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 the horizontal - component of relative velocity for vertical relative - hit

Let the velocity of belt A be \(v_A = 5\ ft/s\) and the velocity of belt B be \(v_B\). The horizontal component of the velocity of the sand as it leaves belt A is \(v_{Ax}=5\ ft/s\). The horizontal component of the velocity of belt B is \(v_{Bx}=v_B\cos15^{\circ}\). For the relative velocity of the sand hitting belt B to be vertical, the horizontal components of the velocities of the sand and belt B must be equal. So \(v_A = v_B\cos15^{\circ}\).
\[v_B=\frac{v_A}{\cos15^{\circ}}\]
We know that \(\cos15^{\circ}=\cos(45 - 30)=\cos45^{\circ}\cos30^{\circ}+\sin45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}\approx0.966\)
\[v_B=\frac{5}{\cos15^{\circ}}=\frac{5}{0.966}\approx5.18\ ft/s\]

Step2: Analyze for minimum relative - velocity

The relative velocity of the sand with respect to belt B, \(\vec{v}_{r}=\vec{v}_{A}-\vec{v}_{B}\). The magnitude of the relative velocity \(v_{r}=\sqrt{(v_A - v_B\cos15^{\circ})^2+(0 - v_B\sin15^{\circ})^2}\). To find the minimum of \(v_{r}\) with respect to \(v_B\), we take the derivative of \(v_{r}^2=(v_A - v_B\cos15^{\circ})^2+( - v_B\sin15^{\circ})^2=v_A^2-2v_Av_B\cos15^{\circ}+v_B^2\cos^{2}15^{\circ}+v_B^2\sin^{2}15^{\circ}=v_A^2-2v_Av_B\cos15^{\circ}+v_B^2\) with respect to \(v_B\) and set it equal to zero.
\(\frac{d(v_{r}^2)}{dv_B}=- 2v_A\cos15^{\circ}+2v_B = 0\)
\[v_B = v_A\cos15^{\circ}\]
\[v_B=5\times0.966 = 4.83\ ft/s\]

Answer:

(a) \(v_B\approx5.18\ ft/s\)
(b) \(v_B\approx4.83\ ft/s\)