## Engineering Mechanics: Statics & Dynamics (14th Edition)

$v_B=12.6058in/s$ $\theta=65.657^{\circ}$
The required velocity can be determined as follows: We know that $\vec{r_{B/G}}=\frac{1.5}{sin45}\hat i=2.1213\hat i$ and the velocity at point G is $\vec{v_{G}}=-6cos30\hat i+6sin30\hat j=-5.196\hat i+3\hat j$ Now, $\vec{v_B}=\vec{v_G}+\vec{\omega}\times \vec{r_{B/G}}$ We plug in the known values to obtain: $\vec{v_B}=(-5.196\hat i+3\hat j)+(4\hat k)\times(2.1213\hat i)$ $\implies \vec{v_B}=-5.196\hat i+11.4852\hat j$ The magnitude of the velocity is given as $v_B=\sqrt{(v_{Bx})^2+(v_{By})^2}$ We plug in the known values to obtain: $v_B=\sqrt{(-5.196)^2+(11.4852)^2}$ $\implies v_B=12.6058in/s$ and the angle can be determined as $\theta=tan^{-1}|\frac{v_{By}}{v_{Bx}}|$ $\implies \theta=tan^{-1}|\frac{11.4852}{-5.196}|=65.657^{\circ}$