Torque: Types of Levers

 Physics Index

Where are we going with this? The information on this page connects to standards such as:

• Gather evidence to defend the claim of Newton's first law of motion by explaining the effect that balanced forces have upon objects that are stationary or are moving at constant velocity.

•Recognize and communicate information about energy efficiency and/or inefficiency of machines used in everyday life.

Torque: Types of Levers

(Hmm… I thought torque was something in car engines!)

Give me a lever long enough and a fulcrum on which to place it, and I shall move the world.

~Archimedes

(As if!)

But, he's not wrong. Mathematically, that is.

Let's dig into this a little!

The concept of levers and torque are directly connected to motion that rotates around a pivot point. This motion can be considered to be clockwise around the point or counterclockwise (anticlockwise).

Every lever system has three parts: the lever, the effort and the load.

And… every lever has two parts. 

We are going to call one of those parts the lever arm. The lever arm is the beam or whatever that is used. More on this after we define other parts.

The second part of the lever is a fulcrum. The fulcrum (sometimes called the pivot) is the thing around which the lever arm rotates.

The effort is a force that is applied to the lever arm at some distance from the fulcrum.

The load is also a force that will have an effect on the lever arm. 

Generally, the effort is acting on the load. 

Circling back to lever arm…

The lever arm is the part of a lever system that sits on and rotates around the fulcrum and to which is applied the effort so that it can act on the load.

The effort is trying to rotate the lever arm around the fulcrum. The load is resisting this motion.

It is incorrect to say that if the effort is greater than the load the lever will rotate as desired. Force alone is not enough to determine what will happen.

What determines the rotation is something called torque.

Torque is the potential to rotate the lever and is the product of the applied force and the distance from the fulcrum.

Caveat: 

• it is ONLY the vector component of the force that is perpendicular to the lever arm. So, sometimes, you have to deconstruct the force to find the perpendicular component. 

Hence, where T is torque, F is force and d is distance from the fulcrum…

T = Fd

Torque adds up in a lever system. Clockwise torque and counterclockwise torque have opposite signs.

If you think of a positive torque as attempting to cause clockwise rotation around the fulcrum, then negative torque will be attempting to cause counterclockwise rotation.

Where T is torque, the net torque will be the signed sum of all the torques acting on the lever, so that…

T(net) = T1 + T2 + T3…

If you say that clockwise is positive…

…if the result is positive, then you have clockwise rotation.

…if the result is negative, then you have counterclockwise rotation.

Now, if you want to complicate the math, you can say that the fulcrum is at distance zero, then measure left as positive and right as negative, then say that forces "up" are negative and "down" are positive, then, plug all those in so that…

    T(net) = F1d1 + F2d2  + F3d3 …

It will work out…

A more "tangible" approach is to keep up logically with the torques as counterclockwise and clockwise. 

Sometimes, you are looking for a solution that results in equilibrium. That is to say that there is either no rotation or that the rotation is occuring at a constant rate.

This would mean that

T(cw) = T(ccw)

In such a case, it's easy and logical to sort the various forces to the appropriate side of the equation.

Types of Levers

Though there are three types of levers, the math on them is essentially the same:

T(net) = T1 + T2 + T3…

The types of levers are categorized according to the configuration of load, effort, and fulcrum.

Class 1 Lever

"For the Class 1 lever the pivot lies between the effort and load. A see saw in a playground is an example of a Class 1 lever where the effort balances the load."

In the above image, if the effort is counterclockwise, then the load is clockwise. So, if the product of effort and distance is greater than the product of load and distance, the lever arm will rotate counterclockwise.

Class 2 Lever

"For the Class 2 lever the load is between the pivot and the effort (like a wheelbarrow). The effort force needed is less than the load force, so there is a mechanical advantage."

In this image, load would be clockwise and effort counterclockwise.  Since torque is distance times force (effort), a small effort would move a larger load.

Class 3 Lever

"For a Class 3 lever the load is further away from the pivot than the effort. There is no mechanical advantage because the effort is greater than the load. However this disadvantage is compensated with a larger movement. This type of lever system also gives us the advantage of a much greater speed of movement."

In this image, load would be clockwise and effort counterclockwise.  Since torque is distance times force (effort), a large effort would be needed to move a smaller load.

See also: https://www.sciencelearn.org.nz/resources/1924-what-levers-does-your-body-use

___________________________

There can be far more complex arrangements involving multiple efforts and loads. However, the principle remains the same:

Where T is torque, the net torque will be the signed sum of all the torques acting on the lever, so that…

T(net) = T1 + T2 + T3…

Where the roadway is considered the lever arm, each support footing is a fulcrum, each cable is an upward force (effort) and the roadway and every car is a load. It is notable that the upward forces only exist in opposition to the downward forces. 

Momentum: Notes and Such

Physics Index

Where are we going with this? The information on this page connects to standards such as:

• Using experimental evidence and investigations, determine that Newton’s second law of motion defines force as a change in momentum, F = Δp/Δt.

• Develop and apply the impulse-momentum theorem along with scientific and engineering ideas to design, evaluate, and refine a device that minimizes the force on an object during a collision (e.g., helmet, seatbelt, parachute).

•  Assess the validity of the law of conservation of linear momentum (p=mv) by planning and constructing a controlled scientific investigation involving two objects moving in one-dimension.

Momentum: Notes and Such

(Wow! Now, that's a fancy title!)

What is momentum? Well… It's not a force… And it's not energy… Basically, it's a thing that helps us understand motion. And collisions.

(You gotta do better than that!)

A more formal definition would say that momentum is the quantity of motion, the product of an object's mass and its velocity, and it can be quantized as

p = mv

  

where is p is momentum, m is mass, and v is velocity.

This means that the units for momentum are

kg•m/s

Momentum is… boring… unless something changes. Just watching a football player run at a constant velocity isn't very interesting.

Where momentum becomes interesting is when something happens. Momentum concepts can then help us describe it.

The first application is when momentum changes… So, you got a thing with some momentum. Then something happens and there is less mass. Or more. Or the velocity changes.

This is sort of an abstract concept that helps us illustrate that (this is important!)…

within any closed system, momentum is conserved. Unless something different is introduced to the system, momentum is a constant.

Hence, within a closed system, any changes are limited to within that system, so the momentum can be thought of as having a first and second state (and 3rd, 4th, etc… if you so desired). So, we can formulize that to say that

p1 = p2

And since p = mv, then it follows that

m1v1 = m2v2

This will lead to questions where WITHIN THE CLOSED SYSTEM the mass or velocity will change (by means that somehow do not violate the "closed system" concept).

A far more realistic application is one in which something affects the moving object and thus causes a change to the momentum. This gives rise questions that examine the m1v1 = m2v2 in momentum from a first condition to a second (and third, fourth, fifth… if so desired).

This can be represented as…

∆p = p2 - p1

∆p =  m2v2  -  m1v1       <--- plug in                (eq. 1

Frequently, this will be seen when the velocity of an object changes. Since it is the same object

m1 = m2

So the formula can be represented as:

∆p = m(v2 - v1)                         (eq. 2

The astute observer will recognize that 

v2 - v1 = ∆v

Leading to the fancified version of equation 2 shown below

∆p = m∆v                         (eq. 3

A less frequent possibility is where mass changes (or when both mass and velocity change). Equation 1 can be a starting point for those situations.

IMPORTANT NOTE

Momentum is the product of velocity and mass, and velocity has a direction, so therefore, momentum has a direction (it is a vector). You will have to attend to the positive and negative values as directions.

Further, (LIKE ALL VECTORS), momentum need not occur on a single axis. Vectors can be deconstructed into orthogonal components, then all of the components can be recombined into resultant vectors.

Momentum In Action—A First Look: Inelastic Collisions

Key to understanding what's going when two (or more) objects collide is to know that, within a closed system momentum is conserved. 

That means that, at any point within the closed system the sum of all the momentums of all the objects in the system stays the same. You can't end up with more momentum than you start with!

So if there are 2 objects, and one has a momentum of 5 kg•m/s and the other has a momentum of 3 kg•m/s the total momentum will always be 8 kg•m/s. If the two objects interact with each other (collide) the distribution of momentum can change, but the total will always be 8.

Collisions

There are two types of collisions. 

In an elastic collision, two things collide and bounce off each other, going off separately after the collision.

In an inelastic collision, two things collide and stick to each other, going off as a single thing after the collision.

Collisions (Both types)

In a system involving a collision, we look at the two (or more) objects as they interact. We look at them in isolation, generally. That is to say, we only consider how they are interacting with each other.

In this case, we have a closed system in which momentum is conserved. That means that, where case 1 is before the collision and case 2 is after, 

pinitial = pfinal            (Eq. 1

Since the total momentum of a system is the momentum of all the objects in it, for any case we can say that

ptot = p1 + p2 + …

Inelastic Collisions

In an inelastic collision, two things collide and stick to each other, going off as a single thing after the collision.

Recalling that momentum is conserved, it follows that the momentum of the two objects before the collision must be equal to the momentum of the two objects after the collision. This…

pinitial = pfinal 

So, if there are 2 objects, object 1 and object 2, then the initial momentum is

pinitial = p1 + p2 

And since momentum is found as

p = mv

we can substitute in for p1 and p2

pinitial = p1 + p2 

pinitial = m1v1 + m2v2  

Now, for inelastic collisions, m1 and m2 become stuck to each other. They go off as one thing, with only one mass and only one velocity.

Okay… so, we have 2 objects and a before (initial) and after (final) situation. We could pile up subscrips such as…

m1v1i + m2v2i …

OR… we can change the symbol for initial velocity and keep things less messy… Let's do that…

 So, what we get is…

let u be initial velocity and v be final velocity

 m1u1 + m2u2 = (m1+ m2)v            (Eq. 3

I like to think of it as…

m3 = m1 + m2 

This makes the the equation

 m1u1 + m2u2 = m3v            (Eq. 3 alt

___________________________

We will look at inelastic collisions after we examine the relationship between force, time, mass, and momentum.
___________________________

Force and Momentum

Another important consideration of momentum is the forces involved in creating momentum change.

This concept assumes that, for some period of time, a force of some magnitude and direction act on an object causing a change in velocity which causes a change in momentum. Let's have a look!

            p = mv            (Fundamental momentum equation)

∆p = m∆v         (Eq. 1

Now, what causes velocity to change? Acceleration! 

∆v = a∆t        (Eq. 2

 

Now, how is force related to acceleration?

F = ma

a = F/m         (Eq. 3

Plug a from Eq. 3 into Eq. 2 we get

∆v = (F/m)∆t        (Eq. 4

Now, plug Eq. 4 into Eq. 1 to get…

∆p = m • F/m • ∆t        

∆p = F∆t      <--- Cancel the m in above               (Eq. 5

One more version…  Substitute Eq. 1 into Eq. 5:

m∆v = F∆t            (Eq. 6

Different situations will call for the use of different equations, so look at what is given and pick accordingly.

Up to this point, we have examined momentum only from the perspective of a single object. However, the interaction of more than one object leads many interesting outcomes. 

This is usually considered collisions

Elastic Collisions

Since momentum for any object is 

p = mv

then

ptot = m1v1 + m2v2 + … (Eq. 2

If we limit our discussion to two objects, we can roll Equation 2 back into Equation 1 and get

pinitial = pfinal 

m1v1i + m2v2i = m1v1f + m2v2f

Woooo… That's a lot of subscripts!

We can make the formula easier on the eyes if we do this…

let u be initial velocity and v be final velocity

Since mass before and after does not change, then:

pinitial = pfinal 

m1u1 + m2u2 = m1v1 + m2v2            (Eq. 3

 

Equation 3 is easy to solve if you have seven of the eight values given. 

Often, though, you need to find one of the final velocities, but know the value of neither. Yikes!

Through a fairly tedious algebra process, you can arrive at the following formulas:

${\displaystyle }$v1=(m1-m2 ) / ( m1+m2) • u1    +    2m2 / (m1+m2) • u2              (Eq. 4

${\displaystyle }$v2=(m2-m1) / (m1+m2) • u2  +    (2m1 / (m1+m2) • u1                    (Eq 5

You will see this in different forms, if you do research. You might see the second arranged like this:

${\displaystyle }$v2= (2m1 / (m1+m2) • u1   +  (m2-m1) / (m1+m2) • u2                    (Eq 5 alt

It's math! You can rearrange things following the… you know… math rules!

Doing a bit more algebra we can get to simpler forms:

v1 =  (m1-m2 ) • u1  +  2m2 • u2

                       (m1+m2)                            (Eq. 6

 

v2 =  (m2-m1 ) • u2  +  2m1 • u1

                       (m1+m2)                            (Eq. 7 

 

Buoyancy: Notes and Such

Physics Index

Where are we going with this? The information on this page connects to standards that create understanding of the forces and interactions between objects is important for describing an object’s motion and determining the stability in a system. Students should understand that forces between objects arise from four types of interactions (gravitational, electromagnetism, and strong and weak nuclear interactions) and that some physical systems are more stable than others.

Buoyancy: Notes and Such

(Okay, this should be interesting!)

Recalling that the formula for buoyancy is 

Fb  = -ρgV

a little discussion is worthwhile. First off, buoyancy is in the opposite direction of gravity. Hence, the negative sign at the front of the equation.

The rho (ρ) stands for density of the fluid. The g is acceleration due to gravity. The V is the volume of the fluid.

So, in essence, the upward force is equal to the weight of the fluid displaced.

If you displace 10 pounds of water (say you are floating a fireproof box), the upward force is 10 pounds. If you diplace 10 pounds of lava, the upward force is 10 pounds. Naturally, 10 pounds of water has a much higher volume than 10 pounds of lava.

Generally, in physics, we work with kilograms, liters, and Newtons.

When a solid object is placed in/on a fluid, it will either sink or float. If the weight of the fluid displaced before it goes under is greater than the object's weight, it will float. If not, it will sink. So, we can say the force of weight (mg) of a floating object is equal to weight of the displaced fluid on which it is floating.

Fweight = Fbuoyancy

mg = -ρgV

Conveniently, both the fluid and the object are affected by the same gravity, so the above relationship is true regarding mass, as well. The mass of a floating object is equal to the mass of the displaced fluid on which it is floating.

When an object sinks, it still displaces some fluid. Thus, there is an upward force of some amount, but it is less than the force of weight. 

If you were to weigh a can of (not diet) soda in the air, then weigh it sunken in water, the weight would be less. It would decrease by an amount equal to the weight of the water displaced. (Plot twist, a lot of diet sodas in a can will float.)

One more thing before we summarize… When the fluid is water, we can enjoy the fact that 1 liter of water weighs nearly, almost exactly 1 kilogram.

Source: https://www.sengpielaudio.com/calculator-milligram.htm

Okay how about that summary?

Floating Objects

Objects float because they displace fluid with a weight equal to theirs before they go below the surface of the fluid.

• The weight of the displaced fluid is equal to the weight of the floating object.
• The mass of the displaced fluid is equal to the mass of the floating object.

If the fluid is water, the volume of the water displaced is equal to the mass of the water displaced.

Sunken Objects

Objects sink because they don't displace enough fluid. So…

• The weight of the displaced fluid is less than the weight of the sunken object.
• The volume of the displaced fluid is equal to the volume of the sunken object.

The "underwater" weight of the sunken object is less than the "above-water" weight of the object. It is reduced by an amount equal to the weight of the displaced fluid.

If the fluid is water, the volume of the water displaced is equal to the mass of the water displaced.

Applied Concepts: Elastic Thing Launcher

 Physics Index

Where are we going with this? The information on this page connects projectile motion with conservation of energy where in potential energy is converted to how far a projectile travels.

Applied Concepts: Elastic Thing Launcher

(Okay, this should be interesting!)

Projectile motion is a fundamental aspect of the study of motion in physics. There are many ways to impart velocity at some angle to a projectile. 

One way is to attach a "pusher" to some sort of spring or elastic band. Pull it back… BAM!

Think of a slingshot… Now, let the cup ride on rails… 

The projectile, a squishy ball, rides in the cup that is pulled back
and released. 

Four quantities are easily measured in this apparatus: The mass of the projectile, the angle of the "thrower," how far back the elastic band was stretched, and how far (distance) that the projectile travels.

From those, everything else can be calculated! Really…

An interesting exercise would be to reverse math back to the coefficient of elasticity in the elastic band (or spring).

Says who?

Stretching the elastic band stores potential energy in the band. And, of course, there is a formula for that!

PE = 1/2kd²

where PE is elastic potential energy, k is the elastic constant, a function of the material, and d is the distance that the spring / band is stretched. 

Often you will see this as

PE = 1/2k∆d²

where ∆d is the stretch. Of course, other notations exist! Of course.

Okay… so… When you release the band, most of the energy stored is converted to kinetic energy. While some of that energy will be in the "pusher" (cup in image above) if the mass of the cup is small compared to the projectile, then it's probably okay to ignore it. An even smaller amount will go into friction as the cup rubs on the rails.

Suppose the mass of the "pusher" is 7 grams and the mass of the projectile is 14 grams… Hmm… I'm betting someone weighed something! Then we can say that 2/3 of the potential energy goes into the projectile and 1/3 goes into the "pusher."

So, what now?

Measuring the angle of the rails and the distance that the projectile travels leads to being able to find the initial velocity of the projectile:

_____________

SUMMARY

So, if you are given theta and the velocity, here's a checklist sort of process…

• #1 Draw the diagram and label everything.
  
  
• #2 Find the component velocities in the x and y directions:

v • cosθ = vx

v • sinθ = vy

• #3 Find the time up using vy and acceleration due to gravity (probably 9.81 m/s/s)

tup = vy  / 9.81

• #4 Find the total time where…

t = tup + tdown

and

tup = tdown

so 

t = 2•tup

• #5 Use the distance equation to find the horizontal displacement: 

dx = vx•t

where vx was found in step 2 and t was found in step 4.

• #6 BONUS: Find the maximum vertical displacement:

dy = vy•tup + 1/2atup²

where a = -9.81 m/s/s,  vy was found in step 2 and tup was found in step 3. 

_____________

GIANT MESS But actually COOL

dx = vx • t

dx = v • cosθ • t

dx = v • cosθ • 2 • tup

dx = v • cosθ • 2 • vy  / 9.81

dx = v • cosθ • 2 • v • sinθ / 9.81

dx = 2 • v² • cosθ • sinθ / 9.81

            wait for the magic!

dx = 2 • v² • 1/2 • sin2θ / 9.81

dx =  v² • sin2θ / 9.81

Say you wanted to find v for some distance x?

9.81 • dx =  v² • sin2θ

(9.81 • dx ) / sin2θ =  v² 

 √ (9.81 • dx ) / sin2θ   = v

_____________

Okay, so… wow! Is this why we take math classes?

Thus, from the distance and and angle we get the velocity. 

And since the potential energy was converted to kinetic energy…

Whereas

EK = 1/2mv²

and whereas 

EK ≈ EP

We can backtrack to find the elasticity.

Considering the materials used, we earlier decided that 2/3 of the potential energy was transferred to the projectile. Thus, in this specific case, and recalling that 

PE = 1/2kd² 

EK = 2/3 EP

We can state that: 

1/2mv² = 2/3 (1/2kd²)

Thus, after finding v, solving for k becomes an exercise easily completed by the reader.

Rounding Numbers (is easy)

 Intro to Chemistry and Physics Index

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Physics Index

Where are we going with this? Rounding off numbers in math is very important to accurately communicate science information 

Rounding Numbers (is easy)

(Round numbers? Like zero, six, nine, and eight? Is 3 a round number or a crooked number? And what about two?)

The idea of rounding numbers is to express a long decimal number more concisely. For this discussion, the whole principle of significant figures is going to be passed over. We are going to just be looking at the rounding process.

Step 1

Figure out how many decimal places you need. This could be dictated by directions in a problem or proper use of significant figures.

In the examples that follow you will be rounding to ONE decimal place.

Step 2

Look at your number. Find the decimal place. Count over to the digit you are going to round. 

Perhaps, as you develop skills, you should underline that number.

EXAMPLE:   1234.7ABxyz

VITAL TO UNDERSTAND!!!!

The only things that can change are:

1. The digit to which you are rounding (underlined above). 

 

2. The digit that comes before it (italics above). There is a 95% chance you will NOT change this.

 

Step 3

Look at your digit that comes after the one you are rounding off. In the example it's the green A.

EXAMPLE:   1234.7ABxyz

Step 4

If the A is less than 5:

Chop off the A and everything after it. Done. 

1234.7ABxyz

becomes

1234.7

If the A is greater than 5:

Increase the number you are rounding by 1 and chop off the rest. Done. 

1234.7ABxyz

becomes

1234.8

Notice that we don't care what the B is!

SPECIAL CASE:
When the number in the place we are rounding is a 9

EXAMPLE:   1234.9ABxyz

If the A is less than 5:

Chop off the A and everything after it. Done. 

1234.9ABxyz

becomes

1234.9

If the A is greater than 5:

Increase the number you are rounding by 1 (makes it a 10, write a 0), then increase the number before the one you are rounding by 1. Then chop off the rest. Done.

1234.9ABxyz

becomes

1235.0

Notice that we don't care what the B is!

 

Batteries in series and parallel

Physics Index

Where are we going with this? The information on this page introduces how batteries work in series and in parallel.

Batteries in series and parallel

(I don't this will be shocking to very many people!)

Most people are familiar with batteries of some sort. Your wireless mouse and TV remotes might require AA batteries. You probably know your car has a battery. Your cell phone has a battery… so, yeah…

The dry cell battery is a convenient point of reference. You probably will think of a AA, AAA, or D battery. Maybe the C battery.

Current and Plumbing Analogy? Why not?

Voltage can be thought of as how much pressure is in a circuit. It is somewhat analogous to water in pipes. Higher voltage = higher pressure. 

Amperage, however, is not pressure. It is pretty much a measure of the number of electrons coming out of a circuit. Considering the plumbing analogy, it is how many gallons of water are coming out of the pipe at any time. 

So, amperage is the AMOUNT of electrons coming out. Voltage is, in this metaphor, kinda like how fast they are coming out or the pressure at which they are coming out.

The math on amperage is fairly intuitive.

If you have 2 hoses that put out 2 gallons per minute, then you are getting 4 gallons per minute in total. If you have 3 hoses at 2 gallons per minute, you are getting 6.

When batteries are connected in parallel, the amperage produced adds up.

Two batteries producing 2 amps each would put out a total of 4 amps. Three of them would produce 6 amps. Etc.

Voltage in parallel would still come out at the same pressure. 

Connecting batteries in series, though, is like having one of the hoses filling up the tank from which a second hose is being filled. The second hose can only put out a set amount of water per minute. The first hose just keeps refilling the second hose’s tank. Water comes out until both tanks are empty.

With regard to voltage,  more to come

Back to batteries…

When connected in series, the amperage for any given period of time is unchanged (but the total amount of electrical current is increased). You will get the same amperage for a longer period of time.

In series…

Amp(total) = Amps

In parallel…

Amps(total) = Amps(1) + Amps(2) + … 

Batteries in Series and Parallel (Batteries in the circuit are of same amperage and voltage)
	Series	Parallel
Amps	A(tot) = A1 = A2 =…	A(tot) = A1 + A2 +…
Volts	V(tot) = V1 + V2 +…	V(tot) = V1 = V2 =…