## Posts Tagged ‘work’

Wednesday, December 6th, 2017
Last time we developed an engineering formula to calculate the *horsepower* required to accelerate a *flywheel* by way of a *reciprocating steam engine, *which contributes to the storage of kinetic energy inside a flywheel. Today we’ll gain a clearer understanding of how this works when we take *a look inside a reciprocating steam engine.*
__A Look Inside a Reciprocating Steam Engine__
A *reciprocating steam engine* performs the work of transforming steam’s heat energy into the mechanical energy needed to move a piston contained within a cylinder. During a complete operating cycle this piston travels from one end of the cylinder to the other, then back again. This is made possible because during the first half of the cycle pressurized steam enters one end of the cylinder and expands inside it, forcing the piston to move.
This process inside the cylinder results in movement of a piston that’s attached to a piston rod, which in turn is connected to a crankshaft via a connecting rod and crank rod. The crankshaft is a device which converts the reciprocating linear motion of an engine’s piston into rotary motion and in so doing facilitates the powering of any externally mounted rotating machinery attached to it. So long as there’s ample steam to power the internal piston, over time, energy in the form of horsepower will be available to externally mounted devices. The energy in the steam decreases as the steam expands behind the moving piston. So, the engine’s horsepower, will decrease as the piston travels to the end of the cylinder. If the energy in the steam should become depleted, the *reciprocating steam engine *will stall. The engine will no longer be able to perform work.
Next time we’ll see how a crankshaft works when we take a look inside it.
opyright 2017 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: connecting rod, crank rod, crankshaft, energy, engineering, flywheel, kinetic energy, piston rod, power, reciprocating steam engine, work

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Saturday, July 29th, 2017
Last time we determined the value for one of the key variables in the Euler-Eytelwein Formula known as the angle of wrap. To do so we worked with the relationship between the two tensions present in our example pulley-belt assembly, *T*_{1 }and *T*_{2}. Today we’ll use physics to solve for *T*_{2} and arrive at *the* *Mechanical Power Formula,* which enables us to compute the amount of *power *present in our *pulley and belt assembly*, a common engineering task.
To start things off let’s reintroduce the equation which defines the relationship between our two tensions, the Euler-Eytelwein Formula, with the value for *e, *Euler’s Number, and its accompanying coefficients, as determined from our last blog,
*T*_{1} = 2.38T_{2 } (1)
Before we can calculate *T*_{1 }we must calculate *T*_{2}. But before we can do that we need to discuss the concept of *power.*
__The Mechanical Power Formula in Pulley and Belt Assemblies__
Generally speaking, power, *P*, is equal to work, *W*, performed per unit of time, *t*, and can be defined mathematically as,
*P = W ÷** t* (2)
Now let’s make equation (2) specific to our situation by converting terms into those which apply to *a pulley and belt assembly*. As we discussed in a past blog, work is equal to force, *F*, applied over a distance, *d*. Looking at things that way equation (2) becomes,
*P = F ×** d ÷** t* (3)
In equation (3) distance divided by time, or “*d ÷** t*,” equals velocity, *V*. Velocity is the distance traveled in a given time period, and this fact is directly applicable to our example, which happens to be measured in units of feet per second. Using these facts equation (3) becomes,
*P = F ×** V* (4)
Equation (4) contains variables that will enable us to determine the amount of *mechanical power*, *P*, being transmitted in our *pulley and belt assembly*.
The force, *F*, is what does the work of transmitting *mechanical power* from the driving pulley, pulley 2, to the passive driven pulley, pulley1. The belt portion passing through pulley 1 is loose but then tightens as it moves through pulley 2. The force, *F,* is the difference between the belt’s tight side tension, *T*_{1}, and loose side tension, *T*_{2}. Which brings us to our next equation, put in terms of these two tensions,
*P = *(*T*_{1} – T_{2}) ×* V * (5)
Equation (5) is known as the *Mechanical Power Formula** ***in** *pulley and belt assemblies*.
The variable *V*, is the velocity of the belt as it moves across the face of pulley 2, and it’s computed by yet another formula. We’ll pick up with that issue next time.
Copyright 2017 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: belt, distance divided by time, engineering, Euler-Eytelwein Formula, Euler's Number, force, loose side tension, mechanical power, mechanical power formula, power, power transmitted, pulley, tight side tension, velocity, work

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Tuesday, February 28th, 2017
For some time now we’ve been analyzing the helpfulness of the engineering phenomena known as *pulleys* and we’ve learned that, yes, they can be very helpful, although they do have their limitations. One of those ever-present limitations is due to the inevitable presence of *friction* between moving parts. Like an unsummoned gremlin, *friction* will be standing by in any mechanical situation to put the wrench in the works. Today we’ll calculate just *how much friction is present* within the example *compound pulley *we’ve been working with. * *
__So How Much Friction is Present in our Compound Pulley?__
Last time we began our numerical demonstration of the inequality between a compound pulley’s work input, *WI*, and work output, *WO*, an inequality that’s due to friction in its wheels. We began things by examining a friction-free scenario and discovered that to lift an urn with a weight, *W*, of 40 pounds a distance, *d*_{1}, of 2 feet above the ground, Mr. Toga exerts a personal effort/force, *F*, of 10 pounds to extract a length of rope, *d*_{2}, of 8 feet.
In reality our compound pulley must contend with the effects of friction, so we know it will take more than 10 pounds of force to lift the urn, a resistance which we’ll notate *F*_{F}. To determine this value we’ll attach a spring scale to Mr. Toga’s end of the rope and measure his actual lifting force, *F*_{Actual}, represented by the formula,
*F*_{Actual} = *F + F*_{F }(1)
We find that *F*_{Actual} equals 12 pounds. Thus our equation becomes,
12 *Lbs = *10 *Lbs + F*_{F }(2)
which simplifies to,
2 *Lbs =* *F*_{F }(3)
Now that we’ve determined values for all operating variables, we can solve for work input and then contrast our finding with work output,
*WI = *(*F ×* *d*_{2}) *+ *(*F*_{F} *×* *d*_{2}) (4)
*WI = *(10* Lbs ×** *8* feet*) *+ *(2* Lbs ×** *8* feet*) (5)
*WI = *96* Ft-Lbs *(6)
We previously calculated work output, *WO* to be 80 *Ft-Lbs, *so we’re now in a position to calculate the difference between work input and work output to be,
*WI – WO =* 16 *Ft-Lbs* (7)
It’s evident that the amount of work Mr. Toga puts into lifting his urn requires 16 more Foot-Pounds of work input effort than the amount of work output produced. This extra effort that’s required to overcome the pulley’s friction is the same as the work required to carry a weight of one pound a distance of 16 feet. We can thus conclude that work input does not equal work output in a *compound pulley*.
Next time we’ll take a look at a different use for pulleys beyond that of just lifting objects.
Copyright 2017 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: compound pulley, engineering, friction, pulley, work, work input, work output

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Wednesday, November 30th, 2016
We’ve been discussing the *mechanical advantage* that *compound pulleys* provide to humans during lifting operations and last time we hit upon the fact that there comes a point of diminished return, a reality that engineers must negotiate in their *mechanical* designs. Today we’ll discuss one of the *undesirable tradeoffs* that results in a diminished return within a *compound pulley *arrangement when we compute the length of rope the Grecian man we’ve been following must grapple in order to lift his urn. What we’ll discover is a situation of *mechanical overkill* – like using a steamroller to squash a bug.
__Mechanical Overkill__
Just how much rope does Mr. Toga need to extract from our working example compound pulley to lift his urn two feet above the ground? To find out we’ll need to revisit the fact that the compound pulley is a work input-output device.
As presented in a past blog, the equations for work input, *WI*, and work output, *WO*, we’ll be using are,
*WI = F ×** d*_{2}
*WO = W ×** d*_{1}
Now, ideally, in a compound pulley no friction exists in the wheels to impede the rope’s movement, and that will be our scenario today. Our next blog will deal with the more complex situation where friction is present. So for our example today, with no friction present, work input equals output…
*WI = WO*
… and this fact allows us to develop an equation in terms of the rope length/distance factors in our *compound pulley* assembly, represented by *d*_{1} and *d*_{2}, …
*F ×** d*_{2} = W ×* d*_{1}
*d*_{2} ÷* d*_{1 }= W ÷* F*
Now, from our last blog we know that *W* divided by *F* represents the *mechanical* advantage, *MA,* to Mr. Toga of using the *compound pulley, *which was found to be 16, equivalent to the sections of rope directly supporting the urn. We’ll set the distance factors up in relation to *MA*, and the equation becomes…
*d*_{2} *÷* d_{1 }= MA
*d*_{2} = MA ×* d*_{1}
*d*_{2} = 16 *×** *2* feet = *32* feet*
What we discover is that in order to raise the urn 2 feet, our Grecian friend must manipulate 32 feet of rope – which would only make sense if he were lifting something far heavier than a 40 pound urn.
In reality, *WI* does not equal *WO, *due to the inevitable presence of friction. Next time we’ll see how friction affects the *mechanical* advantage in our *compound pulley.*
Copyright 2016 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: compound pulley, engineer, force times distance, lift, mechanical advantage, mechanical design, pulley, rope length, work, work input, work output

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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* input-output 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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Sunday, November 6th, 2016
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, *d*_{1} and *d*_{2}. 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 *d*_{2} , his *work input* is expressed as,
*WI = F ×** d*_{2}
When the compound *pulley* lifts the urn a distance *d*_{1} above the ground against gravity, its *work output* is expressed as,
*WO = W ×** d*_{1}
Next time we’ll compare our *pulley’s* *work input to output* to develop a relationship between *d*_{1} and *d*_{2}. 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

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Monday, June 6th, 2016
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 = F*_{F} ×* 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, *F*_{F}, which hampered its free travel, and found that quantity to be 0.35 kilogram-meters/second^{2}. Thus the *kinetic energy* contained within that piece was *calculated* to be 0.70 kilogram-meters^{2}/second^{2}.
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 ×** v*^{2} (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 ×** v*^{2}
0.70 *kilogram-meters*^{2}/second^{2}= ½ × (0.09 *kilograms*) *×** v*^{2}
now we’ll use algebra to rearrange things and isolate *v* to solve for it,
*v*^{2} = 2 *×* (0.70 *kilogram-meters*^{2}/second^{2}) ÷ (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

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Wednesday, May 25th, 2016
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,* *F*_{F} 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, *F*_{F,} acting against a ceramic piece weighing 0.09 kilograms to be 0.35 kilogram-meters/second^{2} 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/second*^{2} ×* *2* meters*
*KE = *0.70 *kilogram-meters*^{2}/second^{2}
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

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Thursday, May 12th, 2016
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 = F*_{F} ×*d* (1)
where, *d* is the distance the piece travels before it stops, and *F*_{F} 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/second*^{2}. 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 ×** v*^{2} (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, *F*_{F}, acting upon it which causes it to stop within a distance, *d*,
*KE = F*_{F} ×* d*
Next time we’ll use this newly derived formula, and the value we found for *F*_{F} 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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Monday, April 4th, 2016
Humans have been battling the force of *friction* since the first cave man built the wheel. Chances are his primitive tools produced a pretty crude wheel that first go-around and the wheel’s surfaces were anything but smooth, making its usefulness less than optimal. As an engineering expert, I encounter these same dynamics when designing modern devices. What held true for the cave man holds true for modern man, *friction* is often a counterproductive force which design engineers must work to minimize. Today we’ll learn about *frictional force* and see how it impacts our example broken coffee mug’s scattering pieces, and we’ll introduce the man behind *friction’s *discovery, Charles-Augustin de *Coulomb.*
__Charles-Augustin de Coulomb__
Last time we learned that the work required to shatter our mug was transformed into the kinetic energy which propelled its broken pieces across a rough concrete floor. The broken pieces’ energetic transformation will continue as the propelling force of kinetic energy held within them is transmuted back into the work that will bring each one to an eventual stop a distance from the point of impact. This last source of work is due to the *force* *of* *friction.*
In 1785 **Charles-Augustin de ***Coulomb,* a French physicist, discovered that *friction* results when two surfaces make contact with one another, and that *friction* is of two types, static or dynamic. Although Leonardo Da Vinci had studied *friction* hundreds of years before him, it is *Coulomb* who is attributed with doing the ground work that later enabled scientists to derive the formula to calculate the effects of *friction*. Our example scenario illustrates dynamic *friction, *that is to say, the *friction* is caused by one of the surfaces being in motion, namely the mug’s ceramic pieces which skid across a stationary concrete floor.
While in motion, each of the mug’s broken pieces has its own unique velocity and mass and therefore a unique amount of kinetic energy. The weight of each piece acts as a vertical force pushing the piece down “into” the floor, this due to the influence of Earth’s gravitational pull, that is, the force of gravity.
* Friction* is created by a combination of factors, including the ceramic pieces’ weights and the surface roughness of both the pieces themselves and the concrete floor they skid across. At first glance the floor and ceramic mug’s surfaces may appear slippery smooth, but when viewed under magnification it’s a whole different story.
Next time we’ll examine the situation under magnification and we’ll introduce the formula used to calculate *friction* along with a rather odd sounding variable, *mu. *
Copyright 2016 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: calculating frictional force, Charles-Augustin de Coulomb, contacting surfaces, dynamic friction, energetic transformation, engineering expert witness, friction, frictional force, kinetic energy, Leonardo Da Vinci, roughness, static friction, work

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