Question

determine the required velocity of the belt b if the relative velocity with which the sand hits belt b is to be (a) vertical, (b) as small as possible. v_a = 5 ft/s

Explanation:

Step1: Analyze relative - velocity components

Let the velocity of belt A be $v_A = 5\ ft/s$ and the velocity of belt B be $v_B$. The velocity of the sand as it leaves the end of belt A is $v_A$. We resolve the velocities into horizontal and vertical components.

Step2: For relative - velocity to be vertical (a)

The horizontal component of the relative velocity $v_{r,x}=v_A - v_B\cos15^{\circ}=0$. So, $v_B=\frac{v_A}{\cos15^{\circ}}$. Since $\cos15^{\circ}=\frac{\sqrt{6}+\sqrt{2}}{4}\approx0.966$, and $v_A = 5\ ft/s$, then $v_B=\frac{5}{0.966}\approx5.18\ ft/s$.

Step3: For relative - velocity to be minimum (b)

The relative velocity $\vec{v}_r=\vec{v}_A-\vec{v}_B$. The magnitude of the relative velocity $v_r=\sqrt{(v_A - v_B\cos15^{\circ})^2+(v_B\sin15^{\circ})^2}$. To minimize $v_r$, we take the derivative of $v_r^2$ with respect to $v_B$ and set it equal to zero.
\[

$$\begin{align*} \frac{d}{dv_B}[(v_A - v_B\cos15^{\circ})^2+(v_B\sin15^{\circ})^2]&=0\\ 2(v_A - v_B\cos15^{\circ})(-\cos15^{\circ})+2v_B\sin^{2}15^{\circ}&=0\\ - 2v_A\cos15^{\circ}+2v_B\cos^{2}15^{\circ}+2v_B\sin^{2}15^{\circ}&=0\\ - 2v_A\cos15^{\circ}+2v_B(\cos^{2}15^{\circ}+\sin^{2}15^{\circ})&=0 \end{align*}$$

\]
Since $\cos^{2}\theta+\sin^{2}\theta = 1$, we have $v_B = v_A\cos15^{\circ}=5\times0.966 = 4.83\ ft/s$.

Answer:

(a) $v_B\approx5.18\ ft/s$
(b) $v_B = 4.83\ ft/s$