FORCES


 * Topic: Review on Forces**
 * Text ref. pg 70-80**
 * Notes by: Josephine Gomes**
 * October 9 2012**

__Types of forces__ __F.B.D (Free Body Diagrams)__ How to draw a diagram Examples
 * 1) Fg à gravity
 * 2) FT à tension
 * 3) FN à normal: perpendicular force applied by a surface to an object
 * 4) Fk à kinetic (moving)
 * 5) FS à static (not moving)
 * Diagram indicating all forces acting on an object
 * 1) use a square or a dot to represent the object
 * 2) draw a polar axis and indicate which direction is positive and negative
 * 3) indicate the forces acting on the object by labelling the diagram (types of force)

__Newton’s First Law of Motion (referred to as the law of inertia)__ “Every body preserves its state of rest or of uniform motion in a right line unless it is compelled to change that state by forces impressed upon it.”
 * Or in other words**…an object at rest stays at rest and an object in uniform motion stays in uniform motion and in the same direction unless it is acted upon by an external unbalanced force.


 * Topic: __Forces in 2 Dimensions__**
 * Notes by: Benedict Batoy**
 * October 10, 2012**

__Newton's Second Law of Motion__
 * If the external net force of an object is not zero, the object accelerates in the direction of the net force.
 * Acceleration is directly proportional to the the net force
 * Acceleration is inversely proportional to the object's mass

There are two boxes and one box has a mass that is double the mass of the other box. I apply an equal net force to each of the two boxes. What will their acceleration be?
 * Example**

The acceleration of box with double the mass of the other box will have half of the acceleration of the other box. Since acceleration is inversely proportionate to mass, then as mass increases, the acceleration decreases.

Writing the second law in equation form:
 * 1) Begin with the proportionality statements in the second law
 * 2) Combine these statements
 * 3) Convert the statement into an equation
 * 4) Rearrange the equation to solve for net force



The unit for mass is in **kg** and unit for acceleration is **m/(s^2)**. Therefore the unit for force is **(kg)(m)/(s^2)**. It is more commonly known as newton **(N),** named after Sir Isaac Newton.

In other words, it requires 1 N of force to give a 1-kg object an acceleration of 1m/s^2.

How does the first law relate to the second law? First law states if there are no forces acting on the object, the object stays at rest or at constant velocity. Second law states that Force=mass x acceleration. If there is no net force acting on the object (F=0), then the acceleration would be zero. Therefore, the object is either at rest or at constant velocity.

**Topic: Forces, More on Newton's Second Law, and Applying Newton's Laws of Motion**

 * Text Ref. pg. 81-83, pg.88-89**
 * Myles Sinyard**
 * October 15th /12**
 * So far we've recapped what we learned in forces last year and have also taken down the fundamentals of Netwon and his very important laws of physics.**
 * In-class today we had an opurtunity to extend our knowledge on forces even further. Here are the in-class notes and some very general examples:**
 * And for even more info and examples:**
 * Now, some of us (including myself) may still be struggling with this portion of physics. Forces is not an easy subject, so thank god for pgs. 88-89.**
 * Not only do these pages provide a perfect example of a problem not studied in class, but they also offer step by step instructions on solving force related problems in a short** **and sweet way. For those of us who hate breaking out the old dust filled textbook sitting in our closets, I was nice enough to summarize those steps here (though you may want to take a look at the wonderful examples, unless you have dust allergies):**
 * Solving Problems in a Systematic Way:**
 * 1) Read the problem carefully. Use a dictionary or your textbook to aid you with any words that you may not know.**
 * 2) Draw a system diagram (like the moo moo physics cow). Label all relevant information, including any numerical quantities given. (Not required, but recommended)**
 * 3) Draw a free body diagram (FBD), labelling all forces and their known quantities. Make sure to label the positive direction with an arrow and plus sign. (You may also label the acceleration of an object in any given direction simply by drawing a squigly line instead of a straght arrow.)**
 * 4) Calculate as many x and y components of the forces as possible, also labelling them on the FBD.**
 * 5) Write the second-law equations and substitute for the variables on both sides.**
 * 6) Repeat steps 3-5 for any and all other objects experiencing force(s).**
 * 7) Solve the resulting equation(s) algebraically.**
 * 8) ALWAYS remember to check and make sure you have stated your answer using appropriate units, a reasonable magnitude, a logical direction (if required), and the correct number of sig figs (if required). After all, every mark counts!**
 * So thats that folks. You can call yourself experts now (kind of).**
 * Newtons's Third Law**
 * Theourrn Amalathasan**
 * October 17th 2012**
 * Newton's Third Law**
 * For every action, there is an equal and opposite reaction.**
 * Newtons's Third Law**
 * Theourrn Amalathasan**
 * October 17th 2012**
 * Newton's Third Law**
 * For every action, there is an equal and opposite reaction.**
 * Types of Forces**
 * 1) Contact forces (i.e road contacts car)**
 * 2) noncontact forces: force that acts at a distance (i.e gravity pulling)**

**Topic: Newton's Third Law & Solving Contact Force**

 * Noyell Sakthikumar**
 * October 18th, 2012**
 * Newton's 3rd Law : 2 forces equal in magnitude, opposite in direction and acting on two objects.**
 * Solving questions the sample questions:**
 * 1.**
 * 2. solving with an angle**

Patricia Kousoulas Friday, October 19th, 2012
 * Topic: Ramp Challenge**

The class began on an interesting note; there was a wooden ramp on an angle of 20 degrees sitting in the front of our classroom and beside it was a scale. Balthazar sacrificed himself, to be the class volunteer. He was asked to go on it, and soon realized that he weighed a total of 68 kg. However, the awesome part of the experiment was to figure out how much would he weigh if he was standing on the ramp with the angle of 20 degrees!

We then drew a free body diagram to understand the question. So to figure out our new mass; we needed to find the force normal first! (Vertical Components) Using this normal force; we then were able to solve for Balthazar's new mass. Therefore, Balthazar should indeed lose 4 kg by standing on the scale that is placed on this ramp.However, since our experiment wasn't exactly accurate, and since we were using angles; Balthazar ended up weighing 66 kg. But at least he lost some weight! And as Ms. Price said, "It's not perfect, but it did work."

After this experiment; we were later placed into our regular groups. We were asked to pick an object that has a mass of around 300 g. With this object, we were to figure out at what angle does our white board need to be positioned at, so that our object can slide down our ramp.

With only a mass given, there were so many unknown variables. The key to the experiment was to draw a simple FBD and then solve for all possible values. Then you could apply your new numbers to solve for the angle. By doing the experiment you would have noticed that trig identities were used (**__SIN theta/COS theta =__ TAN theta).**

In case you haven't finished your lab yet, make sure to hand it in for Monday. And to all the fabulous AP students; an extension was given for the Fluids package, THANKS MS.PRICE!

=**TOPIC: Forces Quiz and Building Challenge!**= Patricia Kousoulas Friday, October 26th, 2012

For those of you who missed class today; you fortunately didn't miss too much other than an awesome quiz and a fun team building challenge.

In our own forces group, we were given three sheets of paper and a piece of tape that was one metre in length. Our task was to construct the tallest free standing structure. The winning structure was over a metre in length - congratulations!

no homework was given out, except for the AP students to finish their fluids package.

=**TOPIC: Circular Motion**=
 * Sabrina Zuccaro**
 * Tuesday, October** **30, 2012**




 * instantaneous velocity is always directly tangential to the center of the circle

Period (T): the number of seconds it takes to make one cycle Frequency (f): the number of cycles made per second

Period (T)/ Frequency (f) = constant Speed = constant Velocity = changes Acceleration magnitude = constant Acceleration vector = changes Net Force = constant Force Tension = constant


 * An object traveling in a circular path with a constant speed and radius therefore has uniform circular motion.*

**CENTRIPETAL MOTION:**


Centripetal Motion:
 * Circular motion
 * acceleration = ac (c = subscript on a)
 * force = mac = Fc (in both cases c is a subscript)
 * net force is always directed towards the center of the circle (as in the acceleration) -- Newton's Second Law

In General:

When considering very small changes in time, (triangle) t: therefore: acceleration = __the change in velocity__ the change in time
 * velocity, v, is directed towards the center of the circle
 * acceleration must also be directed towards the center*

**Equations Specific for Circular Motion:**

 * note: speed = 2(pi)(radius) / period

**Force Equations:**


Fc (circular force) = net force
 * all the forces on the line that connects the center of the circle (all forces in that path)


 * NOTE: Positive direction is always towards the center of the circle!!!!**


 * Example: A 150 gram ball on one end of a string which is 1.15 m long is swinging in a horizontal circle (parallel to the floor) and makes two revolutions in one second. Calculate:**


 * a) speed**
 * b) centripetal acceleration**
 * c) centripetal force**
 * in my equations the letter 'c' is ALWAYS a subscript***

a) m = 150 g = 0.15 kg r = 1.15 m f = # of revolutions / time = 2/1 = 2 Hz T = time / # of revolutions = 1/2 = 0.5 s

v = (2 x pi x r) / T v = (2 x 3.14 x 1.15) / 0.5 v = 14.4 m/s

b) ac = (4 x pi^2 x r) / (T^2) ac = (4 x 3.14^2 x 1.15) / (0.5^2) ac = 181.6 m/s^2

c) Fc = m x ac Ft = m x ac Ft = 0.15 x 181.6 Ft = 27.24 N

=Centrifugal motion (continued) - by Eddi (W Oct 31, 2012)=

I forgot I was doing notes today, so here's what's off the top of the head.

We learned about horizontal centrifugal motion versus vertical centrifugal motion.

We also learned about //minimum velocity//. This value represents the value needed to maintain centrifugal motion. That is to say, the velocity must be greater than this for the motion to maintain its radius (a constant radius defines centrifugal motion) as well as its force of tension (must always exist in centrifugal motion) in vertical centrifugal motion. We haven't done any problems on minimum velocity with horizontal centrifugal motion, to my awareness.

To calculate minimum velocity, consider the calculation at the peak of the centrifugal motion, where velocity can be be at its lowest (if lowest velocity was calculated somewhere else, there would not be enough velocity and therefore tensional force to maintain the motion upward).

We know the equation of centripetal force (the sum of all the acting forces in centrifugal motion) :

Fc = ( mv^2 ) / r

In vertical centrifugal motion:

Fc = Ft + Fg

Therefore, in vertical centrifugal motion:

Ft + Fg = ( mv^2 ) / r

To calculate minimum velocity, we must first calculate the instance where centrifugal motion fails due to loss of velocity: when the tensional force stops existing (becomes 0) (this calculated value is called a limit). So:

0 + Fg = ( mv^2 ) / r

Fg = ( mv^2 ) / r

Thanks to Luis, I can identify a simplified version in finding the limit.

Fg = ( mv^2 ) /r

mg = ( mv^2 ) / r

g = (v^2) / r

v^2 = gr

v = √(gr)

If we have a situation where the minimum velocity is needed, and r = 40 m :

v = √(gr) = √(10)(40) = √400  = 20 m/s

Therefore, we can say that the velocity required for perpetual centrifugal motion (at the peak of the motion) must be **greater than** 20 m/s. Remember that the calculation accounts for a limit, not the actual minimum velocity.

That's what I recall. Correct me if I'm wrong in anything!

=Centrifugal motion (continued even more) - by Eddi (Th Nov 1, 2012)=

So far, we've dealt with minimum-velocity situations, where centrifugal motion would stop existing when the velocity isn't sufficient to overcome gravity.

The new concept today was maximum velocity, in situations where friction is the force that enables the constant radius. Unlike minimum velocity, maximum velocity doesn't require calculating a limit. The Fs (the force of static friction perpendicular to the moving object, i.e. rolling friction) must be overcome in order to break centrifugal motion, therefore, matching Fs with the situation's opposing force (or opposing momentum? I have no idea, I'm so sorry) directly dependent on the velocity would not break the centripetal force (this is the maximum velocity, after all).



Sum of all perpendicular forces = 0 Fg - Fn = 0 Fn = Fg = ma =m(10)

a, = centripetal acceleration

Fc = m**a,** M**Fn** = m(v^2) / r M**m**(10) = **m**(v^2) / r 0.8 x 10 = v^2 / 20 v = √(0.8 x 10 x 20) = 12.65 m/s

I don't understand this, someone who does understand why Fcar isn't factored in, please take the liberty to create an addendum in this post.

Other things learned:

Remember, when solving for a minimum velocity by finding a limit, use the term "assumption", or "when". Add a therefore statement afterwards.

e.g. In a roller coaster, Fn at the peak determines an unknown minimum velocity.

ASSUMPTION: Fn = 0 ... or WHEN: Fn = 0 ...

... v = 15 m/s

Therefore, the minimum amount of speed to continue centrifugal motion in the roller coaster must be greater than 15 m/s.

= Artificial Curves & Banked Curves = Notes By: Patricia Kousoulas Friday, November 2nd, 2012

In case you're interested about space and everything - this was definitely a good lesson for you!

1. Artificial Gravity
The real question is; what is artificial gravity?


 * Artificial gravity ** is the increase or decrease of apparent [|gravity] via artificial means, particularly in space, but also on the Earth. It can be achieved by the use of some different forces, mostly by the centrifugal force.

so now imagine a situation at a space station, where a force normal is being applied to a person; and a diameter of 1.6 km is given. Your challenge is to figure out how to do it!


 * HINT: if you want to "feel" the earth's pull FN = Fg = mg** *

INTERESTING FACT: Each person is giving off a force of gravity no matter where you are (including space) onto each person and object. Now go ahead and share this info with your friends!

2. BANKED CURVES
*These problems are horizontal circles - this question is without any friction! Some do have friction

EIGHT SIMPLE RULES TO SOLVE:

 * 1) Draw a free body diagram of the situation
 * 2) State your givens
 * 3) Make any necessary Conversions
 * 4) Calculate all forces acting in the vertical direction (ma = 0)
 * 5) Using your answer from the vertical calculations; solve for the centripetal force in the horizontal direction
 * 6) Rearrange and solve for the variable that you are looking for!
 * 7) If this question included friction; solve for the angle first and then for friction next :)
 * 8) it's that simple!

How about if we apply these simple rules to a real example;

Question 4: Circular Motion Sheet

Step 1: Draw a FBD

Step 2: State your givens: Step 3: Conversions* m = 1500kg r = 325m v = 200km/hr = 55.5 m/s * angle = ?

Step 4: Make calculations in vertical direction

Step 5: Solve for centripetal force in horizontal direction Step 6: Solve for your unknown variable (theta)

The final answer will be 43 degrees.
Don't forget to do some more circular motion questions; the more practice - the better!

In friction examples; the process is identical! The only difference is that their is an extra vertical and horizontal component that needs to be added into the equation.

Universal Gravitation Notes by: Mikaila Hirschler - For objects experiencing a force of gravity near __Earth’s__ Surface Fg=mg.