also learned that the actual amount of effort, or force, required to lift an object is a combination of the portion of the force which is hampered by friction and an idealized scenario which is frictionfree. Today we’ll begin our exploration into how friction results in reduced work input, manifested as heat energy lost to the environment. The net result is that work input does not equal work output and some of Mr. Toga’s labor is unproductive.
Friction Results in Heat and Lost Work Within a Compound Pulley
In a past blog, we showed how the actual force required to lift our urn is a combination of F, an ideal frictionfree work effort by Mr. Toga, and F_{F} , the extra force he must exert to overcome friction present in the wheels,
F_{Actual} = F + F_{F} (1)
Mr. Toga is clearly working to lift his turn, and generally speaking his work effort, WI, is defined as the force he employs multiplied by the length, d, of rope that he pulls out of the compound pulley during lifting. Mathematically that is,
WI = F_{Actual} × d (2)
To see what happens when friction enters the picture, we’ll first substitute equation (1) into equation (2) to get WI in terms of F and F_{F},
WI = (F + F_{F} ) × d (3)
Multiplying through by d, equation (3) becomes,
WI = (F × d )+ (F_{F} × d) (4)
In equation (4) WI is divided into two terms. Next time we’ll see how one of these terms is beneficial to our lifting scenario, while the other is not.
Copyright 2017 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: compound pulley, engineering, friction, heat energy, lost work, mechanical advantage, pulley, reduced work, work input, work output
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Monday, December 19th, 2016
Tags: engineering expert, engineering expert witness, pulleys
Posted in Courtroom Visual Aids, Engineering and Science, Expert Witness, Forensic Engineering, Innovation and Intellectual Property, Personal Injury, power plant training, Product Liability, Professional Malpractice  Comments Off on Pulleys Make Santa’s Job Easier
Friday, November 18th, 2016
We’re all familiar with the phrase, “too much of a good thing.” As a professional engineer, I’ve often found this to be true. No matter the subject involved, there inevitably comes a point when undesirable tradeoffs occur. We’ll begin our look at this phenomenon in relation to compound pulleys today, and we’ll see how the pulley arrangement we’ve been working with encounters a rope length tradeoff. Today’s arrangement has a lot of pulleys lifting an urn a short distance.
We’ll be working with two distance/length factors and observe what happens when the number of pulleys is increased. Last time we saw how the compound pulley is essentially a work inputoutput device, which makes use of distance factors. In our example below, the first distance/length factor, d_{1}, pertains to the distance the urn is lifted above the ground. The second factor, d_{2}, pertains to the length of rope Mr. Toga extracts from the pulley while actively lifting. It’s obvious that some tradeoff has occurred just by looking at the two lengths of rope in the image below as compared to last week. What we’ll see down the road is that this also affects mechanical advantage.
The compound pulley here consists of 16 pulleys, therefore it provides a mechanical advantage, MA, of 16. For a refresher on how MA is determined, see our preceding blog.
Rope Length Tradeoff in a Compound Pulley
With an MA of 16 and the urn’s weight, W, at 40 pounds, we compute the force, F, Mr. Toga must exert to actively lift the urn higher must be greater than,
F > W ÷ MA
F > 40 Lbs. ÷ 16
F > 2.5 Lbs.
Although the force required to lift the urn is a small fraction of the urn’s weight, Mr. Toga must work with a long and unwieldy length of rope. How long? We’ll find out next time when we’ll take a closer look at the relationship between d_{1 }and d_{2}.
Copyright 2016 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: compound pulley, effor, force, mechanical advantage, professional engineer, pulley, rope length, weight force, work
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Thursday, October 27th, 2016
Last time we saw how compound pulleys within a dynamic lifting scenario result in increased mechanical advantage to the lifter, mechanical advantage being an engineering phenomenon that makes lifting weights easier. Today we’ll see how the mechanical advantage increases when more fixed and movable pulleys are added to the compound pulley arrangement we’ve been working with.
More Pulleys Increase Mechanical Advantage
The image shows a more complex compound pulley than the one we previously worked with. To determine the mechanical advantage of this pulley, we need to determine the force, F_{5}, Mr. Toga exerts to hold up the urn.
The urn is directly supported by four equally spaced rope sections with tension forces F_{1}, F_{2}, F_{3}, and F_{4}. The weight of the urn, W, is distributed equally along the rope, and each section bears one quarter of the load. Mathematically this is represented by,
F_{1} = F_{2} = F_{3} = F_{4 }= W ÷ 4
If the urn’s weight wasn’t distributed equally, the bar directly above it would tilt. This tilting would continue until equilibrium was eventually established, at which point all rope sections would equally support the urn’s weight.
Because the urn’s weight is equally distributed along a single rope that’s threaded through the entire pulley arrangement, the rope rule, as I call it, applies. The rule posits that if we know the tension in one section of rope, we know the tension in all rope sections, including the one Mr. Toga is holding onto. Therefore,
F_{1} = F_{2} = F_{3} = F_{4 }= F_{5 }= W ÷ 4
Stated another way, the force, F_{5} , Mr. Toga must exert to keep the urn suspended is equal to the weight force supported by each section of rope, or one quarter the total weight of the urn, represented by,
F_{5 }= W ÷ 4
If the urn weighs 40 pounds, Mr. Toga need only exert 10 pounds of bicep force to keep it suspended, and today’s compound pulley provides him with a mechanical advantage, MA, of,
MA = W ÷ F_{5}
MA = W ÷ (W ÷ 4)
MA = 4
It’s clear that adding the two extra pulleys results in a greater benefit to the man doing the lifting, decreasing his former weight bearing load by 50%. If we added even more pulleys, we’d continue to increase his mechanical advantage, and he’d be able to lift far heavier loads with a minimal of effort. Is there any end to this mechanical advantage? No, but there are undesirable tradeoffs. We’ll see that next time.
Copyright 2016 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: compound pulley, engineering, fixed pulley, lifting, mechanical advantage, movable pulley, rope section, tension force
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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,
F_{F} = μ × m × g
where the frictional force is denoted as F_{F}, 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
Using these numerical values we calculate the force of friction to be,
F_{F} = μ × m × g
F_{F} = (0.4) × (0.09 kilograms) × (9.8 meters/sec^{2})
F_{F} = 0.35 kilogram meters/sec^{2}
F_{F} = 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 WorkEnergy 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
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Tags: coefficient of friction, Earth's gravity, engineering expert, force of friction, friction, frictional force, frictional force formula, gravitational acceleration constant, kinetic energy, Newtons, weight force, workenergy theorem
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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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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 galacticsized 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, F_{g,} acting upon the Earth, which has a mass of m. It is,
F_{g} = 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) ÷ r^{2} (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.
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,
F_{g} = m × [(G × M) ÷ r^{2}]
then, using algebra to rearrange terms, we’ll set up the combined equation to solve for M, the sun’s mass:
M = (F_{g }× r^{2}) ÷ (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, F_{g}, 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.
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Tags: acceleration of gravity, Earth's gravity, engineering expert witness, falling objects, force of gravity, forensic engineer, gravity, Halley, Henry Cavendish, Newton's Second Law of Motion, Newton's Universal Law of Gravitation, parallax, Sir Isaac Newton, the mass of the sun, the sun's gravity
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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.
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/sec^{2}.
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.
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Tags: acceleration of gravity, Christiaan Huygens, Earth's gravity, engineering expert witness, falling objects, forensic engineer, gravity, Isaac Newton, Newton's gravitational formula, pendulum, period of swing, ship navigation
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Tuesday, July 15th, 2014
We’ve been discussing gear trains for some time now, and last time we posed the question: Why even bother using a gear train and performing complex computations to arrive at a desired torque for an application? Why not just use a bigger motor to start with? Today we’ll see why.
First, we must acknowledge that sometimes higher torque is achieved by simply using a more powerful motor. But sometimes this isn’t possible or practical.
To begin our discussion, we must first understand how torque is related to motor power, the amount of mechanical work a motor can perform. Torque is in fact a function of how much mechanical power a motor produces. In the United States motor power is typically measured in units of horsepower.
The following equation illustrates the relationship between torque, horsepower, and motor speed:
T = [HP ÷ n] × 63,025
where T is the motor shaft’s torque in units of inchpounds, HP is the motor’s horsepower, and n is the speed of the motor shaft in revolutions per minute (RPM). The number 63,025 in the equation is a constant used to convert the units of horsepower and RPM into units of torque (inch pounds). This equation applies to all sources of mechanical power. Its versatility enables design engineers to easily determine if a mechanical power source can deliver the torque required to drive a particular piece of machinery.
The torque equation above tells us that in order to get a higher torque T for a given speed n, you’ll have to get a motor with a higher HP. Put another way, if your speed remains constant and you use a motor with higher horsepower, you’ll get more torque for your application simply by increasing the horsepower.
Next time we’ll plug numbers into our equation and see how it all works.
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Tags: electric motor, engineering expert witness, forensic engineer, gear train, horsepower, machinery, mechanical power, motor shaft, motor shaft torque, motor speed, torque
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Wednesday, May 14th, 2014
Last time we analyzed the angular relationship between the Force and Distance vectors in this simple gear train. Today we’ll discover a commonality between the two gears in this train which will later enable us to develop individual torque calculations for them.
From the illustration it’s clear that the driving gear is mechanically linked to the driven gear by their teeth. Because they’re linked, force, and hence torque, is transmitted by way of the driving gear to the driven gear. Knowing this we can develop a mathematical equation to link the driving gear Force vector F_{1} to the driven gear Force vector F_{2}, then use that linking equation to develop a separate torque formula for each of the gears in the train.
We learned in the previous blog in this series that F_{1} and F_{2} travel in opposite directions to each other along the same line of action. As such, both of these Force vectors are situated in the same way so that they are each at an angle value ϴ with respect to their Distance vectors D_{1} and D_{2. } This fact allows us to build an equation with like terms, and that in turn allows us to use trigonometry to link the two force vectors into a single equation:
F = [F_{1 }× sin(ϴ)] – [F_{2 }× sin(ϴ)]
where F is called a resultant Force vector, so named because it represents the force that results when the dead, or inert, weight that’s present in the resisting force F_{2 }cancels out some of the positive force of F_{1}.
Next week we’ll simplify our gear train illustration and delve into more math in order to develop separate torque computations for each gear in the train.
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Tags: distance vector, driven gear, driving gear, engineering expert witness, force, force vector, forensic engineer, gear expert, gear teeth, gear tooth, gear train, line of action, mathematical link, mechanical engineer, mechanical link, theta, torque, trigonometry, vector
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