Kevin Tran
Lab Partners: Kevin Nguyen, Jose Rodriguez
March 26, 2017

Friction Forces

Purpose: To find out how to find the static and kinetic friction of an object. Students performed several experiments in order to see what role friction plays on the object that is sliding over different types of surfaces.

Introduction/Theory: My lab partners and I have done 5 experiments to determine both static and kinetic frictions at certain cases. We calculated friction by using a two-mass system and by inclining a wooden board.

(1) 

First, we performed an experiment by using a two-mass system in order to help us determine the coefficient of static friction between the block and the table. (picture below). In order to do so, we use the equation

Coefficient of static friction = (maximum force of static friction)/ (normal force)

To find the coefficient of static friction, we need to find the maximum force of static friction and the normal force exerted on the object.

We chose two masses: a block(0.173kg) and a hanging mass(0.55kg).

During this experiment, we tested how much weight will be needed on the hanging mass in order to determine the static friction of the block. We have done 6 trials: block itself, block with 200g, block with 400g, block with 600g, block with 800g, and block with 1000g. With each trial, we continuously added weight onto the hanging mass until the block moves.

Overall, these are the trials for each mass of block and mass required to move the block in order to move.

With these results, we created a mass(x-block) and mass(y-hanging mass) graph on loggers pro in meters so that when we make a line fit of the plot, the slope gives us the coefficient of static force.

In this graph, we were able to find the slope of the graph, which also represents μs(static friction). The slope of the graph is 0.6025.

(2)

In the 2nd experiment, we were required the measure the coefficient of kinetic friction between the block and table. We use this equation to do so:

Coefficient of kinetic friction = (Force of kinetic friction)/ (Normal force on the object)

In the 2nd experiment, we were told to find the kinetic friction. To find kinetic friction, we had to move the block with a dual-range force sensor. Both of these objects will be connected to a string for more accuracy.

We have done 4 trials while performing this experiment. We used the following weights: block(0.173kg), block with 200g, block with 400g, and block with 600g. We set the force sensor on the 10-N range.

Since were pulling the block in a constant velocity, our acceleration will be 0m/s^2

We also have the data for the 4 trials. We used the mean value from the white boxes.

With this information, we were able to construct a weight and force(tension) graph. We took the mean value the of the pulling force in the interval when the block moved horizontally at constant speed. We plotted the average kinetic friction force. We used the 4 different masses for the x-axis(also our normal force) and force pulled on the x-axis and made a line fit of the plot in order to find the slope of the line, which will give us the coefficient of kinetic friction.

With this graph, we were able to find the slope of the coefficient of kinetic friction(Mu), which results in 0.2786.(slope)

Below is a closer look of the white box.

(3)
The 3rd experiment was to incline a wooden board to know what angle will make the block slide down. We used an angle app on our phones and taped it to the wooden board to get the angle.

In the end, the block starts to slide down the board when the board is at a incline of 26 degrees.

We also calculated in order to find static friction(Mu) of the surface at 26 degrees and we ended up getting 0.488.

(4)
Experiment 4 of the lab is similar to experiment 3 but this time, we are looking for kinetic friction when the block sliding down the incline at 26 degrees. When the block is sliding down, there is a motion detector at the top of the incline steep enough that a block will accelerate down the incline.

We have created a position/time graph and a velocity/time graph in order to find the acceleration of the block when it's accelerating. From the velocity/time graph, we were able to find the slope of when the block begins to accelerate. In the end, we end up getting 1.607 m/s^2.



We also found the kinetic friction of the surface as the block slides down. Our kinetic friction(Mu) value happens to be 0.295

(5)
Experiment 5 was to predict the acceleration of a two-mass system. We set up the apparatus similar to part 1 of the lab. Except we placed the motion sensor behind the object to record the acceleration for the block. We solved for the kinetic friction by setting the direction of acceleration of the block as the x-axis and set the y-axis perpendicular to the x-axis. We needed to solve for the vertical and horizontal forces in order to find the kinetic friction force and normal force. We apply the formula for kinetic friction by dividing the kinetic friction force by the normal force in order to find the coefficient of kinetic friction.

This is the set up for experiment 5. (picture below). The reasoning for the white tape that is connected to the block is because it makes the block to detect for the motion sensor when the block accelerates away from the motion detector.

Mass of block= 0.188 kg
Mass of hanging mass= 0.09 kg

For this experiment, we created a position/time graph and a velocity/time graph in order to give us the acceleration of the block when it is moving. From the velocity/time graph, we found that the acceleration of the two-mass system is 0.7627 m/s^2 from the slope.

Below, we have found the kinetic friction(Mu) of the two-mass system of 0.3704



Conclusion:
This lab allowed us to understand how static and kinetic friction works. The value of static Mu from experiment 1 is greater than the value from experiment 3. The difference between these results came up because the experiment 1 tends to be inconsistent. Even though the mass of the block remains the same, it falls with different masses added on the hanging mass. In the comparison to experiment 2 and 4, the value of kinetic Mu is close to each other but not exact. Also the value of kinetic Mu in experiment 2 is closer to experiment 4's result than experiment 5. Even though neither static Mu nor kinetic Mu values are similar, the results of experiment 1 and 3 were both greater than the results from experiments 2, 4, and 5, allowing the lab to be valid.

Kevin Tran
Lab partners: Kevin Nguyen, Jose Rodriguez
March 18, 2017

Trajectories

        Purpose: To use your understanding of projectile motion to predict the impact point of a ball on an inclined board.

       Introduction: We will place a metal ball on an inclined aluminum "v-channel" that is connected to an apparatus and as it rolls down, it will roll on an aluminum "v-channel" parallel to the table and free fall due to gravity onto a metal board after leaving the aluminum "v-channel". First, we will be using a piece of carbon paper to mark where the metal will land each time for five trials. Later on, we will add a metal board into the experiment and our theory is that once the ball makes first contact with a metal board, that will be the distance that the ball has traveled in the x direction.

Here is the visual display of the lab: (picture below).

First, we measured the height of the exit where the ball will drop from. (picture below).

Now, in order to find the distance from the exit to the paper, we needed to use a string to find the origin of where the x distance will begin from.

Once, we placed our string, we begun the 5 trials of the metal ball on the incline. (picture of starting point).

After 5 trials, the metal ball in our case landed on the same exact spot. We measured the distance from the table to the location that the ball has landed.

After doing for the first part of the lab, we have come to a conclusion that we have the following results:

Height: 0.943m +/- 0.001m

Distance: 0.911m +/- 0.001m

With the results, we calculated the velocity of 2.07 m/s (picture below).

Now onto part 2 of the lab, which is to insert a metal board onto the table like so (picture below).

First, we tried finding the angle of the metal board. By using an angle app from our phones and a tool to find an angle, we have found an angle of 24.6 degrees.

We redone the 5 trials for the metal balls, but we included a metal board. As a result, the metal ball has landed on the same spot five times. The distance from the table to the mark is 0.44m +/- 0.001m

(work displayed below).

Now, I will display my work to find uncertainty for d. (work displayed below).

Overall, with uncertainty, the distance that the object traveled is 0.44m +/- 0.01m

Conclusion: The lab allowed us to understand how we can calculate how far an object has traveled after free falling. Also, this lab helps us understand how gravity works as it is in moving throughout the air.

Kevin Tran
Lab Partners: Kevin Nguyen, Jose Rodriguez
March 17, 2017

Modeling the fall of an object falling with air resistance

Purpose: Determine the relationship between air resistance force and speed.

Introduction: My partners and I went to the Design Technology building to video capture our teacher dropping coffee filters with different masses. He will be dropping coffee filters with 1 stacks, 2 stacks, 3 stacks, 4 stacks, and 5 stacks. We will record 5 different drops as the coffee filter drops down due to gravity and air resistance pulling the object in the opposite direction to slow the object down. In theory, if the object is heavier, the velocity will increase and so will the air resistance.

The formula for air resistance is F=kv^n 

F=Air resistance

k= shape and area of the object

v= velocity

n= trials

The location where the lab occurred. (picture below). Our teacher dropped the 5 different masses during this experiment.




Now, I will display all the position time graph in the order of coffee filters (1-5) with air resistance. Let's first note that each coffee filter is 0.878g +/- 0.02g

Coffee Filter 1 stack:

Coffee Filter 2 stacks:



Coffee Filter 3 Stacks:



Coffee Filter 4 stacks:

Coffee Filter 5 stacks:



These 5 graphs represent the position( x and y included) and time graph.

In order to find the values for k (The shape and area of the object) and n (trials) , we have to create a weight and velocity graph with all the 5 different coffee filter weights. (picture below).

Below, we created a weight and terminal velocity graph in order to find the k and n values.


k in this case would be "A:". Where k is 0.003054N(s/m) +/- 0.001011 N(s/m)
n in this case would be "B:". Where n is 2.555 +/- 0.3504


Now, I will be displaying the numerical results of each of the 5 weights in their corresponding order.

Coffee Filter 1 stacks:



Coffee Filter 2 stacks:

Coffee Filter 3 stacks:


























Coffee Filter 4 stacks:



Coffee Filter 5 stacks:



























Note: The yellow mark represents its terminal velocity.

In the 5 graphs we have the variables: delta t (seconds), mass (kg), gravity (m/s^2), k (N*s/m), and n.

With this information and the position/time graph, we are able to find each of the variables: time(seconds), velocity(m/s), net force(N), Acceleration(m/s^2), delta v, average velocity, and delta x.

This model works because we are given all the approximate results of all the variables that we acquire to determine how much air resistance can affect velocity as it is moving.

In order to derive k and n, we do the following (work below).



Now, we will find the uncertainty of the coffee filter.

For the weight of the coffee filter, we have an relative uncertainty of 2.28% ((0.02g / 0.878g) x 100) and a relative error of -2.22%((0.878g -0.898g)/ 0.898g x 100)

Conclusion: The lab has taught us how effective air resistance is to the velocity of a moving object. Our theory was correct, the faster the object moves, the more air resistance will push against the object. In addition, the heavier the object, the faster the object will go, which will lead to more air resistance going against it.

Kevin Tran
Lab Partners: Kevin Nguyen, Jose Rodriguez
March 11, 2017
         

Non- Constant acceleration problem/Activity

          Purpose: Find out how far the elephant goes before coming to rest.

          Introduction: A 5000-kg elephant is going 25 m/s on frictionless roller skates. It will go down a hill and arrive at ground level. A 1500-kg that is attached to the elephant's back will generate a constant 8000 N thrust opposite to the elephants direction of motion. In addition the rocket is burning fuel at a rate of 20 kg/s. To find the distance that the elephant traveled before rest, we need to find the time at which the velocity of the elephant is zero.

Newton's 2nd law gives us the acceleration of the elephant + rocket as a function of time (Picture on the left).

In the picture below are our results of all the calculations for each variable.

We will now look into the data more carefully. We used the numerical approach to get the distance that the elephant traveled.

First, we used a set of numbers in the time column, but we decided to use the first 24 seconds to guess where the elephant might be within the 24 second range.

In the next column(picture below), we found our accelerations for each second interval within the 24 seconds. By applying the acceleration formula that we found (formula at top of the page), we are able to find the acceleration for each second. Since we know that the rocket is trying to push the elephant in the opposite direction, the acceleration will be negative.

Next, we have the chart for delta v and velocity within the 24 seconds time interval. 

In order to find delta v, we use the formula to find delta v. (picture below). To find delta v, we use the accelerations from one time to another. 

Here is an example to find delta v1 (picture below). I use the collected data from my acceleration chart and applied it into the equation.

To find velocity, we add up delta v and velocity. Since the elephant has an initial velocity of 25 m/s, the velocity at t=0 is still 25 m/s. As every second progress, we add the current velocity at each second interval with the delta v at each time interval.

For example:

Next, we have the average acceleration chart at each time interval of 24 seconds. (picture below). To find average acceleration, we use the formula provided below.

Next, we have the chart for delta x and position for the first 24 seconds. Since we know that the elephant is at the starting point, when t= 0, x= 0 m

To find delta x, we use the formula to find delta x. (picture below).

Here is an example to find delta x1 (picture below). I use the collected data from my velocity chart and applied it into the equation.

To find distance, we add up delta x and distance. Since the elephant is at the starting point, it is starting at x=0. As every second progress, we add the current distance at each second interval with the delta x at each time interval.

For example: 

Next, we have the average velocity chart at each time interval within the 24 seconds. To find average velocity, we use the formula provided below. 

To recap, in order to find the distance that the elephant traveled, we have the know when velocity=0. In between the 19th and 20th row on the velocity column, we can see the velocity is a little above 0 (0.90392744) and below 0(-0.405405). Now, we look into the x column. In between 19 and 20 seconds, we took both x values and divided by 2 to find an approximate distance and we have come to a conclusion that the elephant traveled approximately 248.50 meters.

Analysis:

1. In comparison to our numerical approach and the analytical approach, we have calculated and got a result of 248.50 meters by using the numerical approach. By analytical approach, we have got the result of 248.7 meters. In comparison to the two results, they were .20 off from each other but incredibly close to each other.

2. In order to know when the time interval we chose for doing the integration is "small enough" it can be found when time is closest to 0. If we did not have our analytical results, we would have to reduce the time interval from 1 second to something smaller like 0.01 seconds.

3.When the elephant initial mass is 5500kg, has a rocket mass of 1500 kg, has the fuel burn rate of 40 kg/s and the thrust force is 13000 N at 25 m/s, the elephant distance is 164.03 m by doing the analytical approach. After doing all the steps, I have plugged in my time of 12.96 seconds into the expression of x(t). (picture below).

          In conclusion, we were able to find out how far the elephant has traveled. Around 19 seconds, the elephant has traveled 248.50 meters according to our numerical approach. This lab has taught us how to find the distance of anything with a certain amount information. We were able to apply numerous formulas to find the results that we needed.