Force

The Weirdness of Empty Space – Casimir Force

(This is an excerpt from The Book of Forces – enjoy!)

The forces we have discussed so far are well-understood by the scientific community and are commonly featured in introductory as well as advanced physics books. In this section we will turn to a more exotic and mysterious interaction: the Casimir force. After a series of complex quantum mechanical calculations, the Dutch physicist Hendrick Casimir predicted its existence in 1948. However, detecting the interaction proved to be an enormous challenge as this required sensors capable picking up forces in the order of 10^(-15) N and smaller. It wasn’t until 1996 that this technology became available and the existence of the Casimir force was experimentally confirmed.

So what does the Casimir force do? When you place an uncharged, conducting plate at a small distance to an identical plate, the Casimir force will pull them towards each other. The term “conductive” refers to the ability of a material to conduct electricity. For the force it plays no role though whether the plates are actually transporting electricity in a given moment or not, what counts is their ability to do so.

The existence of the force can only be explained via quantum theory. Classical physics considers the vacuum to be empty – no particles, no waves, no forces, just absolute nothingness. However, with the rise of quantum mechanics, scientists realized that this is just a crude approximation of reality. The vacuum is actually filled with an ocean of so-called virtual particles (don’t let the name fool you, they are real). These particles are constantly produced in pairs and annihilate after a very short period of time. Each particle carries a certain amount of energy that depends on its wavelength: the lower the wavelength, the higher the energy of the particle. In theory, there’s no upper limit for the energy such a particle can have when spontaneously coming into existence.

So how does this relate to the Casimir force? The two conducting plates define a boundary in space. They separate the space of finite extension between the plates from the (for all practical purposes) infinite space outside them. Only particles with wavelengths that are a multiple of the distance between the plates fit in the finite space, meaning that the particle density (and thus energy density) in the space between the plates is smaller than the energy density in the pure vacuum surrounding them. This imbalance in energy density gives rise to the Casimir force. In informal terms, the Casimir force is the push of the energy-rich vacuum on the energy-deficient space between the plates.

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(Illustration of Casimir force)

It gets even more puzzling though. The nature of the Casimir force depends on the geometry of the plates. If you replace the flat plates by hemispherical shells, the Casimir force suddenly becomes repulsive, meaning that this specific geometry somehow manages to increase the energy density of the enclosed vacuum. Now the even more energy-rich finite space pushes on the surrounding infinite vacuum. Trippy, huh? So which shapes lead to attraction and which lead to repulsion? Unfortunately, there is no intuitive way to decide. Only abstract mathematical calculations and sophisticated experiments can provide an answer.

We can use the following formula to calculate the magnitude of the attractive Casimir force FCS between two flat plates. Its value depends solely on the distance d (in m) between the plates and the area A (in m²) of one plate. The letters h = 6.63·10^(-34) m² kg/s and c = 3.00·10^8 m/s represent Plank’s constant and the speed of light.

FCS = π·h·c·A / (480·d^4) ≈ 1.3·10^(-27)·A / d^4

(The sign ^ stands for “to the power”) Note that because of the exponent, the strength of the force goes down very rapidly with increasing distance. If you double the size of the gap between the plates, the magnitude of the force reduces to 1/2^4 = 1/16 of its original value. And if you triple the distance, it goes down to 1/3^4 = 1/81 of its original value. This strong dependence on distance and the presence of Plank’s constant as a factor cause the Casimir force to be extremely weak in most real-world situations.

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Example 24:

a) Calculate the magnitude of the Casimir force experienced by two conducting plates having an area A = 1 m² each and distance d = 0.001 m (one millimeter). Compare this to their mutual gravitational attraction given the mass m = 5 kg of one plate.

b) How close do the plates need to be for the Casimir force to be in the order of unity? Set FCS = 1 N.

Solution:

a)

Inserting the given values into the formula for the Casimir force leads to (units not included):

FCS = 1.3·10^(-27)·A/d^4
FCS = 1.3·10^(-27) · 1 / 0.0014
FCS ≈ 1.3·10^(-15) N

Their gravitational attraction is:

FG = G·m·M / r²
FG = 6.67·10^(-11)·5·5 / 0.001²
FG ≈ 0.0017 N

This is more than a trillion times the magnitude of the Casimir force – no wonder this exotic force went undetected for so long.  I should mention though that the gravitational force calculated above should only be regarded as a rough approximation as Newton’s law of gravitation is tailored to two attracting spheres, not two attracting plates.

b)

Setting up an equation we get:

FCS = 1.3·10^(-27)·A/d^4
1 = 1.3·10^(-27) · 1 / d^4

Multiply by d4:

d4 = 1.3·10^(-27)

And apply the fourth root:

d ≈ 2·10^(-7) m = 200 nanometers

This is roughly the size of a common virus and just a bit longer than the wavelength of violet light.

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The existence of the Casimir force provides an impressive proof that the abstract mathematics of quantum mechanics is able to accurately describe the workings of the small-scale universe. However, many open questions remain. Quantum theory predicts the energy density of the vacuum to be infinitely large. According to Einstein’s theory of gravitation, such a concentration of energy would produce an infinite space-time curvature and if this were the case, we wouldn’t exist. So either we don’t exist (which I’m pretty sure is not the case) or the most powerful theories in physics are at odds when it comes to the vacuum.

All about the Gravitational Force (For Beginners)

(This is an excerpt from The Book of Forces)

All objects exert a gravitational pull on all other objects. The Earth pulls you towards its center and you pull the Earth towards your center. Your car pulls you towards its center and you pull your car towards your center (of course in this case the forces involved are much smaller, but they are there). It is this force that invisibly tethers the Moon to Earth, the Earth to the Sun, the Sun to the Milky Way Galaxy and the Milky Way Galaxy to its local galaxy cluster.

Experiments have shown that the magnitude of the gravitational attraction between two bodies depends on their masses. If you double the mass of one of the bodies, the gravitational force doubles as well. The force also depends on the distance between the bodies. More distance means less gravitational pull. To be specific, the gravitational force obeys an inverse-square law. If you double the distance, the pull reduces to 1/2² = 1/4 of its original value. If you triple the distance, it goes down to 1/3² = 1/9 of its original value. And so on. These dependencies can be summarized in this neat formula:

F = G·m·M / r²

With F being the gravitational force in Newtons, m and M the masses of the two bodies in kilograms, r the center-to-center distance between the bodies in meters and G = 6.67·10^(-11) N m² kg^(-2) the (somewhat cumbersome) gravitational constant. With this great formula, that has first been derived at the end of the seventeenth century and has sparked an ugly plagiarism dispute between Newton and Hooke, you can calculate the gravitational pull between two objects for any situation.

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(Gravitational attraction between two spherical masses)

If you have trouble applying the formula on your own or just want to play around with it a bit, check out the free web applet Newton’s Law of Gravity Calculator that can be found on the website of the UNL astronomy education group. It allows you to set the required inputs (the masses and the center-to-center distance) using sliders that are marked special values such as Earth’s mass or the distance Earth-Moon and calculates the gravitational force for you.

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Example 3:

Calculate the gravitational force a person of mass m = 72 kg experiences at the surface of Earth. The mass of Earth is M = 5.97·10^24 kg (the sign ^ stands for “to the power”) and the distance from the center to the surface r = 6,370,000 m. Using this, show that the acceleration the person experiences in free fall is roughly 10 m/s².

Solution:

To arrive at the answer, we simply insert all the given inputs into the formula for calculating gravitational force.

F = G·m·M / r²
F = 6.67·10^(-11)·72·5.97·10^24 / 6,370,000² N ≈ 707 N

So the magnitude of the gravitational force experienced by the m = 72 kg person is 707 N. In free fall, he or she is driven by this net force (assuming that we can neglect air resistance). Using Newton’s second law we get the following value for the free fall acceleration:

F = m·a
707 N = 72 kg · a

Divide both sides by 72 kg:

a = 707 / 72 m/s² ≈ 9.82 m/s²

Which is roughly (and more exact than) the 10 m/s² we’ve been using in the introduction. Except for the overly small and large numbers involved, calculating gravitational pull is actually quite straight-forward.

As mentioned before, gravitation is not a one-way street. As the Earth pulls on the person, the person pulls on the Earth with the same force (707 N). However, Earth’s mass is considerably larger and hence the acceleration it experiences much smaller. Using Newton’s second law again and the value M = 5.97·1024 kg for the mass of Earth we get:

F = m·a
707 N = 5.97·10^24 kg · a

Divide both sides by 5.97·10^24 kg:

a = 707 / (5.97·10^24) m/s² ≈ 1.18·10^(-22) m/s²

So indeed the acceleration the Earth experiences as a result of the gravitational attraction to the person is tiny.

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Example 4:

By how much does the gravitational pull change when the  person of mass m = 72 kg is in a plane (altitude 10 km = 10,000 m) instead of the surface of Earth? For the mass and radius of Earth, use the values from the previous example.

Solution:

In this case the center-to-center distance r between the bodies is a bit larger. To be specific, it is the sum of the radius of Earth 6,370,000 m and the height above the surface 10,000 m:

r = 6,370,000 m + 10,000 m = 6,380,000 m

Again we insert everything:

F = G·m·M / r²
F = 6.67·10^(-11)·72·5.97·10^24 / 6,380,000² N ≈ 705 N

So the gravitational force does not change by much (only by 0.3 %) when in a plane. 10 km altitude are not much by gravity’s standards, the height above the surface needs to be much larger for a noticeable difference to occur.

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With the gravitational law we can easily show that the gravitational acceleration experienced by an object in free fall does not depend on its mass. All objects are subject to the same 10 m/s² acceleration near the surface of Earth. Suppose we denote the mass of an object by m and the mass of Earth by M. The center-to-center distance between the two is r, the radius of Earth. We can then insert all these values into our formula to find the value of the gravitational force:

F = G·m·M / r²

Once calculated, we can turn to Newton’s second law to find the acceleration a the object experiences in free fall. Using F = m·a and dividing both sides by m we find that:

a = F / m = G·M / r²

So the gravitational acceleration indeed depends only on the mass and radius of Earth, but not the object’s mass. In free fall, a feather is subject to the same 10 m/s² acceleration as a stone. But wait, doesn’t that contradict our experience? Doesn’t a stone fall much faster than a feather? It sure does, but this is only due to the presence of air resistance. Initially, both are accelerated at the same rate. But while the stone hardly feels the effects of air resistance, the feather is almost immediately slowed down by the collisions with air molecules. If you dropped both in a vacuum tube, where no air resistance can build up, the stone and the feather would reach the ground at the same time! Check out an online video that shows this interesting vacuum tube experiment, it is quite enlightening to see a feather literally drop like a stone.

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(All bodies are subject to the same gravitational acceleration)

Since all objects experience the same acceleration near the surface of Earth and since this is where the everyday action takes place, it pays to have a simplified equation at hand for this special case. Denoting the gravitational acceleration by g (with g ≈ 10 m/s²) as is commonly done, we can calculate the gravitational force, also called weight, an object of mass m is subject to at the surface of Earth by:

F = m·g

So it’s as simple as multiplying the mass by ten. Depending on the application, you can also use the more accurate factor g ≈ 9.82 m/s² (which I will not do in this book). Up to now we’ve only been dealing with gravitation near the surface of Earth, but of course the formula allows us to compute the gravitational force and acceleration near any other celestial body. I will spare you trouble of looking up the relevant data and do the tedious calculations for you. In the table below you can see what gravitational force and acceleration a person of mass m = 72 kg would experience at the surface of various celestial objects. The acceleration is listed in g’s, with 1 g being equal to the free-fall acceleration experienced near the surface of Earth.

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So while jumping on the Moon would feel like slow motion (the free-fall acceleration experienced is comparable to what you feel when stepping on the gas pedal in a common car), you could hardly stand upright on Jupiter as your muscles would have to support more than twice your weight. Imagine that! On the Sun it would be even worse. Assuming you find a way not get instantly killed by the hellish thermonuclear inferno, the enormous gravitational force would feel like having a car on top of you. And unlike temperature or pressure, shielding yourself against gravity is not possible.

What about the final entry? What is a neutron star and why does it have such a mind-blowing gravitational pull? A neutron star is the remnant of a massive star that has burned its fuel and exploded in a supernova, no doubt the most spectacular light-show in the universe. Such remnants are extremely dense – the mass of several suns compressed into an almost perfect sphere of just 20 km radius. With the mass being so large and the distance from the surface to the center so small, the gravitational force on the surface is gigantic and not survivable under any circumstances.

If you approached a neutron star, the gravitational pull would actually kill you long before reaching the surface in a process called spaghettification. This unusual term, made popular by the brilliant physicist Stephen Hawking, refers to the fact that in intense gravitational fields objects are vertically stretched and horizontally compressed. The explanation is rather straight-forward: since the strength of the gravitational force depends on the distance to the source of said force, one side of the approaching object, the side closer to the source, will experience a stronger pull than the opposite side. This leads to a net force stretching the object. If the gravitational force is large enough, this would make any object look like a thin spaghetti. For humans spaghettification would be lethal as the stretching would cause the body to break apart at the weakest spot (which presumably is just above the hips). So my pro-tip is to keep a polite distance from neutron stars.

The Difference Between Mass and Weight

In general, it is acceptable to use weight as a synonym for mass. However, in a very strict physical sense this is incorrect. Weight is the gravitational force experienced by an object and accordingly measured in Newtons and not kilograms. An object of mass m has the weight F:

F = m · g

with the gravitational acceleration g. On Earth the value of the gravitational acceleration at the surface is g = 9.81 m/s². So a typical adult with a mass of m = 75 kg has a weight of:

F = 75 kg · 9.81 m/s² = 735.75 N

On the moon (or any other point of the universe), the mass would remain at m = 75 kg. But since the gravitational acceleration on the moon is much lower (g = 1.62 m/s²), the weight changes to:

F = 75 kg · 1.62 m/s² = 121.5 N

Keep this distinction in mind. Mass is a fundamental property of an object that does not depend on the conditions outside the object, while weight is a variable that changes with the strength of surrounding gravitational field.

(This was an excerpt from Physics! In Quantities and Examples)

Released Today for Kindle: Physics! In Quantities and Examples

I finally finished and released my new ebook … took me longer than usual because I always kept finding new interesting topics while researching. Here’s the blurb, link and TOC:

This book is a concept-focused and informal introduction to the field of physics that can be enjoyed without any prior knowledge. Step by step and using many examples and illustrations, the most important quantities in physics are gently explained. From length and mass, over energy and power, all the way to voltage and magnetic flux. The mathematics in the book is strictly limited to basic high school algebra to allow anyone to get in and to assure that the focus always remains on the core physical concepts.

(Click cover to get to the Amazon Product Page)

cover

Table of Contents:

Length
(Introduction, From the Smallest to the Largest, Wavelength)

Mass
(Introduction, Mass versus Weight, From the Smallest to the Largest, Mass Defect and Einstein, Jeans Mass)

Speed / Velocity
(Introduction, From the Smallest to the Largest, Faster than Light, Speed of Sound for all Purposes)

Acceleration
(Introduction, From the Smallest to the Largest, Car Performance, Accident Investigation)

Force
(Introduction, Thrust and the Space Shuttle, Force of Light and Solar Sails, MoND and Dark Matter, Artificial Gravity and Centrifugal Force, Why do Airplanes Fly?)

Area
(Introduction, Surface Area and Heat, Projected Area and Planetary Temperature)

Pressure
(Introduction, From the Smallest to the Largest, Hydraulic Press, Air Pressure, Magdeburg Hemispheres)

Volume
(Introduction, Poisson’s Ratio)

Density
(Introduction, From the Smallest to the Largest, Bulk Density, Water Anomaly, More Densities)

Temperature
(Introduction, From the Smallest to the Largest, Thermal Expansion, Boiling, Evaporation is Cool, Why Blankets Work, Cricket Temperature)

Energy
(Introduction, Impact Speed, Ice Skating, Dear Radioactive Ladies and Gentlemen!, Space Shuttle Reentry, Radiation Exposure)

Power
(Introduction, From the Smallest to the Largest, Space Shuttle Launch and Sound Suppression)

Intensity
(Introduction, Inverse Square Law, Absorption)

Momentum
(Introduction, Perfectly Inelastic Collisions, Recoil, Hollywood and Physics, Force Revisited)

Frequency / Period
(Introduction, Heart Beat, Neutron Stars, Gravitational Redshift)

Rotational Motion
(Extended Introduction, Moment of Inertia – The Concept, Moment of Inertia – The Computation, Conservation of Angular Momentum)

Electricity
(Extended Introduction, Stewart-Tolman Effect, Piezoelectricity, Lightning)

Magnetism
(Extended Introduction, Lorentz Force, Mass Spectrometers, MHD Generators, Earth’s Magnetic Field)

Appendix:
Scalar and Vector Quantities
Measuring Quantities
Unit Conversion
Unit Prefixes
References
Copyright and Disclaimer

As always, I discounted the book in countries with a low GDP because I think that education should be accessible for all people. Enjoy!

Car Dynamics – Sliding and Overturning

In this post we will take a look at car performance in curves. Of central importance for our considerations is the centrifugal force. Whenever a body is moving in a curved path, this force comes into play. You probably felt it many times in your car. It is the force that tries to push you out of a curve as you go through it.

The centrifugal force C (in N) depends on three factors: the velocity v (in m/s) of the car, its mass m (in kg) and the radius r (in m) of the curve. Given these quantities, we can easily compute the centrifugal force using this formula:

C = m · v² / r

Note the quadratic dependence on speed. If you double the car’s speed, the centrifugal force quadruples. With this force acting, there must be a counter-force to cancel it for the car not to slide. This force is provided by the sideways friction of the tires. The frictional force F (in N) can be calculated from the so called coefficient of friction μ (dimensionless), the car mass m and the gravitational acceleration g (in m/s²).

F = μ · m · g

The coefficient of friction depends mainly on the road type and condition. On dry asphalt we can set μ ≈ 0.8, on wet asphalt μ ≈ 0.6, on snow μ ≈ 0.2 and on ice μ ≈ 0.1. At low speeds the frictional force exceeds the centrifugal force and the car will be able to go through the curve without any problems. However, as we increase the velocity, so does the centrifugal force and at a certain critical velocity the forces cancel each other out. Any increase in speed from this point on will result in the car sliding.

We can compute the critical speed s (in m/s) by equating the expressions for the forces:

m · s² / r = μ · m · g

s = sqrt (μ · r · g)

This is the speed at which the car begins to slide. Note that there’s no dependence on mass anymore. Since both the centrifugal as well as the frictional force grow proportionally to the car’s mass, it doesn’t play a role in determining the critical speed for sliding. All that’s left in terms of variables is the coefficient of friction (lower friction, lower critical speed) and the radius of the curve (smaller radius, more narrow curve, smaller critical speed).

However, sliding is not the only problem that can occur in curves. Under certain circumstances a car can also overturn. Again the centrifugal force is the culprit. Assuming the center of gravity (in short: CG) of the car is at a height of h (in m), the centrifugal force will produce a torque T acting to overturn the car:

T = h · C = m · v² · h / r

On the other hand, there’s the weight of the car giving rise to an opposing torque T’ that grows with the width w (in m) and mass m of the car:

T’ = 0.5 · m · g · w

At low speeds, the torque caused by the centrifugal force will be lower than the one caused by the gravitational pull. But at a certain critical speed o (in m/s), the torques will cancel each other and any further increase in speed will result in the car overturning. Equating the above expressions, we get:

m · o² · h / r = 0.5 · m · g · w

o = sqrt (0.5 · r · g · w / h)

Aside from the coefficient of friction, the determining factor here is the ratio of width to height. The larger it is, the harder it will be for the centrifugal force to overturn the car. This is why lowering a car when intending to go fast makes sense. If you lower the CG while keeping the width the same, the ratio w / h, and thus the critical speed for overturning, will increase.

Let’s look at some examples before drawing a final conclusion from these truly great formulas.

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According to caranddriver.com the center of gravity of a 2014 BMW 435i is h = 0.5 m above the ground. The width of the car is about w = 1.8 m. Calculate the critical speed for sliding and overturning in a curve of radius r = 300 m on a dry asphalt road (μ ≈ 0.8).

Nothing to do but to apply the formulas:

s = sqrt (0.8 · 300 m · 9.81 m/s²)

s ≈ 49 m/s ≈ 175 km/h ≈ 108 mph

So with normal driving behavior you certainly won’t get anywhere near sliding. But note that sudden steering in a curve can cause the radius of the your car’s path to be considerably lower than the actual curve radius.

Onto the critical overturning speed:

o = sqrt (0.5 · 300 m · 9.81 m/s² · 3.6)

o ≈ 73 m/s ≈ 262 km/h ≈ 162 mph

Not even Michael Schumacher could bring this car to overturn.

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How would the critical speeds change if we drove the 2014 BMW 435i through the same curve on an icy road? In this case the coefficient is considerably lower (μ ≈ 0.1). For the critical sliding speed we get:

s = sqrt (0.1 · 300 m · 9.81 m/s²)

s ≈ 17 m/s ≈ 62 km/h ≈ 38 mph

So even this sweet sport car is in danger of sliding relatively quickly under these conditions. What about the overturning speed? Well, it has nothing to do with the friction of the tires, so it will still be at 73 m/s.

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This was an excerpt from More Great Formulas Explained. Interested in more car dynamics? Take a look at my post on How to Compute Maximum Car Speed. For other interesting physics articles, check out my BEST OF. I hope you enjoyed and drive safe!

Law Of The Lever – Explanation and Examples

Imagine a beam sitting on a fulcrum. We apply one force F'(1) = 20 N on the left side at a distance of r(1) = 0.1 m from the fulcrum and another force F'(2) = 5 N on the right side at a distance of r(2) = 0.2 m. In which direction, clockwise or anti-clockwise, will the beam move?

Great Formulas II_html_m2eb595ec

(Before reading on, please make sure that you understand the concept of torque)

To find that out we can take a look at the corresponding torques. The torque on the left side is:

T(1) = 0.1 m · 20 N = 2 Nm

For the right side we get:

T(2) = 0.2 · 5 N = 1 Nm

So the rotational push caused by force 1 (left side) exceeds that of force 2 (right side). Hence, the beam will turn anti-clockwise. If we don’t want that to happen and instead want to achieve equilibrium, we need to increase force 2 to F'(2) = 10 N. In this case the torques would be equal and the opposite rotational pushes would cancel each other. So in general, this equation needs to be satisfied to achieve a state of equilibrium:

r(1) · F'(1) = r(2) · F'(2)

This is the law of the lever in its simplest form. Let’s see how and where we can apply it.

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A great example for the usefulness of the law of the lever is provided by cranes. On one side, let’s set r(1) = 30 m, it lifts objects. Since we don’t want it to fall over, we stabilize the crane using a 20,000 kg concrete block at a distance of r(2) = 2 m from the axis. What is the maximum mass we can lift with this crane?

First we need to compute the gravitational force of the concrete block.

F'(2) = 20,000 kg · 9.81 m/s² = 196,200 N

Now we can use the law of the lever to find out what maximum force we can apply on the opposite site:

r(1) · F'(1) = r(2) · F'(2)

30 m · F'(1) = 2 m · 196,200 N

30 m · F'(1) = 392,400 Nm

Divide by 30 m:

F'(1) = 13,080 N

As long as we don’t exceed this, the torque caused by the concrete block will exceed that of the lifted object and the crane will not fall over. The maximum mass we can lift is now easy to find. We use the formula for the gravitational force one more time:

13,080 N = m · 9.81 m/s²

Divide by 9.81:

m ≈ 1330 kg

To lift even heavier objects, we need to use either a heavier concrete block or put it at a larger distance from the axis.

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The law of the lever shows why we can interpret a lever as a tool to amplify forces. Suppose you want use a force of F'(1) = 100 N to lift a heavy object with the gravitational pull F'(2) = 2000 N. Not possible you say? With a lever you can do this by applying the smaller force at a larger distance to the axis and the larger force at a shorter distance.

Suppose the heavy object sits at a distance r(2) = 0.1 m to the axis. At what what distance r(1) should we apply the 100 N to be able to lift it? We can use the law of the lever to find the minimum distance required.

r(1) · 100 N = 0.1 m · 2000 N

r(1) · 100 N = 200 Nm

r(1) = 2 m

So as long as we apply the force at a distance of over 2 m, we can lift the object. We effectively amplified the force by a factor of 20. Scientists believe that the principle of force amplification using levers was already used by the Egyptians to build the pyramids. Given a long enough lever, we could lift basically anything even with a moderate force.

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This was an excerpt from More Great Formulas Explained.

Check out my BEST OF for more interesting physics articles.

What is Torque? – A Short and Simple Explanation

Often times when doing physics we simply say “a force is acting on a body” without specifying which point of the body it is acting on. This is basically point-mass physics. We ignore the fact that the object has a complex three-dimensional shape and assume it to be a single point having a certain mass. Sometimes this is sufficient, other times we need to go beyond that. And this is where the concept of torque comes in.

Let’s define what is meant by torque. Assume a force F (in N) is acting on a body at a distance r (in m) from the axis of rotation. This distance is called the lever arm. Take a look at the image below for an example of such a set up.

Great Formulas II_html_m7f65e4ec

(Taken from sdsu-physics.org)

Relevant for the rotation of the body is only the force component perpendicular to the lever arm, which we will denote by F’. If given the angle Φ between the force and the lever arm (as shown in the image), we can easily compute the relevant force component by:

F’ = F · sin(Φ)

For example, if the total force is F = 50 N and it acts at an angle of Φ = 45° to the lever arm, only the the component F’ = 50 N · sin(45°) ≈ 35 N will work to rotate the body. So you can see that sometimes it makes sense to break a force down into its components. But this shouldn’t be cause for any worries, with the above formula it can be done quickly and painlessly.

With this out of the way, we can define what torque is in one simple sentence: Torque T (in Nm) is the product of the lever arm r and the force F’ acting perpendicular to it. In form of an equation the definition looks like this:

T = r · F’

In quantitative terms we can interpret torque as a measure of rotational push. If there’s a force acting at a large distance from the axis of rotation, the rotational push will be strong. However, if one and the same force is acting very close to said axis, we will see hardly any rotation. So when it comes to rotation, force is just one part of the picture. We also need to take into consideration where the force is applied.

Let’s compute a few values before going to the extremely useful law of the lever.

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We’ll have a look at the wrench from the image. Suppose the wrench is r = 0.2 m long. What’s the resulting torque when applying a force of F = 80 N at an angle of Φ = 70° relative to the lever arm?

To answer the question, we first need to find the component of the force perpendicular to the lever arm.

F’ = 80 N · sin(70°) ≈ 75.18 N

Now onto the torque:

T = 0.2 m · 75.18 N ≈ 15.04 Nm

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If this amount of torque is not sufficient to turn the nut, how could we increase that? Well, we could increase the force F and at the same time make sure that it is applied at a 90° angle to the wrench. Let’s assume that as a measure of last resort, you apply the force by standing on the wrench. Then the force perpendicular to the lever arm is just your gravitational pull:

F’ = F = m · g

Assuming a mass of m = 75 kg, we get:

F’ = 75 kg · 9.81 m/s² = 735.75 N

With this not very elegant, but certainly effective technique, we are able to increase the torque to:

T = 0.2 m · 735.75 N = 147.15 Nm

That should do the trick. If it doesn’t, there’s still one option left and that is using a longer wrench. With a longer wrench you can apply the force at a greater distance to the axis of rotation. And with r increased, the torque T is increased by the same factor.

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This was an excerpt from my Kindle ebook More Great Formulas Explained.

Check out my BEST OF for more interesting physics articles.

Acceleration – A Short and Simple Explanation

The three basic quantities used in kinematics are distance, velocity and acceleration. Let’s first look at velocity before moving on to the main topic. The velocity is simply the rate of change in distance. If we cover the distance d in a time span t, than the average velocity during this interval is:

v = d / t

So if we drive d = 800 meters in t = 40 seconds, the average speed is v = 800 meters / 40 seconds = 20 m/s. No surprise here. Note that there are many different units commonly used for velocity: kilometers per hour, feet per second, miles per hour, etc … The SI unit is m/s, so unless otherwise stated, you have to input the velocity in m/s into a formula to get a correct result.

Acceleration is also defined as the rate of change, but this time with respect to velocity. If the velocity changes by the amount v in a time span t, the average acceleration is:

a = v / t

For example, my beloved Mercedes C-180 Compressor can go from 0 to 100 kilometers per hour (or 27.8 meters per second) in about 9 seconds. So the average acceleration during this time is:

a = 27.8 meters per second / 9 seconds = 3.1 m/s²

Is that a lot? Obviously we should know some reference values to be able to judge acceleration.

The one value you should know is: g = 9.81 m/s². This is the acceleration experienced in free fall. And you can take the word “experienced” literally because unlike velocity, we really do feel acceleration. Our inner ear system contains structures that enable us to perceive it. Often times acceleration is compared to this value because it provides a meaningful and easily relatable reference value.

So the acceleration in the Mercedes C-180 Compressor is not quite as thrilling as free fall, it only accelerates with about 3.1 / 9.81 = 0.32 g. How much higher can it go for production cars? Well, meet the Bugatti Veyron Super Sport. It goes from 0 to 100 kilometers per hour (or 27.8 meters per second) in 2.2 seconds. This translates into an acceleration of:

a = 27.8 meters per second / 2.2 seconds = 12.6 m/s²

This is more than the free fall acceleration! To be more specific, it’s 12.6 / 9.81 = 1.28 g. If you got $ 4,000,000 to spare, how about getting one of these? But even this is nothing compared to what astronauts have to endure during launch. Here you can see a typical acceleration profile of a Space Shuttle launch:

(Taken from http://www.russellwestbrook.com)

Right before the main engine shutoff the acceleration peaks at close to 30 m/s² or 3 g. That’s certainly not for everyone. How much can a person endure by the way? According to “Aerospace Medicine” accelerations of around 5 g and higher can result in death if sustained for more than a few seconds. Very short acceleration bursts can be survivable up to about 50 g, which is a value that can be reached and exceeded in a car crash.

One more thing to keep in mind about acceleration: it is always a result of a force. If a force F (measured in Newtons = N) acts on a body, it responds by accelerating. The stronger the force is, the higher the resulting acceleration. This is just Newton’s Second Law:

a = F / m

So a force of F = 210 N on a body of m = 70 kg leads to an acceleration of a = 210 N / 70 kg = 3 m/s². The same force however on a m = 140 kg mass only leads to the acceleration a = 210 N / 140 kg = 1.5 m/s². Hence, mass provides resistance to acceleration. You need more force to accelerate a massive body at the same rate as a light body.

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