nt Forces
Abstract:
Within the experiment my lab partner and I recorded data on the different weights/ masses acting on a ring and how it correlates with the forces acting upon it. When recording we found for the first part that the 3 forces 5.39N@0(Q1), 9.80N@100(Q2), 1.048N@250(Q3), their net components for X=3.33and Y=8.67. Making the resultant force of 9.29N@68.97(Q1). When recorded the 4 forces acting on the ring they were .7056N@0(Q1), 1.387N@100(Q2), 1.274N@210(Q3), .980N@310(Q4). Making the resultant force of 0.0086@-11.44(Q2).
Theory:
In the lab experiment the lab had us record data on forces acting on a ring. We were supposed to balance the ring or put it into “equilibrium”. We added weights of different masses onto the different points for both parts of the experiment. We used those masses to calculate the forces acting and used the angles they were at to find the resultant force to see if the net forces or net component is close to 0.
Given Equations: F=ma Cx=C*Cos Cy=C*Sin m=mass in kg=kilograms C=√((Cx^2)+(Cy^2)) a=acceleration =gravity=9.8m/s^2 (meters per second squared)
Force in newton’s Q=quadrant Angles in degrees
Procedure:
Part1 1. Select three forces, A,B,C 2. Choose different angles for each, for convenience choose one to be at 0 3. Balance the weights so that the ring in the middle is at “equilibrium” 4. Record down the masses that you have at each location 5. Change these masses into forces and record the data into the tables 6. Create a graphical representation of the vector forces acting in the system

Part 2 Repeat all steps; however add a fourth force to the system so you have 4 different weights and 4 different angles.
Force Table 1 | Magnitude | Direction/angle | Y-component | X-component | | A | 5.39N | 0 | 0 | 5.39 | | B | 9.80N | 100 | 9.65 | -1.701 | | C | 1.048N | 250 | -.984 | -.358 | | Net Component | | 68.97 ( Q1) | 8.67 | 3.33 | |
Force Table 2 | Magnitude | Direction/angle | Y-component | X-component | A | .7056N | 0 | 0 | .7056 | B | 1.387N | 100 | 1.365 | -.2408 | C | 1.274N | 210 | -.637 | -1.12 | D | .980N | 310 | -.7507 | 0.6299 | Net Component | | -11.44 | 0.0017 | -0.0084 |

Resultants for Force table 1
Cx=3.33 Cy=8.67
C^2=(3.33^2)+( 8.67^2)= 9.29
Tan^-1(8.67/3.33)=68.97(Q1)
Resultants for Force table 2
Cx= -0.0084 Cy=0.0017 C^2=(-0.0084^2)+(0.0017^2)= 0.0086 Tan^-1(0.0017/-0.0084)=-11.44
Analysis:
Table 1 | Mass | Magnitude | Direction | Y component | X component | Force A | 55g=.055kg | .055kg*9.8=5.39N | 0 | 5.39N*sin(0)=0 | 5.39N*Cos(0)=5.39 | Force B | 100g=.100kg | .100kg*9.8=9.80N | 100 | 9.80N*Sin(100)=9.65 | 9.80N*Cos(100)=-1.701 | Force C | 107g=.107kg | .107kg*9.8=1.048N | 250 | 1.048N*Sin(250)=-0.984 | 1.048N*Cos(250)=-0.358 |

Net Component:
Y-Component: 0+9.65+(-0.984)=8.666
X-Component:5.39+(-1.701)+(-0.358)=3.331
Table 2 | Mass | Magnitude | Direction | X component | Y componet | Force A | 72g=.072kg | .072*9.8=.7056N | 0 | 0.7056*cos(0)=.7056 | 0.7056*sin(0)=0 | Force B | 141.5g=.1415kg | .1415*9.8=1.3867N | 100 | 1.386*cos(100)=-0.2406 |