Fill the table and verify the equation \(\vec{A}+\vec{B}+\vec{C}=\vec{0}\).
Calculate the magnitude of A
Magnitude of A = \(0.200 kg \times 9.8 m/s^2 = 1.96 N\)
Calculate the magnitude of B
Magnitude of B = \(0.150 kg \times 9.8 m/s^2 = 1.47 N\)
Calculate the x and y components of A
\(A_x = 1.96 N \cos(0^\circ) = 1.96 N\)
\(A_y = 1.96 N \sin(0^\circ) = 0 N\)
Calculate the x and y components of B
\(B_x = 1.47 N \cos(90^\circ) = 0 N\)
\(B_y = 1.47 N \sin(90^\circ) = 1.47 N\)
Calculate the x and y components of C
For \(\vec{A} + \vec{B} + \vec{C} = \vec{0}\), we have:
\(C_x = -A_x - B_x = -1.96 N - 0 N = -1.96 N\)
\(C_y = -A_y - B_y = -0 N - 1.47 N = -1.47 N\)
Calculate the magnitude of C
Magnitude of C = \(\sqrt{C_x^2 + C_y^2} = \sqrt{(-1.96)^2 + (-1.47)^2} = \sqrt{3.8416 + 2.1609} = \sqrt{6.0025} \approx 2.45 N\)
Calculate the mass of C
Mass of C = \(2.45 N / 9.8 m/s^2 \approx 0.25 kg\)
Calculate the direction of C
Direction of C = \(\arctan(\frac{C_y}{C_x}) = \arctan(\frac{-1.47}{-1.96}) = \arctan(0.75) \approx 36.9^\circ\)
Since both \(C_x\) and \(C_y\) are negative, the angle is in the third quadrant. Therefore, the direction of C is \(180^\circ + 36.9^\circ = 216.9^\circ\)
\begin{tabular}{|c|c|c|c|c|c|}
\hline vector & mass, kg & magnitude, N & direction & x-component & y-component \\
\hline A & 0.200 & 1.96 & \(0^{\circ}\) & 1.96 & 0 \\
\hline B & 0.150 & 1.47 & \(90^{\circ}\) & 0 & 1.47 \\
\hline C & 0.25 & 2.45 & \(216.9^{\circ}\) & -1.96 & -1.47 \\
\hline
\end{tabular}
Verification: \(A_x + B_x + C_x = 1.96 + 0 - 1.96 = 0\) and \(A_y + B_y + C_y = 0 + 1.47 - 1.47 = 0\), therefore \(\vec{A}+\vec{B}+\vec{C}=\vec{0}\).
\begin{tabular}{|c|c|c|c|c|c|}
\hline vector & mass, kg & magnitude, N & direction & x-component & y-component \\
\hline A & 0.200 & 1.96 & \(0^{\circ}\) & 1.96 & 0 \\
\hline B & 0.150 & 1.47 & \(90^{\circ}\) & 0 & 1.47 \\
\hline C & 0.25 & 2.45 & \(216.9^{\circ}\) & -1.96 & -1.47 \\
\hline
\end{tabular}
\(A_x + B_x + C_x = 0\) and \(A_y + B_y + C_y = 0\), therefore \(\vec{A}+\vec{B}+\vec{C}=\vec{0}\).
Answered
