March 9th, 2017
Last time we ended our blog series on *pulleys* and their application within engineering as aids to lifting. Today we’ll embark on a new focus series, *pulleys *used in mechanical devices. We begin with some *history**,* a peek at Swiss scientist and mathematician *Leonhard Euler,* *a historical figure* credited to be perhaps the greatest mathematician of the 18^{th} Century.
__Leonhard Euler, a Historical Figure in Pulleys__
*Euler* is so important to math, he actually has two numbers named after him. One is known simply as *Euler’s* Number, 2.7182, most often notated as *e*, the other *Euler’s* Constant, 0.57721, notated *γ*, which is a Greek symbol called gamma. In fact, he developed most math notations still in use today, including the infamous function notation, *f(x)*, which no student of elementary algebra can escape becoming intimately familiar with.
*Euler *authored his first theoretical essays on the science and mathematics of *pulleys* after experimenting with combining them with belts in order to transmit mechanical power. His theoretical work became the foundation of the formal science of designing pulley and belt drive systems. And together with German engineer Johann Albert Eytelwein, *Euler *is credited with a key formula regarding pulley-belt drives, the Euler-Eytelwein Formula, still in use today, and which we’ll be talking about in depth later in this blog series.
We’ll talk more about Eytelwein, another important historical figure who worked with *pulleys,* next time.
Copyright 2017 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: belt and pulley drive systems, belts, engineering, Euler's Constant, Euler's Number, Johann Albert Eytelwein, Leonhard Euler, mechanical power transmission, pulleys

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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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February 15th, 2017
Last time we began *work* on a *numerical* demonstration and engineering analysis of the inequality of work input and output as experienced by our example persona, an ancient Greek lifting an urn. Today we’ll get two steps closer to demonstrating this reality as we *work a compound pulley’s numerical puzzle, *shuffling equations like a Rubik’s Cube to arrive at values for two variables crucial to our analysis, *d*_{2}, the length of rope he extracts from the *pulley* while lifting, and *F, *the force/effort required to lift the urn in an idealized situation where no friction exists. * *
__A Compound Pulley’s Numerical Puzzle is Like a Rubik’s Cube__
We’ll continue manipulating the work input equation, *WI,* as shown in Equation (1), along with derivative equations, breaking it down into parts, and handle the two terms within parentheses separately. Term one, (*F *× *d*_{2}), corresponds to the force/effort/work required to lift the urn in an idealized no-friction world. It’ll be our focus today as it provides a springboard to solving for variables *F* and *d*_{2}.
*WI = *(*F ×* *d*_{2}) *+ *(*F*_{F} *×* *d*_{2}) (1)
Previously we learned that when friction is present, work output, *WO,* is equal to work input minus the work required to oppose friction while lifting. Mathematically that’s represented by,
*WO = WI – *(*F*_{F} ×* d*_{2}) (2)
We also previously calculated *WO* to equal 80 *Ft-Lbs*. To get *F* and *d*_{2} into a relationship with terms we already know the value for, namely *WO*, we substitute Equation (1) into Equation (2) and arrive at,
80 *Ft-Lbs = *(*F ×* *d*_{2}) *+ *(*F*_{F} *×* *d*_{2})* – *(*F*_{F} *×* d_{2}) (3)
simplified this becomes,
80 *Ft-Lbs =* *F ×* *d*_{2} (4)
To find the value of *d*_{2}, we’ll return to a past equation concerning compound pulleys derived within the context of *mechanical advantage, MA*. That is,
*d*_{2} ÷ *d*_{1 }= _{ }MA_{ }(5)
And because in our example four ropes are used to support the weight of the urn, we know that *MA* equals 4. We also know from last time that* d*_{1} equals 2 feet. Plugging these numbers into Equation (5) we arrive at a value for *d*_{2},
*d*_{2} ÷ 2* ft*_{ }= _{ }4 (6)
*d*_{2} *= *4 *×* 2* ft *(7)
*d*_{2} = 8* ft *(8)
Substituting Equation (8) into Equation (4), we solve for *F,*
80 *Ft-Lbs =* *F ×* 8* ft *(9)
*F = 10 Pounds *(10)
Now that we know *F* and *d*_{2} we can solve for *F*_{F}, the amount of extra effort required by man or machine to overcome friction in a *compound pulley* assembly. It’s the final piece in the *numerical puzzle* which will then allow us to compare work input to output.
Copyright 2017 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: compound pulley, engineering, friction, friction force, lifting force, work input, work output

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February 6th, 2017
Last time we performed an engineering analysis of a *compound pulley* which resulted in an equation *comparing* the amount of true work effort, or *work input,* *WI*, required by machine or human to lift an object, in our case a toga’d man lifting an urn. Our analysis revealed that, in real world situations, work input does not equal *work output,** WO,* due to the presence of friction. Today we’ll begin to numerically demonstrate their inequality by first solving for work output, and later *work input.*
__Comparing Work Input to Output in a Compound Pulley__
To solve for the *work output* of our compound pulley, we’ll use an equation provided previously that is in terms of the variables *W* and *d*_{1},
*WO = W ×* *d*_{1} (1)
In our example Mr. Toga lifts an urn of weight, *W,* equal to 40 pounds to a height, or distance off the floor, *d*_{1}, of 2 feet. Inserting these values into equation (1) we arrive at,
*WO = *40* pounds ×* * *2* feet* *= 80 Ft-Lbs* (2)
where, *Ft-Lbs* is a unit of *work* which denotes *pounds* of force moving through *feet* of distance.
Now that we’ve calculated the* ***work output****,** we’ll turn our attention to the previously-derived equation for *work input**,* shown in equation (3). Interrelating equations for *WO* and *WI* will enable us to solve for unknown variables, including the force, *F,* required to lift the urn and the length of rope, *d*_{2}, extracted during lifting.** **Once *F* and *d*_{2} are known, we can solve for the additional force required to overcome friction, *F*_{F, }then finally we’ll solve for *WI*.
Once again, the equation we’ll be working with is,
*WI = *(*F ×* *d*_{2}) *+ *(*F*_{F} *×* *d*_{2}) (3)
To calculate *F*, we’ll work the two terms present within parentheses separately, then use knowledge gained to further work our way towards a numerical *comparison* of *work input* and *work output*. We’ll do that next time.
Copyright 2017 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: compound pulley, engineering analysis, friction, weight, work input, work output

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January 28th, 2017
In our blog series on *pulleys* we’ve been discussing the effects of *friction,* subjects also studied by Leonardo *da Vinci,* a *historical* figure whose genius contributed so much to the worlds of art, engineering, and science. The *tribometre* shown in his sketch here is one of history’s earliest recorded attempts to understand the phenomenon of *friction*. Tribology, according to the Merriam-Webster Dictionary, is “a study that deals with the design, friction, wear, and lubrication of interacting surfaces in relative motion.” Depicted in *da Vinci’s* sketch are what appear to be *pulleys *from which dangle objects in mid-air.
__da Vinci’s Tribometre; a Historical Look at Pulleys and Friction__
Copyright 2017 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: engineering, friction, Leonardo Da Vinci, pulleys, tribometre

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January 16th, 2017
We left off last time with an engineering analysis of energy factors within a compound pulley scenario, in our case a Grecian man lifting an urn. We devised an equation to quantify the amount of work effort he exerts in the process. That equation contains two terms, one of which is beneficial to our lifting scenario, the other of which is not. Today we’ll explore these two terms and in so doing show how there are situations when* work input does not equal work output.*
__Work Input Does Not Equal Work Output __
Here again is the equation we’ll be working with today,
*WI = *(*F ×* *d*) *+ *(*F*_{F} *×* *d*) (1)
where, *F* is the entirely positive force, or *work,* exerted by human or machine to lift an object using a compound pulley. It represents an ideal but not real world scenario in which no friction is present within the pulley assembly.
The other force at play in our lifting scenario, *F*_{F,} is less obvious to the casual observer. It’s the force, or *work,* which must be employed over and above the initial positive force to overcome the friction that’s always present between moving parts, in this case a rope moving through pulley wheels. The rope length extracted from the pulley to lift the object is *d*.
Now we’ll use this equation to understand why *work input*, *WI,* *does not equal* *work output*, *WO*, in a compound pulley arrangement where friction is present.
The first term in equation (1), (*F ×* *d*), represents the *work input* as supplied by human or machine to lift the object. It is an idealistic scenario in which 100% of energy employed is directly conveyed to lifting. Stated another way, (*F ×* *d*) is entirely converted into beneficial *work* effort, *WO*.
The second term, (*F*_{F} *×* *d*), is the additional *work input* that’s needed to overcome frictional resistance present in the interaction between rope and pulley wheels. It represents lost *work* effort and makes no contribution to lifting the urn off the ground against the pull of gravity. It represents the heat energy that’s created by the movement of rope through the pulley wheels, heat which is entirely lost to the environment and contributes nothing to *work output*. Mathematically, this relationship between *WO, WI,* and friction is represented by,
*WO = WI –* (*F*_{F} *×* *d*) (2)
In other words, *work input* is *not equal* to *work output* in a real world situation in which pulley wheels present a source of friction.
Next time we’ll run some numbers to demonstrate the inequality between *WI* and *WO*.
Copyright 2017 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: compound pulley, engineering, friction, heat energy, pulley, work input, work output

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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 *friction*-free. 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 *friction*-free 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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December 19th, 2016
Tags: engineering expert, engineering expert witness, pulleys

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December 13th, 2016
The presence of *friction* in *mechanical *designs is as guaranteed as conflict in a good movie, and engineers inevitably must deal with the conflicts *friction* produces within their *mechanical* designs. But unlike a good movie, where conflict presents a positive, engaging force, *friction’s *presence in *pulleys* results only in impediment, wasting energy and *reducing* *mechanical advantage*. We’ll investigate the math behind this phenomenon in today’s blog.
__Friction Reduces Pulleys’ Mechanical Advantage__
A few blogs back we performed a work input-output analysis of an idealized situation in which no *friction* is present in a compound pulley. The analysis yielded this equation for *mechanical advantage*,
*MA = d*_{2} ÷* d*_{1} (1)
where *d*_{2} is the is the length of rope Mr. Toga extracts from the *pulley* in order to lift his urn a distance *d*_{1} above the ground. Engineers refer to this idealized frictionless scenario as an *ideal mechanical advantage*, *IMA*, so equation (1) becomes,
*IMA = d*_{2} ÷* d*_{1} (2)
We also learned that in the idealized situation *mechanical advantage* is the ratio of the urn’s weight force, *W,* to the force exerted by Mr. Toga, *F,* as shown in the following equation. See our past blog for a refresher on how this ratio is developed.
*IMA = W ÷** F* (3)
In reality, friction exists between a *pulley’s* moving parts, namely, its wheels and the rope threaded through them. In fact, the more *pulleys* we add, the more *friction *increases.
The actual amount of lifting force required to lift an object is a combination of *F*_{F }, the friction-filled force, and *F*, the idealized friction-free force. The result is *F*_{Actual} as shown here,
*F*_{Actual} = *F + F*_{F} (4)
The real world scenario in which *friction* is present is known within the engineering profession as *actual mechanical advantage*, *AMA, *which is equal to,
*AMA = W ÷** F*_{Actual} (5)
To see how *AMA* is affected by *friction* force *F*_{F}, let’s substitute equation (4) into equation (5),
*AMA = W ÷** *(*F + F*_{F}) (6)
With the presence of *F*_{F} in equation (6), *W* gets divided by the sum of *F *and *F*_{F} . This results in a smaller number than *IMA,* which was computed in equation (3). In other words, *friction* *reduces* the actual* ***mechanical advantage** of the compound *pulley*.
Next time we’ll see how the presence of *F*_{F} translates into lost work effort in the compound *pulley,* thus creating an inequality between the work input, *WI *and work output *WO*.
Copyright 2016 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: actual mechanical advantage, AMA, compound pulley, engineering, friction, friction force, ideal mechanical advantage, IMA, mechanical advantage, mechanical design, pulley, pulley friction, pulley work input, pulley work output, weight force

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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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