Question

The three displacement vectors in the drawing have magnitudes of A = 5.55 m, B = 6.16 m, and C = 3.30 m. Find the resultant
((a) magnitude and (b) directional angle) of the three vectors by means of the component method. Express the directional angle as an angle above the positive or negative x axis.

+1 20.0° 60.0°

Answer 1

A = $- 5.55 cos 20 \hat{i} + 5.55 sin 20 \hat{j} = -5.215 \hat{i} + 1.898 \hat{j}$

B = $6.16 cos 60 \hat{i} + 6.16 sin 60 \hat{j} = 3.08 \hat{i} + 5.335 \hat{j}$

C = $-3.3 \hat{j}$

Resultant, S = A+B+C = $-2.045 \hat{i} + 3.933 \hat{j}$

Magniture of resultant = $\sqrt{-2.045^{2} + 3.933^{2}}$ = 4.433 m

Angle = $tan^{-} (3.933/2.045) = 62.53^{\circ}$ above negative x axis

Answer 2

A = $- 5.55 cos 20 \hat{i} + 5.55 sin 20 \hat{j} = -5.215 \hat{i} + 1.898 \hat{j}$

B = $6.16 cos 60 \hat{i} + 6.16 sin 60 \hat{j} = 3.08 \hat{i} + 5.335 \hat{j}$

C = $-3.3 \hat{j}$

Resultant, S = A+B+C = $-2.045 \hat{i} + 3.933 \hat{j}$

Magniture of resultant = $\sqrt{-2.045^{2} + 3.933^{2}}$ = 4.433 m

Angle = $tan^{-} (3.933/2.045) = 62.53^{\circ}$ above negative x axis