Questions: Lab 2 Intro to Vectors B. Graphical Addition of Vectors ( 30 pts ) Table 2: Graphical Force Data (11 pts) Scale 7^7 cm=1.0 N Force Magnitude of Force (N) Magnitude of Force converted (cm) Direction of Force (deg) F1 .98 6.86 0 F2 1.96 15.72 145 F3 1.2793 8.95 300 FR 1.2793 6.9 20 1. Use the graphical head to tail method of vector addition to add the two force vectors F1 and F2. Show conversions using scale given in Table 2. Record the magnitude and direction of the graphical resultant force, FR, in Table 2. ( 11 pts ) 2. On the same sheet of graph paper and using the same scale as in the previous exercise, draw an arrow that represents the balancing or equilibrium force, F3 and record in Table 2. (2 pts) 3. How well do the magnitude and direction of the experimentally determined resultant FR (in the fourth row of Table 1) agree with the magnitude and direction of the graphical head to tail resultant FR (in fifth row of Table 1). Show a percent difference for each. ( 6 pts )

Lab 2 Intro to Vectors B. Graphical Addition of Vectors ( 30 pts ) Table 2: Graphical Force Data (11 pts) Scale 7^7 cm=1.0 N Force Magnitude of Force (N) Magnitude of Force converted (cm) Direction of Force (deg) F1 .98 6.86 0 F2 1.96 15.72 145 F3 1.2793 8.95 300 FR 1.2793 6.9 20 1. Use the graphical head to tail method of vector addition to add the two force vectors F1 and F2. Show conversions using scale given in Table 2. Record the magnitude and direction of the graphical resultant force, FR, in Table 2. ( 11 pts ) 2. On the same sheet of graph paper and using the same scale as in the previous exercise, draw an arrow that represents the balancing or equilibrium force, F3 and record in Table 2. (2 pts) 3. How well do the magnitude and direction of the experimentally determined resultant FR (in the fourth row of Table 1) agree with the magnitude and direction of the graphical head to tail resultant FR (in fifth row of Table 1). Show a percent difference for each. ( 6 pts )
Transcript text: Lab 2 Intro to Vectors B. Graphical Addition of Vectors ( 30 pts ) Table 2: Graphical Force Data (11 pts) \begin{tabular}{|c|c|c|c|} \hline Scale & \multicolumn{3}{|l|}{${ }_{7}^{7 c m=1.0 N}$} \\ \hline Force & \begin{tabular}{l} Magnitude of Force \\ (N) \end{tabular} & \begin{tabular}{l} Magnitude of Force \\ converted (cm) \end{tabular} & \begin{tabular}{l} Direction of Force ${ }^{\dagger}$ \\ (deg) \end{tabular} \\ \hline$\vec{F}_{1}$ & .98 & 6.86 & 0 \\ \hline$\vec{F}_{2}$ & 1.96 & 15.72 & 145 \\ \hline$\vec{F}_{3}$ & 1.2793 & 8.95 & 300 \\ \hline$\vec{F}_{R}$ & 1.2793 & 6.9 & 20 \\ \hline \end{tabular} 1. Use the graphical head to tail method of vector addition to add the two force vectors $\vec{F}_{1}$ and $\vec{F}_{2}$. Show conversions using scale given in Table 2 . Record the magnitude and direction of the graphical resultant force, $\vec{F}_{R}$, in Table 2. ( 11 pts ) 2. On the same sheet of graph paper and using the same scale as in the previous exercise, draw an arrow that represents the balancing or equilibrium force, $\vec{F}_{3}$ and record in Table 2. (2 pts) 3. How well do the magnitude and direction of the experimentally determined resultant $\vec{F}_{R}$ (in the fourth row of Table 1) agree with the magnitude and direction of the graphical head to tail resultant $\vec{F}_{R}$ (in fifth row of Table 1). Show a percent difference for each. ( 6 pts )
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Solution

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Solution Steps

Step 1: Convert Forces to Vector Components

To add the vectors \(\vec{F}_{1}\) and \(\vec{F}_{2}\) using the head-to-tail method, we first convert them into their vector components.

For \(\vec{F}_{1}\):

  • Magnitude: \(0.98 \, \text{N}\)
  • Direction: \(0^\circ\)

\[ \vec{F}_{1x} = 0.98 \cos(0^\circ) = 0.98 \] \[ \vec{F}_{1y} = 0.98 \sin(0^\circ) = 0 \]

For \(\vec{F}_{2}\):

  • Magnitude: \(1.96 \, \text{N}\)
  • Direction: \(145^\circ\)

\[ \vec{F}_{2x} = 1.96 \cos(145^\circ) = -1.6051 \] \[ \vec{F}_{2y} = 1.96 \sin(145^\circ) = 1.1285 \]

Step 2: Calculate Resultant Vector Components

Add the components of \(\vec{F}_{1}\) and \(\vec{F}_{2}\) to find the resultant vector \(\vec{F}_{R}\).

\[ \vec{F}_{Rx} = \vec{F}_{1x} + \vec{F}_{2x} = 0.98 - 1.6051 = -0.6251 \] \[ \vec{F}_{Ry} = \vec{F}_{1y} + \vec{F}_{2y} = 0 + 1.1285 = 1.1285 \]

Step 3: Calculate Magnitude and Direction of Resultant Vector

Calculate the magnitude and direction of the resultant vector \(\vec{F}_{R}\).

Magnitude: \[ |\vec{F}_{R}| = \sqrt{(-0.6251)^2 + (1.1285)^2} = 1.2881 \, \text{N} \]

Direction: \[ \theta_R = \tan^{-1}\left(\frac{1.1285}{-0.6251}\right) = 118.3^\circ \]

Final Answer

The magnitude of the graphical resultant force \(\vec{F}_{R}\) is \(1.2881 \, \text{N}\) and the direction is \(118.3^\circ\).

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