Download - Force Table Lab
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.