# 012 Resultant of two velocity vectors

**Problem 012**

Find the resultant vector of vectors **A** and **B** shown in Fig. P-012.

**Solution 012: Component Method**

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$R_x = 44 \cos 50^\circ - 17 \sin 70^\circ$

$R_x = 12.31 \, \text{ m/sec to the right}$

$R_y = \Sigma F_y$

$R_y = -44 \sin 50^\circ - 17 \cos 70^\circ$

$R_y = -27.89 \, \text{ m/sec}$

$R_y = 39.52 \, \text{ m/sec downward}$

$R = \sqrt{{R_x}^2 + {R_y}^2} = \sqrt{12.31^2 + 39.52^2}$

$R = 41.39 \, \text{ m/sec}$

$\tan \theta_x = \dfrac{R_y}{R_x} = \dfrac{39.52}{12.31}$

$\theta_x = 72.70^\circ$

The resultant vector R = 41.39 m/sec downward to the right at θ_{x} = 72.70°.

**Another Solution: Vector Method**

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${\bf A} = -15.97{\bf i} - 5.81{\bf j} \, \text{ m/sec}$

${\bf B} = 44 \cos 50^\circ{\bf i} - 44 \sin 50^\circ{\bf j}$

${\bf B} = 28.28{\bf i} - 33.70{\bf j} \, \text{ m/sec}$

${\bf R} = {\bf A} + {\bf B}$

${\bf R} = (-15.97 + 28.28){\bf i} + (-5.81 - 33.70){\bf j}$

${\bf R} = 12.31{\bf i} - 39.51{\bf j} \, \text{ m/sec}$

$R = \sqrt{{R_x}^2 + {R_y}^2} = \sqrt{12.31^2 + (-39.51)^2}$

$R = 41.39 \, \text{ m/sec}$ (*okay!*)

$\tan \theta_x = \dfrac{R_y}{R_x} = \dfrac{-39.52}{12.31}$

$\theta_x = -72.70^\circ$

$\theta_x = 72.70^\circ$ downward to the right (*ok!*)

**Another Solution: Geometry Method**

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$R^2 = 17^2 + 44^2 - 2(17)(44) \cos 70^\circ$

$R = 41.39 \, \text{ m/sec}$ (*ok!*)

By Sine Law

$\dfrac{R}{\sin 70^\circ} = \dfrac{17}{\sin \alpha}$

$\sin \alpha = \dfrac{17 \sin 70^\circ}{R} = \dfrac{17 \sin 70^\circ}{41.39}$

$\alpha = 22.70^\circ$

$\theta_x = 50 + \alpha = 50 + 22.70^\circ$

$\theta_x = 72.70^\circ$ (*okay!*)

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