Solution

Engineering Mechanics (Dynamics)

Chapter 12

Exercise 231

vw​=vs​+vw/s​

$$10\cos 30^{\circ }-10\sin 30^{\circ }=20\cos 45^{\circ }+20\sin 45^{\circ }+{\mathbf{v}}_{w/s}$$

$$\bold{v}_{w/s} = \{-5.48\bold{i} - 19.14\bold{j}\}\mathrm{~m/s}$$

So that,

$$v_{w/s} = \sqrt{(-5.48)^2 + (-19.14)^2} = \boxed{\color{#4257b2}19.91\mathrm{~m/s}}$$

$$\tan\theta = \frac{(v_{w/s})_y}{(v_{w/s})_x} = \frac{19.14}{5.48}$$

$$\boxed{\color{#4257b2}\theta = 74.02^\circ~~~c}$$

Answer