Questions: HW 04 - Class 08 Vectors Problems 1-3: The first three problems below exist in the 2D rectangular coordinate system (xy-plane), with a horizontal x-axis and vertical y-axis. Positive angles measure clockwise from the (+) x-axis. All forces are in Newtons (N). a) Draw a sketch of the applied forces and resultant force vector R. b) Give your result for R in both rectangular and polar notations. 1. Find the magnitude and direction of the force vector R that results from two applied forces: A=8 N at 90 degrees B=12 N at 0 degrees

Solution Steps

First, we need to decompose each force vector into its rectangular components (x and y components).

• For \(\overrightarrow{\boldsymbol{A}} = 8 \, \text{N}\) at \(90^\circ\):

  • \(A_x = 8 \cos(90^\circ) = 0 \, \text{N}\)
  • \(A_y = 8 \sin(90^\circ) = 8 \, \text{N}\)

• For \(\overrightarrow{\boldsymbol{B}} = 12 \, \text{N}\) at \(0^\circ\):

  • \(B_x = 12 \cos(0^\circ) = 12 \, \text{N}\)
  • \(B_y = 12 \sin(0^\circ) = 0 \, \text{N}\)

Add the components of the two vectors to find the components of the resultant vector \(\overrightarrow{\boldsymbol{R}}\).

• \(R_x = A_x + B_x = 0 + 12 = 12 \, \text{N}\)
• \(R_y = A_y + B_y = 8 + 0 = 8 \, \text{N}\)

Use the Pythagorean theorem to find the magnitude of \(\overrightarrow{\boldsymbol{R}}\).

\[ R = \sqrt{R_x^2 + R_y^2} = \sqrt{12^2 + 8^2} = \sqrt{144 + 64} = \sqrt{208} \approx 14.4222 \, \text{N} \]

The direction \(\theta\) of \(\overrightarrow{\boldsymbol{R}}\) is given by:

\[ \theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) = \tan^{-1}\left(\frac{8}{12}\right) = \tan^{-1}\left(\frac{2}{3}\right) \approx 33.6901^\circ \]

The polar notation of \(\overrightarrow{\boldsymbol{R}}\) is given by its magnitude and direction:

\[ \overrightarrow{\boldsymbol{R}} = 14.4222 \, \text{N} \text{ at } 33.6901^\circ \]

Final Answer