## Posts Tagged ‘friction’

Monday, March 20th, 2017
They say necessity is the mother of invention, and today’s look at an influential historical figure in engineering bears that out. Last week we introduced Leonhard Euler and touched on his influence to the science of pulleys. Today we’ll introduce his contemporary and partner in science, *Johann Albert Eytelwein*, a German mathematician and visionary, a true *engineering trailblazer* whose contributions to the blossoming discipline of engineering led to later studies with pulleys.
__Johann Albert Eytelwein, Engineering Trailblazer__
*Johann Albert Eytelwein’s* experience as a civil *engineer* in charge of the dikes of former Prussia led him to develop a series of practical mathematical problems that would enable his subordinates to operate more effectively within their government positions. He was a *trailblazer* in the field of applied mechanics and their application to physical structures, such as the dikes he oversaw, and later to machinery. He was instrumental in the founding of Germany’s first university level *engineering* school in 1799, the Berlin Bauakademie, and served as director there while lecturing on many developing *engineering* disciplines of the time, including machine design and hydraulics. He went on to publish in 1801 one of the most influential *engineering* books of his time, entitled *Handbuch der Mechanik* (Handbook of the Mechanic), a seminal work which combined what had previously been mere *engineering *theory into a means of practical application.
Later, in 1808, *Eytelwein* expanded upon this work with his *Handbuch der Statik fester Koerper* (Handbook of Statics of Fixed Bodies), which expanded upon the work of Euler. In it he discusses friction and the use of pulleys in mechanical design. It’s within this book that the famous Euler-*Eytelwein* Formula first appears, a formula *Eytelwein* derived in conjunction with Euler. The formula delves into the usage of belts with pulleys and examines the tension interplay between them.
More on this fundamental foundation to the discipline of *engineering* next time, with a specific focus on pulleys.
Copyright 2017 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: bel, engineering, friction, Johann Albert Eytelwein, mechanical power transmission, pulley, pulleys

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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, 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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Monday, 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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Saturday, 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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Monday, 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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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 *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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Tuesday, 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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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, 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 Work-Energy 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, work-energy theorem

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