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.