Objective: To determine the relationship among the forces acting on an object which is remaining at rest for a long time.

Diagrams: Diagram is attached in the back of the lab report

Data:
Table 1 | Magnitude (N) | Angle (degrees) | Orientation | Uncert. in magnitude (N) | Uncert. in angle(degrees) | F1 | 2.2 | 5.1 | Counter clockwise from +x | ± 0.1 | ± 1.15 | F2 | 1.6 | 50.2 | Clockwise from -x | ± 0.1 | ± 1.15 | F3 | 1.8 | 46.2 | Clockwise from -y | ± 0.1 | ± 1.15 |

Table 2 | x components (N) | y components (N) | F1 | 2.2 | 0.2 | F2 | -1.0 | 1.2 | F3 | -1.3 | -1.2 | Sum of two forces F1 + F2 | 1.2 | 1.4 | The force – F3 | 1.3 | 1.2 | Sum of two forces F1 + F3 | 0.9 | -1.0 | The force – F2 | 1.0 | -1.2 | Sum of two forces F2 + F3 | -2.3 | 0.0 | The force –F1 | -2.2 | -0.2 |
Analysis:
1. To calculate the uncertainty in magnitude we divided the smallest scale of the instrument (scale) and divided it by 2. Which gives us the following,
Smallest scale of a scale is 0.2 N,
0.22= ± 0.1 N

2. Calculate the uncertainty in angle we divided the smallest scale of the instrument (protractor) and divided it by 2. To that, we added the uncertainty of ruler, the uncertainty of position (± 0.5 N) and the uncertainty of the scale (± 0.1 N). Which gives us the following,
Smallest scale of a protractor is 1°,
12=± 0.5 ° Smallest scale of a ruler is 0.1 cm,
0.12= ± 0.05 cm
The uncertainty of angle is,
0.5 + 0.05 + 0.5 + 0.1 = ± 1.15 °

3. To calculate the x component of F1, F2 and F3, it consisted of multiplying the magnitude of the force 1, 2 and 3 by sinus or cosines, depending of the angle that will be used to calculate the component, and by the value of the angle.
For example, to calculate F3x we used this equation:
F3sinθ=
1.8sin46.2=
-1.3 N

4. To calculate the y component of F1, F2 and F3, it consisted of multiplying the magnitude of the force 1, 2 and 3 by sinus or cosines, depending of the angle that will be used to calculate the component, and by the value of the angle.
For example, to calculate F3y we used this equation:
F3cosθ=
1.8cos46.2 =
-1.2 N

5. To calculate the sum of two forces either for F1 + F2 or F1 + F3 or F2 +F3, we simply added both magnitudes together.
For example, the sum of F1 + F2 would have to be,
2.2 + (-1.0) =
1.2 N

Results: 1. As for the x components, the sum of forces F1 + F2 = 1.2 N. However, the x component of the force –F3 is equal to 1.3 N. 2. As for the y components, the sum of forces F1 + F2 = 1.4 N. However, the y component of the force –F3 is equal to 1.2 N. 3. As for the x components, the sum of forces F1 + F3 = 0.9 N. However, the x component of the force –F2 is equal to 1.0 N. 4. As for the y components, the sum of forces F1 + F3 = -1.0 N. However, the y component of the force –F2 is equivalent to -1.2 N. 5. As for the x components, the sum of forces F2 + F3 = -2.3 N. However, the x component of the force –F1 is equivalent to -2.2 N. 6. As for the y components, the sum of forces F2 + F3 = 0.0 N. However, the y component of the force –F1 is equal to -0.2 N.

As we can see, it seems like the difference between the sum of two forces and the opposite of the resultant force is about 0.1 N for the x components. As for the y components the difference is about -0.2 N.

Conclusion/Discussion:
To conclude, the purpose of this lab was to determine the relationship between the 3 forces and also the object when it’s at rest. We noticed that in table 2 of the Data section, the sum of the x component and y component of F1and F2 are equal to the components of F3 but negative. So we can confirm that F1+F2=-F3. This rule also applies for F1+F3=-F2 and F2+F3=-F1. This also explains why the object looked like it wasn’t moving. All these 3 forces make the object balance.

Sources of errors: