Fan Cart Post

Question: In this challenge, the question we were trying to answer was, “Where will two fan carts with different accelerations collide .”

What we did: Each individual group used a motion sensor and a fan cart running at a constant desired acceleration, and performed a test to find out the initial velocity and acceleration, (which was the slope of a velocity time graph), by letting the cart run down a table towards the sensor. We ran the test twice with the fan at the highest speed just to make sure the values we collected were similar in range, but we only ended up using one of the trials for the final prediction since we knew the data from the motion sensor would be accurate. 

Data: For this particular experiment, all that was needed from each group to make the prediction was the acceleration of the fan cart. My group used Cart A, and the acceleration of the first trial was .184, and the acceleration of the second trial was .0175. We found initial velocities as well (.0765 and .182), but they weren't needed to make our prediction and they were extremely small as well. The group we partnered with, Serenity and Connor, used Cart C, and their accelerations were .0219 and .0205.

How we made the prediction: Our groups decided that we would use the formula 1/2at^2 to predict the point at which the carts would collide since that is the distance/displacement formula ({vi}{t}) was not needed because the initial velocity was practically 0). 

The first step we took was to calculate the time at which the carts would collide using the first velocities stated above from both groups as the acceleration. 2 meters or 200 centimeters represents the distance the carts would be placed from each other on opposite sides of the origin:

2=1/2at^2+1/2at^2

2=1/2(.184)(t)^2+1/2(.0219)

2=(.10295)t^2

2/.10295=(.10295)t^2/.10295

t^2=19.46 (take the root)

t=4.41 seconds= the time it will take the carts to collide

To make the prediction for where the carts would collide we used the same formula only with 4.40 seconds as our time:

1/2at^2=distance cart traveled

CART C: 1/2(.0219)(4.41)^2=.21295m or 21.295 cm

CART A: 1/2(.184)(4.41)^2=-1.79m or 179 cm

Prediction: We predicted that the Cart C would travel 21.295 cm from the starting point of 100cm towards the origin, and Cart A would travel 179 cm from the starting point of 100 cm towards the origin from the other side to collide after traveling those distances.

Results: Our predictions were pretty accurate as the carts collided right around the distances predicted. Cart C traveled 28cm and Cart A traveled 175cm. Our percent error calculated for the predicted distance is:

 179-172/172x100%= 4.06% error

CAPM Unit 3 Summary

In this unit entitled CAPM, or Constant Acceleration Particle Model, we were finally able to learn about acceleration and how it relates to velocity,  a topic we had held out on all year. This unit, in a way, was a more advanced continuation of the CVPM Model we did at the beginning of the year, as we took the same concepts and added in more elements to fully connect everything together. We covered new topics such as acceleration, instantaneous velocity, acceleration vs. time graphs,  and finding displacement from a velocity time graph. 

CAPM.1: 

Instantaneous velocity by definition is the velocity of an object in motion at a specific point in time. So that means that one would find the instantaneous velocity at 2 seconds to find out how fast the object was going RIGHT AT 2 seconds, not at any other time. 

Instantaneous velocity can be calculated 2 ways:

1. when using a position vs. time graph, one can find the slope of the tangent at a given point

2. use the formula: v final= at+ v initial

EXAMPLES (as related to the methods above): 

1.In the position vs. time graph pictured below,  the object is shown with constant acceleration, meaning it is speeding up, and traveling in the positive direction (how I know this will be explained later).  Because this is a position not a velocity graph,  you must calculate the instantaneous velocity using the slopes of the tangent, which are the points on either side of the desired midtime. 

SOLVING:

In this particular example, I am calculating the instantaneous velocity at 2 seconds, which then becomes the midtime. I will then use the x,y      coordinates at 3 and 1seconds, plug it into the formula change in x divided by change in time and get an answer in m/s (meters per second), which will be the instantaneous velocity at 2 seconds. The work is shown in the photo below. Notice that the smaller number is always subtracted from the larger number.

2:  The second method that can be used to solve for instantaneous velocity is when you simply have to plug values into the formula:                     v (velocity)= (a {acceleration})(t {time}) + vi (initial velocity).                                                                                                                   THIS EQUATION CAN ONLY BE USED WHEN THERE IS A STRAIGHT LINE!!!!!!  

SOLVING: (Acceleration has not been described, nor has solving it been displayed yet in this blog, but for the sake of this problem it will              be briefly overviewed.) 

One has the same goal when using this formula: to find the specific velocity of an object at a given time. Using the graph pictured below, I will show how to find the instantaneous velocity at 6 seconds.

a. the first step is to solve for the acceleration of the object over the course of the whole period using the formula acceleration=                    change in v divided by change in t (time). Basically,  written out longhand it is final velocity- initial velocity/ final time- initial time.  Using the graph above as a reference and the picture below to see the work written out, the final velocity of the object was 0 m/s, the initial was -15 m/s, all over the final time which was 8 seconds - the initial time, 0 seconds.  This got me an answer of about 1.87 m/s^2, which are the units for acceleration. 

b. calculating the acceleration would have to be the hardest part, because now all that is left is to plug the given values into the formula v=at+vi. One must be careful to pay special attention to the exact value the problem is asking for because it is very easy to plug in the wrong time. For this problem, since we are finding the instantaneous velocity at 6 seconds, 1.87 is plugged into a, 6 is plugged into t, and -15 is added at the end as the initial velocity of the object. In all the formula would look like this: (1.87)(6)+(-15). I would get the answer of -3.78m/s as the instantaneous velocity at 6 seconds.

CAPM.2:  To continue our study on displacement, which is the shortest distance from the initial to the final position of an object, we learned 2 new methods when working with velocity time graphs. 

1. finding the area under a velocity vs time curve

2. using the mathematical model x= 1/2at^2 +(v initial)(time)

EXAMPLES (as related to the methods above): 

1: When using a velocity vs. time graph, one can calculate the area under the line in 2 ways using the formula l (length) x w (width), or the formula 1/2 ( b {base})(h {height}) + (length)(width).  One would only use just length times width when there is a straight line, meaning constant velocity on the graph, because that is the formula for area for a rectangle, which is the only shape made from a straight line. The other formula would be used when there is a slanted line, and a triangle, which used the formula 1/2bh to find the area, and a rectangle can be formed, which will be shown in both examples below. 

a. FINDING DISPLACEMENT USING L x W USING A RECTANGLE

The graph below shows a straight line, meaning the object is traveling at a constant velocity. As you can see, I have drawn a straight 

line down to the axis to show myself that only a rectangle is present, meaning that I only have to multiply length times width to find the displacement. The length of the rectangle is 5 and the length width is 2, multiplied together gives me 10 meters, which is the displacement.

b. FINDING DISPLACEMENT USING 1/2bh USING A TRIANGLE

This example is just like the one above except this time. only a triangle is formed under the curve because the line touches the X axis, so          only the formula 1/2bh needs to be used to find the displacement.  For this particular graph one would use 1/2(2)(5) which would get 5m for the displacement.

c. FINDING DISPLACEMENT USING 1/2bh + L x W USING A TRIANGLE AND RECTANGLE 

The graph to the right shows a slanted line, meaning that the object is not traveling in one constant velocity, it is speeding up. That means that I have to use the formula 1/2bh + L x W because both a rectangle and a triangle can be formed under the curve as you can see labeled. Now all that is left is finding the values to plug into the formulas. For the triangle, the formula would look like this: 1/2(5)(3), and for the rectangle it would look like this: (5)(1). Altogether, the formula would be: 1/2(5)(3) + (5)(1)= 12.5 meters which is the displacement. 

2. You can also use the formula 1/2(a{acceleration})(t{time})^2+(v initial)(time). This formula can be used when the acceleration is constant.

SOLVING: Using the graph to the right as a reference, the displacement will be found using the new formula. As overviewed earlier and will be overviewed again until acceleration is fully described later on, the first step to completing this problem is finding the acceleration of the object. Once again, use the formula final velocity- initial velocity divided by final time - initial time.  As can be seen in the work below, I used 0 -(-3) divided by 5-0 which is 3 over 5 which gave me .6m/s^2 as the acceleration. Now I need to plug that value into the rest of the equation along with the other numbers. I plugged the numbers in like this: 1/2(acceleration=.6)(total time=5)^2+(initial velocity=-3)(total time=5), without the explanations in the equation it would look like: 1/2(.6)(5)^2+(-3)(5). The answer is -7.5 which is the displacement.

CAPM.3: 

By definition, acceleration is the rate of change of velocity of an object. In this class we have looked at acceleration as whether or not an object is speeding up. There are a few ways to find acceleration based on graphs, but the main ways we have used in class include finding the slope of a v vs t graph which is basically the same as using the mathematical model that was used earlier which was r a=change in v (final velocity-initial)/ change in t(final time-initial).  The example below shows the work and the graph of a problem that asks for acceleration. 

Other ways of finding acceleration including rearranging the mathematical models that were used in the previous example: x=1/2at^2 +(vi)(t), and final velocity= (a)(t)+ v initial. You simply manipulate the equation so you are solving for acceleration. 

EXAMPLE: 

1. x=1/2at^2 +(vi)(t)- subtract (vi)(t) and move it to the other side, then divide both sides by 1/2(t)^2 so a is isolated

2. velocity= (a)(t)+ v initial- subtract v initial from both sides as well as divide both sides by t which leaves a alone on the right side

CAPM.4/CAPM.5:  These 2 sections have been combined because they both involve position-time graphs, velocity-time graphs, and acceleration-time graphs, and how they relate with each another. 

POSITION VS. TIME GRAPH: Below is a picture of all the different position time graphs we have learned about throughout the course of this unit and CVPM. The corresponding numbers will describe what the object is doing in each graph.

1. the object is at rest, there is no movement 

                                                          2. the object travels in the positive direction at a constant velocity

                                                          3. the object travels in the negative direction at a constant velocity  

                                                          4. the object is speeding up in the positive direction

                                                          5. the object is slowing down in the positive direction

                                                          6. the object is speeding up in the negative direction

                                                         7. the object is slowing down in the negative direction

A line with a positive slope on a position time graph is always going forward, one with a negative slope is going backward, and the steepness of the slope determines how fast the object is going with a horizontal line meaning the object is at rest.

REMINDERS:

• finding instantaneous velocity: use the tangent of the slope, or one point on either side of the desired midtimet using the formula: x. Finding average velocity is the same as finding the instantaneous velocity of the midtime using the same formula, change in x over change in t!!

VELOCITY VS. TIME GRAPH:  below is a picture of the different types of velocity time graphs.  Above the x axis represents the positive direction, and below the axis represents the negative direction, which corresponds with the slope of a position time graph.  As the object moves farther from the x axis, it begins to speed up, and when it moves closer to the x axis, it slows down, when the motion is constant the graph has a straight line. The very first graph below shows an object going in the positive direction speeding up, (moving away from the axis), the second one shows an object going in the positive direction slowing down, (going towards the axis), the the third dhows an object going in the negative direction slowing down, and the fourth is an object in the negative direction speeding up. Using this information, one can construct a velocity time graph that corresponds with the position time graph, but one can also construct a position time graph from the velocity time graph information as well, the key is simply to remember the rules from each graph. 

CONSTANT VELOCITY:  

REMINDERS: 

• determining acceleration: use the same formula as before, v final- v initial/ t final- t initial, but the only difference is, this time you can pull the velocities straight from the graph without having to calculate velocity like from the position time graph 
• determining instantaneous velocity: because this is a velocity time graph, there is no need to take the tangent of the line because there are already velocities on the graph, so if you would like to find the instantaneous velocity at 2 seconds, for example, you would just find the specific velocity at the corresponding y point on the graph
• determining displacement for a given time interval: one would use the formula 1/2(a)(t)^2+vi. After calculating the acceleration and the initial velocity, the time is the difference between the time intervals, for example, between 2 and 6 seconds would be 4 seconds. Once you have figured out all those values, you would then plug them into the formula as pictured below:

ACCELERATION  VS. TIME GRAPH:  acceleration-time graphs are drawn based off of velocity-time graphs and it tells us whether the acceleration is positive or negative. Once again, above the x axis is positive acceleration and below is negative acceleration, and where the line is drawn is based on the SLOPE of the velocity-time graph, not the direction or location of the line, which is a common mistake. The acceleration line is always straight, it will never be diagonal, and if the object is traveling at constant velocity or if it’s not moving at all, the line will be drawn on the x axis representing 0 acceleration. 

MOTION MAPS: The motion maps we completed for this unit were much less specific than the ones we have done in the past. These maps are more focused on the speed of the object represented by the length of the vectors, rather than the specific location the object is located. When constructing these graphs, however, it is important to label which side is positive and which is negative and draw the direction of the vectors specific to the graph. When the object has positive acceleration, you would draw the vectors going towards the positive end of the map getting longer as they go, and vice versa if there was negative acceleration. 2 example pictures are shown below. So far, we have only learned about constant acceleration, so the motion map for acceleration will always have equal vectors.

CAPM.6: Below are a few examples of how to construct position time, velocity time, acceleration time, and motion maps when given written information. 

1. The object travels forward in the positive direction at constant acceleration speeding up

2. The object travels backward at constant velocity

3. The object travels backward in the negative direction slowing down 

CAPM.7:  When you want to solve for the amount of time it took for an object undergoing constant acceleration using the formula x=1/2at^2 +(v initial)(time), you simply manipulate the formula to isolate t which would look like the picture below.

CAPM Challenge 1

CAPM Challenge 1

a). 

The sketch above shows the setup for this experiment. We worked on a slanted table propped up with 2 thick textbooks. We aligned 2 meter sticks along the length of the table and used a metal cart with wheels as the object in which we would collect the data from. A piece of chalk was used to mark the position of the cart every half second which was timed out using a metronome. We performed a few test runs to make sure we were comfortable with where and when to put the marks as the cart rolled down the table. After adjusting, we ran the cart down the table 5 times so we would have 5 measurements originally in centimeters eventually converted to meters, for each half second that we would average together to have the most accurate results. 

b). 

Above is an image of our raw data table. The X column represents the time every half second starting at 1 second (we originally began at 1/2 second but soon realized that we skipped a half second before marking the position), and the Y column is the position which shows 5 measurements, and the final column represents the average of the 5 measurements. 

c).  

This graph was formed using the raw data. As can be seen, the line is not linear, therefore it does not allow us to make a prediction, and so the graph below represents the data after it was linearized, which was achieved by squaring the X values (time).

The x values are now squared in this graph making it linear, so we were able to find the equation of the line which will be explained more in depth in the next section. We used this equation to make our prediction for acceleration which will also be explained next. 

d). The equation we derived from the line is y=0.0785x+ 0.1135 or position=0.0785(time)^2+ 0.1135. We learned in a previous study that the slope of a velocity vs. time graph is the acceleration, but on a position vs. time^2, the slope is equal to half the slope (which is the acceleration) of the velocity time graph. In order to find the acceleration on a position vs. time^2 graph, we used the formula slope=1/2a. With the values plugged in, we had: .0785=1/2a. To isolate a we multiplied both sides by the reciprocal of 1/2 which was just 2, and we ended up with .157m/s^2.

e). RECAP TO FIND ACCELERATION AT 4 SECONDS: slope=1/2a

.0785=1/2a

(1/2a)(2)=(.0785)(2)

acceleration= .157 m/s^2

f). The velocity of the object at 4 seconds is equal to the acceleration just solved for in the previous step, therefore, the velocity is .157m/s. This is because we knew that the slope of a velocity time graph is equal to the acceleration of the object which was found in the previous step by multiplying the slope by 2. 

g). The formula for percent error is predicted value- actual value/actual value x 100. The actual value was .192m/s.  Strangely enough, in our physics class, all of the groups were a considerable amount off, and no one was within 10%, much unlike the other class. With our values plugged into the formula it would look like this: .192-.157/.192. x 100%= 18% error. Unfortunately, we were not within 10%, but we did complete the lab using all the proper steps and requirements.

BFPM Practicum

BFPM Practicum

This experiment was meant to test our skills in forming free body diagrams as well as using SOH CAH TOA to solve for the missing vector for gravity.

a.     Picture of the diagram:

b. Free body diagram:

c. Making the free body diagram was quite simple, the tensions were both at an angle, causing me to split them up into components in the X and Y directions. All that was left was the force of gravity pointing downward. The work written out to solve the problem can be seen in the picture below. To start off, I knew that all the horizontal and vertical force vectors were congruent because the object was not in motion. That information tells me that the force of the 2 tension components in the Y direction, FT1y and FT2y, added together will equal the weight of the object because they are both half the total weight. We were given 2 preliminary angle measurements, 23 and 75 degrees, each half of a quadrant of the diagram meaning the other part of the angle must add to equal 90 degrees. This easily allowed me to find the measures for the other angle which were 65 and 15 degrees. Because I was also given the force of tension for both vectors I was able to use that value along with the angle to solve for the component in the Y direction. I decided to use cosine, which is adjacent over hypotenuse because I was given the hypotenuse, which was the force of tension, and I was trying to solve for the adjacent value to the angle measurement. I set up COS 65=FT1y/.9 and COS 15= FT2y/2.2. I multiplied the forces of friction by the cosine of the angle and got .380 Newtons for one of them and 2.125 Newtons for the other. To find the gravity I just added the numbers together.

d.  PREDICTED WEIGHT: .380 N + 2.125N= 2.5 Newtons

e. ACTUAL WEIGHT: 

f. PERCENT ERROR:

Unit 2 Summary Blog Post

Unit 2 Summary Blog Post

In this unit, we learned all about forces and their interactions, free body diagrams, and Newton’s 1st and 3rd Laws. We began with the basics, which was an overview of each force including how and where they act, and how they work on a free body diagram. We learned about Newton and his laws, even doing a few experiments to prove their dependency until we got more advanced and could apply them to all types of general life situations such as how seatbelts keep one safe in a car, and how the motion of a skydiver is affected by a parachute. We learned how to calculate weight as well as friction, and definitely, the biggest challenge of all was applying numbers to our free body diagrams to calculate other values from the number using SOH CAH TOA.

BFPM.1: 

Wherever we go and whatever we do, there are always forces interacting around us. In    BFPM.1, we learned how to recognize those forces and make a diagram for the system called a Free Body Diagram.

FORCE OVERVIEW:

F of Gravity: stems from the Earth and is an attraction between and object and a planet towards the center of the planet- it always points down from the surface

F of Friction: occurs as a result of 2 objects rubbing together- it always points in the opposite direction of motion parallel to the surface 

F of Tension: occurs along a rope or string when it is taut (pulled in general direction)- it is always in the direction being pulled 

F of Normal: occurs when two surfaces touch each other that prevent one another from using the same space- occurs perpendicular to the surface 

F of Air: occurs when the force of air is pushing against an object- occurs opposite to the direction of motion

Free Body Diagrams:

By definition, a free body diagram, or a force diagram, is a graphical illustration used to visualize the applied forces, movements, and resulting reactions on a system. We’ve done various types of free body diagrams this unit ranging from basic to advanced, yet they all encompass the same general premises. Basically, a free body diagram represents the forces being done TO THE OBJECT/SYSTEM, not the force the object itself is doing, which is a general misconception and an easy way to mess up the diagram. 

Constructing a BASIC Free Body Diagram with Horizontal Coordinate Axes:

When making a free body diagram, you want to start with a dot in the center of the intersecting X and Y axes. This dot represents, the system (it could be a log, wagon, etc.), and it is the object you will focus on when figuring out what forces are applied on it. The X axis, drawn horizontally, represents a flat surface and the Y axis represents the direction perpendicular to the surface The arrows, also called vectors, can be used to represent the direction the forces are applied, and the size- the longer arrows are the largest, and the shorter ones the smallest. When the vectors are congruent on both sides (either horizontally or vertically), it means the forces are equal/balanced so the object is either at rest or traveling at a constant velocity. When one vector is longer than the other, it means the object has an unbalanced force causing it to slow down, speed up, or change direction.

The diagram above represents a simple free body diagram with normal, friction, applied, and weight (gravity) forces on a system. The X axis represents the flat surface with the normal force pointing upwards, perpendicular to the surface, and the force of gravity is shown pointing downwards from the surface. Horizontally, the friction force is drawn to the left, meaning the object’s direction of movement is to the right, and the F applied force, meaning it could be a pull, push, or any other force that could be applied, is drawn to the right in the direction of motion. Notice all the vectors are equal/congruent, so the object is either traveling at a constant velocity, or not moving at all.

Findng the Equation of Forces in the Horizontal and Vertical Directions:

The equation of forces is basically a statement of what all the forces combined, either along the horizontal or vertical axes (you never combine forces that aren't on the same axes), equal when added together. In the diagram above, the equation for forces in the vertical direction (Y axis), would be F Normal+F Gravity= 0N, and for the horizontal direction the equation would be F Friction and F Applied = 0N. Because the vectors are balanced in both the horizontal and vertical directions, we know that the two forces counteracting each other equal 0 because they cancel each other out. Pretend the F Applied force is drawn longer that the F Friction force displaying that the object is moving to the right, then the equation would be F Friction + F Applied > 0N because the forces aren't canceling out causing movement. 

BFPM.2:

Newton’s 1st Law- an object in motion stays in motion, and an object at rest stays at rest until acted upon by an unbalanced force Example: ball in motion will continue to move at the same constant velocity unless an unbalanced force  speeds it up, slows it down, or changes its direction ball at rest will stay at rest unless a force moves it

Our study on this law first began when we rode hovercrafts in the gym. Using the law we were able to figure out why we stayed in motion once someone pushed us off, and why we would have stayed in motion unless someone stopped us. While traveling, there were no forces keeping us in motion, therefore the forces were balanced which is why were traveled at a constant velocity and would have kept traveling at a constant velocity until an unbalanced force was applied. An unbalanced force causes an object to slow down, speed up, or change direction, and in this case, the unbalanced force that caused us to stop, was a person waiting at the other end with their hands out. They applied the unbalanced force greater than the force of motion which stopped us, and then applied another unbalanced force to push us back in the opposite direction.  

The diagram above represents Newtons’s 1st Law. The car at the top represents an object at rest that will stay at rest unless acted upon by an unbalanced force, and the car at the bottom represents the object in motion that will stay in motion unless acted on by an unbalanced force. 

Newton’s 1st Law and Seatbelts:

Seatbelts keep us safe in a car, because when traveling, you are moving at the same speed as the car. When an accident occurs, a force is applied to the car to make it slow down, but that same force is not applied to you, meaning you will continue to move at the same speed the car had been traveling which is what can cause you to fly through the wind shield. The seat belt prevents this from happening because it applies an unbalanced force (in the horizontal direction) that stops your motion and prevents you from flying forward. 

Unbalanced Forces and Skydivers:

The speed of a parachutist changes throughout his journey from the plane to the ground and it is because of the effect of unbalanced forces. For this particular example you would only focus on the forces in the vertical direction, as the horizontal forces are non-applicable at this point because he is falling downwards. When he first jumps out of the plane, he is speeding up because the direction of the unbalanced force, also the only force acting on the skydiver at the moment which is gravity, is in the same direction of movement. Terminal velocity, or constant velocity, is reached when the forces in the vertical direction are balanced, so the force of air and the force of gravity are both equal. The skydiver would start to slow down when the force of air, stemming from his pulled parachute is greater than that of the force of gravity, and since the unbalanced force is pointing upwards, or in the opposite direction of motion, he will slow down.

KEY FACTS to REMEMBER:

• when the unbalanced force is in the opposite direction of motion the object will speed up
• when the unbalanced force is in the direction of the motion, the object will speed up
• when the forces are balanced/equal/congruent the object is traveling at a constant velocity or it’s not moving at all

BFPM.3:

Calculating Weight: 

When calculating weight, you use the formula w=(g)(m). The w, is the weight in Newtons, the g is the universal gravitational constant which is set at 9.8, but for our purposes we’ve been allowed to round it to 10N/kg, and the m is the mass in kilograms, and it represents the number of atoms or how much matter is present in an object. For example, when trying to calculate the weight of an object that is 750 kg, you would do the following: (750)(10)= 7500N. When trying to find the mass of an object that is 500N you would do the following: 500= (10)(x), divide 500 by 10, and get x which equals 50 kg.  

BFPM.4:

Friction: as stated earlier, friction always opposes the direction of motion, and is often the underlying cause of many phenomena that we see on our every day lives. As will be described later on, friction is the reason one team wins a game of tug of war, the reason why a buggy can pull a cart, and many more. It is also stated that surface friction is equal to the normal force of two objects pressing together. 

Calculating Friction: 

When calculating the friction of an object you simply plug values into the formula: F= (Mk)(weight). F is the friction found in Newtons, Mk is the coefficient of friction, and weight is the  weight in Newtons.

AFFECTS AIR RESISTANCE: speed and surface area

AFFECCTS FRICTION: nature of surface and weight

DOES NOT AFFECT FRICTION: speed or surface area

what affects are resistance does not affect friction 

The coefficient of friction is a constant based on the surface of the 2 interacting objects 

BPFM.5:

Constructing a Free Body Diagram with Shifted Coordinate Axes:

The process for constructing a free body diagram with shifted or tilted coordinate axes is the exact same as making one with horizontal axes except you must be aware of which direction you tilt the X and Y axes so they properly match the picture and you can accurately apply the forces. It is important to remember that the X axis is the surface, so it must be angled in the same direction as the diagram. The reason for shifting the diagram is for visualization purposes: if we are meant to diagram a car driving up a hill or a skier coming down a slope, it is easier to tilt the axes in the direction the system is traveling or located. 

you would shift the axes if the example looks similar to the following:  

notice the skier coming down the slope at an angle- therefore this diagram would require a tilted axis. For this example, the x axis, which represents the surface, is going to run from the top left to the bottom right in the same direction as the “slope” or hypotenuse of the triangle is displayed. For better viewing purposes the axes would look like   the following:        

                           

 here the box is moving down an incline but in a different direction, therefore a tilted axis is also required but the X axis would be tilted in the opposite direction than the example above, pay careful attention to the difference because inaccurately naming the axes could cause major problems down the road. It should look like the following:

                                    

Components 

Components come into play on a free body diagram when vectors are at angles that do not match up with the direction of the axes, whether they be horizontal or tilted. For example, on a tilted axis, the force of gravity has to point downwards, naturally meaning the vector does not align with the shifted diagram, so you must split up the gravity vector into X and Y components that combine to make the single downward pouting vector. Basically, a vector can be seen as having two parts that when combined, make one. We did an example in class where Maren and Ms. Lawrence both pushed a table in different directions, and it moved in a direction that was like a “middle way” between their two pushes. The diagram below shows a visual representation of how components of a single vector work.

BPFM.6:

Newton’s 3rd Law: for every action, there is an equal and opposite reaction 

Example: ground pushes dog up, dog pushes ground down

    girl pulls wagon, wagon pulls girl

How a Team Wins Tug of War:

The common misconception is that the team who pulls the hardest will win a game of tug of war, but that is actually incorrect. Because of Newton’s 3rd Law, we know that both teams pull each other with equal force because they are action-reaction pairs: team a pulls team b while team b pulls team a, so the determining factor for which team will win is how much friction is between their feet and the ground. 

How a Buggy Pulls a Car (Horse and Cart Problem):

Extremely similar to the tug of war example, we completed a horse and buggy challenge problem in class. Relating to Newton’s 3rd Law, the cart pulling the horse and the horse pulling the cart are action reaction pairs, therefore they are equal and opposite, so unlike common assumption of horse pulling the cart with a “greater force” is not the cause for the movement of the cart. Once again, the cause for the cart’s movement is because of friction. As the cart begins to move, the friction on the horse’s hooves becomes greater than the friction on the cart’s wheels which causes it to move. The diagram below displays the forces that are present on both the horse and the cart.



 

Examples of Newton’s 3rd Law Free Body Diagrams:

The diagrams above represent two examples that represent Newton’s 3rd Law. The example on the left is an accurate display of an action reaction pair because the earth pushes the body down, while the body pushes the earth up. The example on the left shows that the body pushes down on the table while the table pushes up on the body, also an action reaction pair. 

KEY TIPS TO REMEMBER WHEN USING NEWTONS’S 3RD LAW:

• remember to draw vectors equal and opposite when making free body diagrams
• forces can be equal and opposite without being an action-reaction pair. For example, this is not an example of an action-reaction pair: earth pulls book down, table pushes book up. This, however, is an example of an action reaction pair: book pushes table down and table pushes book up

a force is one half of the interaction between 2 objects 

SOLVING BALANCED FORCE PROBLEMS USING SOH CAH TOA

Sometimes it is necessary to solve for certain components of a free body diagram using angles, so this is where trigonometry comes into play. Often times a diagram will provide an angle measurement and the weight or force of tension for one of the vectors. You then have to use the information given to solve for a given component using SOH CAH TOA.

The example above is from a lab we completed towards the end of the unit. We were given the values for the forces of tension (notice the components), and we were told to solve for gravity, the unknown value. Because this object is not in motion, we know that the forces must be balanced on each side, so we can infer that the two Y (vertical) components of friction are each half of the total force of gravity, or weight of the object. We were also given two angle measurements: 23 and 75 degrees. Using the laws of geometry, we know that each quadrant is equal to 90 degrees, making it easy to find the other angle measurements of that quadrant. Then considering what values you have been given, you use either Sine, Cosine, or Tangent to solve for FT1y and FT2y. Because you have the adjacent and hypotenuse, you can use the Cosine of angles 65 and 15. You then simply plug the values into the equation and solve for each vector. After solving you add the values together and that is equal to the weight or gravity of the object. The work is shown in the picture.

Application: I found this unit to be extremely applicable to daily life. Whether we realize it or not, there are always forces acting on each other that create some sort of effect, and learning about how they work greatly changed my perspective. It was super interesting for me to learn that a team wins tug of war not because of how hard they pull, but because of friction, a theory that is extremely counterintuitive. I also found Newton’s Laws to be extremely relevant because they disprove common misconceptions about the way we view interactions between forces.