Balanced Torques and Center of Gravity

Purpose: To investigate the conditions for rotational equilibrium of a rigid bar and to determine the center of gravity of a system of masses.

Clamp 1: .02305 kg
Clamp 2: .0212 kg
Clamp 3: .0195 kg
Clamp 4: .0215 kg
m actual = .1469 kg
Meter stick: .091 kg
Balance point: .486 m

The meter stick balanced

2) C2 + .1 kg at .1 m                  C3 + .15 kg at .762 m

Meter stick with the two masses balanced

g*length*(C2+mass) = g*length*(C3+mass)

(9.8)(.386)(.0212+.1) + (9.8)(.276)(.0195+.15)

.45848N = .45846N

% difference = .0002%

- It is important to add the mass of the clamp into the calculations because they are not small enough to be negligible.

3) C3 + .15 kg at .19 m              C2 + .1 kg at .38 m                   C4 + .2 kg at .769 m

Meter stick balanced with three masses

g*length*(C3+mass) + g*length*(C2+mass) = g*length*(C2+mass)

(9.8)(.296)(.0195+.15) + (9.8)(.106)(.0212+.1) = (9.8)(.283)(.0215+.2)

.4917N + .1259N = .6143N

.6176N = .6143N

% difference = .0033%

4) C3 + .15 kg at .19 m              C2 + .1 kg at .38 m                 C4 + m at .86 m

Meter stick balanced looking for mass of meter stick

g*length*(C3+mass) + g*length*(C2+mass) = g*length*(C4+m)

(9.8)(.296)(.0195+.15) + (9.8)(.106)(.0212+.1) = (9.8)(.374)(.215+m)

.4917 +.1259 = (9.8)(.374)(.0215+m)

.6176 = (3.6652)(.0215+m)

.1685 = .0215+m

.147 = m

% difference = .07%

5) BP at .486 m                          New fulcrum at .78 m       C2 + 200 at .9 m

New fulcrum work and picture

- The clamp holding the meter stick shouldn't be included in the mass because it is holding it balanced not cause more or less mass on it to cause it to change the balance point.

Work and picture for masses at set points

Conclusion: This lab we worked with torque. We were able to find an equilibrium for different scenarios with a meter stick. I learned that there are several ways to add the clamps into the amount of mass on the clamp. Also, I learned that torque is force*lever arm. For us it was simple because there wasn't an angle to deal with. Some of the source of error would be the balance support. The piece supporting the stick wasn't exactly level by itself. Another is that there was a hole drilled into the stick. This changes where the balance point should be. Also, the table and the block of wood that the system was on could not have been balanced. To improve this lab, we would need a better meter stick and to make sure everything was level and working correctly.

Human Power

Purpose: To determine the power output of a person.

Vertical distance between the ground floor and the second floor: 4.29 m
Time 1: 13.47 s

Time 2: 13.31 s   

average time: 13.39 s

F = 618 N

delta PE = mgh

delta PE = 618*4.29

delta PE = 2651.22

Power = 2651.22/13.39

Power = 198 watts

Horse power = 198*.00134102209

Horse power = .2655

Stair diagram

Questions:

1. Is it okay to use your hands and arms on the hand railing to assist you in your climb up the stairs? Explain why or why not.

- It is not okay because it can change the time it takes you to climb the stairs. By changing the time, it changes the power exerted. 

2. Discuss some of the problems with the accuracy of this experiment.

- One problem could be the measurement of the stairs. It could have been off a little. Also, the time. Since the person climbing the stairs wasn't timing themselves, they might not have started walking at the exact time the person started the timer. Along with that, when stopping, the timer may not have stopped exactly when they got to the top of the stairs.

Conclusion questions:

1. Two people of the same mass climb the same flight of stairs. Hinrik climbs the stairs in 25 seconds. Valdis takes 35 seconds. Which person does the most work? Which person expends the most power? Explain your answers.

H: (600)(4.29)/25 = 108.14w(.00134102209) = .145 hp

V: (600)(4.29)/35 = 77.47w(.00134102209) = .104 hp

- Work doesn't depend on time so the work is the same. Hinrik exerts more power because it was a shorter amount of time. Power = work/time

2. A box that weighs 1000 newtons is lifted a distance of 20 meters straight up by a rope and pulley system. The work is done in 10 seconds. What is the power developed in watts and kilowatts?

(1000)(20)/10 = 2000w(1kw/1000w) = 2kw

3. Brynhildur climbs up a ladder to a height of 5 meters. If she is 64 kg:

a) what work does she do?

work = mgh

work = (64)(9.8)(5)

work = 3136J

b) what is the increase in the gravitational potential energy of the person at this height?
- It is depended on the potential energy

c) where does the energy come from to cause this increase in PE?
- Chemical energy

4. Which requires more work: lifting a 50 kg box vertically for distance of 2 m, or lifting a 25 kg box vertically for a distance of 4 meters?

A: work = (50)(9.8)(2) = 980J

B: work = (25)(9.8)(4) = 980J

- The work is the same.

Conclusion: In this lab i learned how much power it takes to climb the stairs. I learned that the amount of time it takes is what controls how much power it is. One problem could be the measurement of the stairs. It could have been off a little. Also, the time. Since the person climbing the stairs wasn't timing themselves, they might not have started walking at the exact time the person started the timer. Along with that, when stopping, the timer may not have stopped exactly when they got to the top of the stairs. To make this lab improved, it would be smart to have each person time themselves.

Motion in One Dimension with Air Drag

Purpose: To analyze how changing force affects motion in one dimension.

Introduction Questions:
1) By unit analysis, show that the above equation (Vnew = Vold +aavg*delta(t)) is valid.

m/s = m/s + (m/s^2)*s = m/s

2) Why do we use aavg in equation (1)?

- We assume that acceleration is changing

3) Come up with an analogous equation relating Ynew to Yold.

Ynew = Yold + deltaY

4) What is the benifit of choosing a small delta t?

- For more accuracy

Questions: Fd = -kv (where k is a proportionality constant)

1) Draw a detailed motion diagram of the object falling down.

1a) Now draw a force diagram for the object falling down. Include vectors for all forces, and write a statement of Newton's Second Law. Solve for the acceleration.               Force Diagram

- For every action there is an equal and opposite reaction.

Fnet = Fd - Fg

ma = Fd - Fg

a = (Fd - Fg)/m

a = (-kv - mg)/m

a = (-kv)/m - g

1b) Give the condition for the object at terminal velocity (hit: a = 0). Using the condition solve for k.

0 = (-kv)/m - g

g = -kv/m

mg = -kv

k = -mg/v(terminal)

1c) Substitue k into your expression for the acceleration from part a.

a = -(-gm/v(terminal))v/m - g

a = gv/v(terminal) - g

a = -g(v/v(terminal) +1)

2a) What are the assumptions?

- t = 0 s

- g = 9.8 m/s^2

- v = 100 m/s

- r = 1000 m

- v (terminal) = 40 m/s

- delta t = .1 s

2b) What is v-halfstep?

- Taking the average velocity at two times to find the actual velocity

2c) What is a-halfstep?

- Taking the average acceleration at two times to find the actual acceleration

3) Using the graph paper provided, draw scaled graphs of position vs. time, velocity vs. time, and acceleration vs. time for the object, first assuming no air drag. Make predictions on another sheet of paper about how position and velocity would change if you include air drag.

The position vs. time graph that we predicted is very similar to the actual position vs. time graph. However, the velocity vs. time graph that we predicted is no where near the actual graph. I believe that we accounted for drag wrong and we predicted the graph without drag wrong too.

Now look at a drag force that is dependent on the square of the velocity. Assuming a drag force, Fd = |kv^2|, find the new formula for the acceleration.

Conclusion: This lab was complicated. Even though we did it more as a class, it was still difficult. There could have been a lot of different kinds of error. The fact that most of us didn't really understand it, is cause for a lot of error. I think the only way to do better on this lab is to do it over again.

Centripetal Force

Purpose: To verify Newton's second law of motion for the case of uniform circular motion.

 Table of Trials and Times

The table above shows the five trails for two different masses for the bob. The radius was the same for both masses. To find the time we spun the bob around in a circle 50 times and that gave us our time. The frequency was found by dividing the revolutions (50) buy the time. The time was different for each trial. To find the velocity it is (2pi*50*radius)/time. To find acceleration, it's just velocity/time. To find force we did the (bob mass*velocity^2)/radius.
Once we did that, we found the hanging mass. To find that we hung weight from the bob. The weight was determined by where the weight leveled out the bob to the radius. After found the hanging mass we could find weight, which is just hanging mass*gravity. Then we found the difference between the weight and the force for each trial. As you can tell, for the first bob mass we have under 6% difference for each. Then for the second bob mass, we have under 6.5% difference.

This is the force diagram for the spring. The spring is what the bob is connected to. There are only two forces acting on it. The force of static friction and the force of tension. The forces are equal.

This is the diagram for the hanging weight that held the bob in place at the radius. There are two forces acting on it. The force of tension and the force of gravity. These two forces are equal.















Conclusion: I learned that the hanging weight should be the objective of the force that we found. For sources of error, we could have timed things wrong, measured things wrong, or any other possible human error. To improve we could do more trials and take the best five. When adding mass to the hanging bob, the centripetal force stays the same, but the radius the bob is at changes.

Drag Force on a Coffee Filter

Purpose: To study the relationship between air drag forces and the velocity of a falling body.

This is the prediction for the position versus time graph. Before the fall, there is a constant position because it is being held in the same place. Once it falls, the coffee filters decrease in position moving to the floor (origin). The once it hits the floor, it stays in the same place because there is nothing causing it to come back up.

These are all of our trials for the different number of coffee filters. We started with nine filters and did five trials. Then from there we took one filter away after five trials with the same number of filters. We ended with one coffee filter.

This is the position versus time graph. This is the best one out of all of our trials. We found the best fit line and our slope was -2.004. The actual would be -2.

This is the curve from out plotted points. The y-axis is the number of coffee filters, and the x-axis is the absolute values of slope averages of the five trials for each value of filters. With the power fit line, the A = k and the B = n in the drag equation we used. Drag = k|v|^n. For our value of n, we have a 2.289. The actual is supposed to be a 2. Our percent error is 14.45%.




The equation for drag moving through a fluid is:

drag = .5*density*v^2*area*drag coefficient 

This is equation is similar to ours where velocity is squared. However, in the fluid equation are is used not surface area. To compare the drag fluid equation to the normal drag equation (drag = 1/4 * Av^2) you can see many more similarities.  

Conclusion: We learned that air slows down the fall of objects. Also, the less mass there is the more the air acts upon it. Our error could be from not dropping from the exact same height each time. Not dropping it at the same height does affect the experiment. The higher we drop the coffee filter, the longer it takes to reach the ground. If we had dropped the filters from the exact same height each time, then we would have better numbers. Next time, we can make sure we drop the filters at the exact same height and even do more trials to have more data.

Working With Spreadsheets

Purpose: To get familiar with electronic spreadsheets by using them in some simple applications.

These are pictures of the first spreadsheet we did. We used f(x)=Asin(Bx+C) as our function. We had x start at zero and then go to 10 by 0.1. In the equation, A = amplitude which is 5 in this equation, B = frequency which is 3, and C = phase which is pi/3.
The first picture has the equation shown. In the second picture, we have the numerical value after excel plugged in the numbers.

This is the graph of the plotted points. The x values are that of which we used in excel sheet. It starts at zero and goes to ten by 0.1. The y values are the numerical values we got after excel plugged the numbers into the equation.
The original equation is f(x)=5sin(3x+(pi/3). After doing the best fit quadratic line, out equation comes out to be f(x)=5sin(3x+1.05). This is fairly correct when pi/3 is rounded to two decimal places. We used 1.047198 for pi/3.








These are the excel spreadsheets from the second equation we did, f(x)=A+Bx+Cx^2. In this equation, A is position (x0) which equals 1000m. B is velocity (v0) which equals 50m/s. Lastly, C is gravity which equals -4.9m/s^2. For the x values, x is represented by delta t which equals 0.2. We interpreted this as the x values start at 0.2 and increase by 0.2 as well. We had it go to ten just like the last equation.
The first spreadsheet shows the equation and the second spreadsheet shows the numerical value after excel plugged the numbers into the equation.

This it the graph that comes from the second equation. The x values are delta t which starts at 0.2 and increase by 0.2. The y values are the numerical values given by plugging the numbers into the equation.
Our original equation was f(x)=1000+50x-4.9x^2. The best fit line we got was f(x)=1000+50x-4.9x^2. The equation is the exact same as the one we used.








Conclusion: This was a good lab to do because we were able to practice or learn how to use excel. It was also good because we could see how the different parts of equations affect the graph. In our group we originally had put acceleration (gravity) as positive 4.9m/s^2. Our graph was completely different from the one that was negative. Our parabola was upside down opening up. That was the only problem we had in this lab. Once we fixed the problem, it all made sense.

Vector Addition of Forces

Purpose: To study vector addition by: 1) Graphical means and 2) Using components. A circular force table is used to check results

Our group started with magnitudes 200cm, 100cm, and 150 cm with degrees of 0, 41, and 132 respectively. This is the graph of the vectors given. Thee scale is 1 cm = 20 g.
Vector D is the resultant force. The angle of vector D and the x-axis is 45 degrees.
This graph also shows the x and y components of each vector.





Vector A) Ax = 200g                                       x = 200+100cos(41)+150cos(132) = 175.1g
                Ay = 0g                                            y = 0+100sin(41)+sin(132) = 177.1g
Vector B) Bx = 100cos(41) = 75.5g
                By = 100sin(41) = 65.6g                 R = 250g at 45 degree
Vector C) Cx = 150cos(132) = -100.4g          Rx = 250cos(45) = 176.8g
                Cy = 150sin(132) = 111.5g             Ry = 250sin(45) = 176.8g
 

This graph is solely the x and y components of each vector. The final vector goes from the tail of the first to the head of the last. That vector is the resultant vector.

These pictures are of the circular force table. Our measurements were correct because the first time we put the weights on at the given angles, it was balanced.

This is the picture from the simulation website. using our vectors, we got the same resultant vector.










Conclusion: In this lab we worked with vectors. We found the resultant vector from three vectors given. With this we found that it is balanced with wights and angles that were given. I learned that vectors are basically trig problems. This problem was fairly simple and we didn't have any errors with calculations. We did calculate our vectors two ways, mathematically and with a graphical approach. In our math we got whole numbers but graphically we got numbers that were decimals but could be rounded to the same numbers as ours. Another source of error comes from using the force table. There could have been friction or it could have been unbalanced (in out favor) which would have thrown off the balance. If out calculations were off and the table wasn't balanced but it said it was is why it would be in our favor.