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Lab 2 Force and Vectors

Physics 20700 Section GH2

Introduction:

In this lab we are conducting experiments around the force table with the goal of understanding vector addition using graphical and analytical methods. We conducted several experiments and calculations with the force table to test our theories on how to maintain/reach equilibrium with masses in different angles of the force table with different variables and concludes with a formula generalizing the mass needed to balance the system.

Experiments:

The first exercise was to gain a basic familiarity of the force table, we started by calculating the tension on the system using the formula T=mg, with mass being the weight and the pan.

EXP.1:

In experiment number 1, two pans, both with an additional 50g weight attached are placed 180 degrees apart, with one on 0 degrees and other on the 180. We then started adding weights to mass two and observed the point in which the center ring shifted. Second part of the experiment we placed a mass 234 degrees counterclockwise from 0 degrees, we then use the resultant, the angle that the center ring is being pulled toward and take the negative value, or plus/minus 180 degrees from the value to gain the equilibrant by using the formula E=equilibrant=-R.

EXP.2:

In experiment number two, we set up the force table using the configuration of mass one with 30 grams at 30 degrees and mass two with 30 grams at 330 degrees. We then balanced out the system by placing our third mass at the opposite of the resultant point, after that we added weight until the system reached equilibrium. We then use the web simulation provided by the lab website and determine the difference between our experiment and the simulation.

EXP.3:

In the third experiment, we collected a series of data to determine the function of mass three by using mass one and two as constant and changing the angle between mass one and two and determine the weight of mass three necessary to reach equilibrium. The next exercise in this experiment is to determine the angle necessary to meet certain conditions in the relationship between the x component of the vector and the y component of the vector. The final exercise of the experiment is to manipulate the forces acting on a plane so that the velocity stays the same. (i.e Fnet=0.)

EXP.4:

In this experiment we are given three vectors and using algebra determine an unknown vector D so that the system maintains equilibrium.

Results:

Ex1

20+50=70g

70g x 1kg/1000g = .07kg

T=F=m/g=.07/9.8=0.007N

EXP1

M1=50g @ 0 degrees

M2= 50g @ 180 degrees

Mass needed to break equilibrium = 5g

Resultant Vs. Equilibrant

Direction of equilibrant

R=300, -R=-300=120degrees

EXP2

M1=30g @ 30degrees

M2=30g @ 330degrees

Mass needs to balance the system 55g @ 180degrees.

Result from simulated force table 0.06N or 60 g.

EXP3

Angle	Mass of Pan System (g)
5	180
10	172
15	165
20	162
25	156
30	143
35	135
40	122
45	110
50	95
55	80
60	64
65	45
70	25
75	10
80	0

P1 cos(b)+P2 cos(b)=P3

P1=m+mpan=P2

2（m+mpan）cos(b)=p3

2(p1+ps)cos(b)=P3

Vector Component

By maniupulating the graph

We determine that ax=2ay when angle = 26.75

And ay=0.5IaI when angle = 30.75

EXP4

 

 

Fa= 30g x 1kg/1000g x 9.8 m/s^2

Fa= 0.294N

Ax= 0.294 cos(5)=0.293 N

Ay = 0.294 sin(5) = 0.0256 N

Fb= 40g x 1kg/1000g x 9.8 m/s^2

Fb= 0.392N

Bx= 0.392 cos(105)= -0.1015 N

By = 0.392 sin(105) = 0.3786 N

Fc= 25g x 1kg/1000g x 9.8 m/s^2

Fc= 0.245N

Cx= 0.294 cos(195)=-0.2367 N

Cy = 0.294 sin(5) = -0.0634 N

A+B+C+D=0

D=-(A+B+C)

Dx=-(0.293+-0.1015+-0.2367)= 0.0452N

Dy=-(0.0256+0.3786+-0.0634)= 0.3408N

Report Questions

1. What factors could contribute to this sensitivity?

Factors that could contribute to this sensitivity are the friction of the wheels and the line.

2. Report the difference between what you've experimentally measured and what the simulation predicted. Are they within the expected sensitivity of the instrument?

There is an 5 g difference between the experiment and the simulation, it is within the sensitivity of the instrument.

3. On one graph plot the experimental data from your table along with the analytical prediction of the function you found. Do they follow the same trend?

They did follow the general downward slope. Within reasonable difference of 5 grams.

 

4. Give the details of this calculation and compare your analytical results with the the experimental results.

Fa= 30g x 1kg/1000g x 9.8 m/s^2

Fa= 0.294N

Ax= 0.294 cos(5)=0.293 N

Ay = 0.294 sin(5) = 0.0256 N

Fb= 40g x 1kg/1000g x 9.8 m/s^2

Fb= 0.392N

Bx= 0.392 cos(105)= -0.1015 N

By = 0.392 sin(105) = 0.3786 N

Fc= 25g x 1kg/1000g x 9.8 m/s^2

Fc= 0.245N

Cx= 0.294 cos(195)=-0.2367 N

Cy = 0.294 sin(5) = -0.0634 N

A+B+C+D=0

D=-(A+B+C)

Dx=-(0.293+-0.1015+-0.2367)= 0.0452N

Dy=-(0.0256+0.3786+-0.0634)= 0.3408N

Draw a vector diagram that shows the table arrangement.

  

Conclusion

After conducting the experiment, we concluded that the sensitivity of the force table is 5g, which could be caused by friction of the wheel and the line. The difference between simulated results and experimental results is 5g which is the sensitivity of the force table. We then use the data collected from experiment to determine the relationship between angle b, which is half the angle between mass one and two, and the mass three that is needed to reach equalibrim for the system is 2(m1+m2)cos(b)=m3. And finally we solve A+B+C+D=0 with D being unknown and reach the result of Dx=0.0452N and Dy=0.3408N.