Name: Briana Blessitt, Alexa Marshall, Narish Singh Period: 9

Physics Section/Group: 9

Mr. Bradshaw

20 January 2012

Force Table Lab

CASE 1

vector Mass kg Magnitude N Direction (deg) x-component y-component

1 .2 1.962 0 1.962 0

2 .2 1.962 120 -.981 1.699

3 .2 1.962 240 -.981 -1.699

TOTAL 0 0

CASE 2

vector Mass kg Magnitude N Direction (deg) x-component y-component

1 .2 1.962 0 1.962 0

2 .1 .981 90 0 .981

3 .24 2.3544 206 -2.116 -1.032

TOTAL -.154 -.051

CASE 3

vector Mass kg Magnitude N Direction (deg) x-

component

y-

component

1 .24 2.3544 0 2.3544 0

2 .26 2.5506 150 -2.2089 1.2753

3 .14 1.3734 270 0 -1.3734

TOTAL .1455 -.0981

CASE 4

vector Mass, kg Magnitude,

N

Direction (deg) x-component y-component

1 .31 3.0411 0 3.0411 0

2 .22 2.1582 140 -1.6533 1.3873

3 .21 2.0601 225 -1.4567 -1.4567

TOTAL -.0689 -.0694

CASE 5

vector Mass, kg Magnitude,

N

Direction (deg) x-component y-component

1 .1 .981 300 .4905 -.8496

2 .1 .981 308 .6040 -.7730

3 .22 2.1582 127 -1.299 1.7236

TOTAL -.2045 .101

CASE 6

vector Mass, kg Magnitude,

N

Direction (deg) x-component y-component

1 .15 1.4715 99 -.2302 1.4534

2 .15 1.4715 21 1.3738 .5273

3 .22 2.1582 240 -1.0791 -1.8691

TOTAL .0645 .1116

Analysis:

1. What do you think your x and y components should have added up to and why?

The x and y components should have a resultant of zero because the net force between

the three vectors should be zero newton. The forces acting on the ring should be both

horizontally and vertically equilibrant. The ring at the center of the force table would be

at equilibrium (or at rest) since there is no force to move the ring around. This can be

represented by the equation A + B + C = R (in which R is the resultant vector).

2. Did they add up to it? Where do you think error might have entered into your

experiment?

The x and y components did not add up to zero. Possible sources or error within this

experiment may have been caused by an unbalanced force table, miscalculations of the

forces added and the angles, and not accounting for the weight of the string

3. This lab was an exercise in vector analysis. Where in everyday life could an

understanding of vector analysis be useful? Specifically address about how the

vectors are used and analyzed. (They don’t have to be force vectors.)

An understanding of vector analysis can be applied to the motion a plane experiences

while in flight. For example, a plane traveling northwest from New York to Canada at

100 m/s has both x and y components. To accurately calculate the plane’s displacement

and distance during flight a Global Positioning System (GPS) is normally used. It graphs

the planes displacement for Point A to Point B by visually indicating the x and y axis.

This is very important because without proper direction and visual representation a pilot

would not be able to arrive at a specific destination.

Conclusion

In this lab, we learned vectors have both direction and magnitude. Vectors can be

used to analyze situations in which forces are displaced. An understanding of vector

analysis can also apply to everyday scenarios such as commutes to schooling, taking the

elevator, and climbing up staircases. We identified the properties of an object or objects

at static equilibrium. However, our data for Cases 2 through 6 indicated that our forces

were not truly equilibrant; hence resulting in to a vector sum not equal to zero. Possible

sources or error within this experiment may have been caused by an unbalanced force

table, miscalculations of the forces added and the angles, and not accounting for the

weight of the string. This lab was very effective in demonstrating the principle of Static

Equilibrium and resultants of multiple vectors.

Works Referenced

Serway, R., & Faughn, J. (2006). Physics. (pp. 81 - 106). Austin, Texas: Holt, Rinehart,

and Winston.