Questions: Find the direction of the vector sum vec(A)+vec(B) Vector A is 17.6 m long in a 32.8° direction. Vector B is 236 m long in a 63.9° direction. direction (degrees)

Solution Steps

First, we need to break down each vector into its horizontal (x) and vertical (y) components using trigonometry.

For vector \(\vec{A}\):

• \(A_x = A \cdot \cos(\theta_A) = 17.6 \cdot \cos(32.8^\circ)\)
• \(A_y = A \cdot \sin(\theta_A) = 17.6 \cdot \sin(32.8^\circ)\)

For vector \(\vec{B}\):

• \(B_x = B \cdot \cos(\theta_B) = 236 \cdot \cos(63.9^\circ)\)
• \(B_y = B \cdot \sin(\theta_B) = 236 \cdot \sin(63.9^\circ)\)

Calculate the components using a calculator:

\[ A_x = 17.6 \cdot \cos(32.8^\circ) \approx 14.7981 \] \[ A_y = 17.6 \cdot \sin(32.8^\circ) \approx 9.5481 \]

\[ B_x = 236 \cdot \cos(63.9^\circ) \approx 104.6783 \] \[ B_y = 236 \cdot \sin(63.9^\circ) \approx 211.6783 \]

Add the components of \(\vec{A}\) and \(\vec{B}\) to find the components of the resultant vector \(\vec{A} + \vec{B}\):

\[ R_x = A_x + B_x = 14.7981 + 104.6783 = 119.4764 \] \[ R_y = A_y + B_y = 9.5481 + 211.6783 = 221.2264 \]

The direction \(\theta_R\) of the resultant vector is given by:

\[ \theta_R = \tan^{-1}\left(\frac{R_y}{R_x}\right) = \tan^{-1}\left(\frac{221.2264}{119.4764}\right) \]

Calculate \(\theta_R\):

\[ \theta_R \approx \tan^{-1}(1.8514) \approx 61.7^\circ \]

Final Answer