Chapter 11.4, Problem 11.127P

Coal discharged from a dump truck with an initial velocity (v_c)₀ = 5 ft/s ⦫ 50° falls onto conveyor belt B. Determine the required velocity v_B of the belt if the relative velocity with which the coal hits the belt is to be (a) vertical, (b) as small as possible.

Fig. P11.127

To determine

The required velocity $\left(v_B\right)$ of the belt if the relative velocity with which the coal hit the belt is to be vertical.

Answer to Problem 11.127P

The required velocity $\left(v_B\right)$ of the belt if the relative velocity with which the coal hit the belt is to be vertical is $\underset{\_}{3.26\text{ft/s}}$ at an angle of $\underset{\_}{10°}$.

Explanation of Solution

Given Information:

The initial ${\left(v_C\right)}_0$ velocity of coal $5\text{ft/s}$ at an angle $\theta$ of $50°$.

The angle $\left(\alpha \right)$ between the horizontal surface and the belt is $10°$.

Calculation:

Calculate the x component initial velocity ${\left({\left(v_C\right)}_0\right)}_x$ of coal using the relation:

${\left({\left(v_C\right)}_0\right)}_x=-{\left(v_C\right)}_0\cos \theta$

Substitute $5\text{ft}/\text{s}$ for ${\left(v_C\right)}_0$ and $50°$ for $\theta$

$\begin{array}{c}{\left({\left(v_C\right)}_0\right)}_x=-5\cos \left(50°\right)\\ =-3.2139\text{ft/s}\end{array}$

Calculate the x component velocity ${\left(v_C\right)}_x$ of coal using the relation:

$\begin{array}{l}{\left(v_C\right)}_x={\left({\left(v_C\right)}_0\right)}_x\\ {\left(v_C\right)}_x=-3.2139\text{ft/s}\end{array}$

Calculate the y component initial velocity ${\left({\left(v_C\right)}_0\right)}_y$ of coal using the relation:

${\left({\left(v_C\right)}_0\right)}_y={\left(v_C\right)}_0\sin \theta$

Substitute $5\text{ft}/\text{s}$ for ${\left(v_C\right)}_0$ and $50°$ for $\theta$.

$\begin{array}{c}{\left({\left(v_C\right)}_0\right)}_y=5\sin \left(50°\right)\\ =3.830\text{ft/s}\end{array}$

Calculate the y component kinematic velocity ${\left(v_C\right)}_y$ of coal using the relation:

${\left(v_C\right)}_y={\left({\left(v_C\right)}_0\right)}_y^2+2a_{y}d$

Here, d is distance between the dump truck and belt.

Substitute $3.830\text{ft}/\text{s}$ for ${\left({\left(v_C\right)}_0\right)}_y$ and $-g$ for $a_y$.

$\begin{array}{l}{\left(v_C\right)}_y^2=3.830{}^2+2\left(-g\right)d\\ {\left(v_C\right)}_y^2=3.830{}^2-2gd\end{array}$

Substitute $32.2{\text{ft}/\text{s}}^2$ for g and $-4\text{ft}$ for d.

$\begin{array}{l}{\left(v_C\right)}_y^2=3.830{}^2-2\left(32.2\right)\left(-4\right)\\ {\left(v_C\right)}_y=\sqrt{272.2689}\\ {\left(v_C\right)}_y=-16.501\text{ft/s}\end{array}$

Calculate the angle $\left(\beta \right)$ between the velocity C and horizontal line.

$\tan \beta =\frac{{\left(v_C\right)}_y}{{\left(v_C\right)}_x}$

Substitute $-16.501\text{ft}/\text{s}$ for ${\left(v_C\right)}_y$ and $-3.2139\text{ft}/\text{s}$ for ${\left(v_C\right)}_x$.

$\begin{array}{c}\tan \beta =\frac{-16.501}{-3.2139}\\ ={\tan }^{-1}\left(5.134\right)\\ =78.99°\end{array}$

Calculate the velocity $\left(v_C\right)$ of coal using the relation:

$v_C=\sqrt{{\left(v_C\right)}_x^2+{\left(v_C\right)}_y^2}$

Substitute $-3.2139\text{ft/s}$ for ${\left(v_C\right)}_x$ and $-16.501\text{ft}/\text{s}$ for ${\left(v_C\right)}_y$.

$\begin{array}{c}{v}_C=\sqrt{{\left(-3.2139\right)}^2+{\left(-16.501\right)}^2}\\ =\sqrt{10.329+272.28}\\ =16.81\text{ft/s}\end{array}$

Write the expression for velocity vector $\left(v_C\right)$ of the coal:

$v_C=\left(-3.2139\right)i+\left(-16.501\right)j$

Write the expression for velocity vector $\left(v_B\right)$ of the belt:

$v_B=v_B\left(-\cos \alpha i+\sin \alpha j\right)$

Substitute $10°$ for $\alpha$.

$\begin{array}{c}{v}_B=v_B\left(-\cos \left(10°\right)i+\sin \left(10°\right)j\right)\\ =-0.985v_{B}i+0.1736v_{B}j\end{array}$

Write the expression for relative velocity vector $\left(v_{\text{C/B}}\right)$:

$v_{\text{C/B}}=v_C-v_B$ (1)

When the belt positioned vertical, then x component of relative velocity ${\left(v_{\text{C/B}}\right)}_x$ become zero.

Write the y component relative velocity vector equation ${\left(v_{\text{C/B}}\right)}_y$ as below:

Substitute $\left(-3.2139\right)i+\left(-16.501\right)j$ for $v_C$ and $-0.985v_{B}i+0.1736v_{B}j$ for $v_B$ in Equation (1).

${\left(v_{\text{C/B}}\right)}_{y}j=\left(-3.2139\right)i+\left(-16.501\right)j-\left(-0.985v_{B}i+0.1736v_{B}j\right)$

Write the y component relative velocity vector equation ${\left(v_{\text{C/B}}\right)}_y$ as below:

Substitute $\left(-3.2139\right)i+\left(-16.501\right)j$ for $v_C$ and $-0.985v_{B}i+0.1736v_{B}j$ for $v_B$ in Equation (1).

${\left(v_{\text{C/B}}\right)}_{y}j=\left(-3.2139\right)i+\left(-16.501\right)j-\left(-0.985v_{B}i+0.1736v_{B}j\right)$

Write the x component relative velocity vector equation ${\left(v_{\text{C/B}}\right)}_x$:

Substitute 0 for ${\left(v_{\text{C/B}}\right)}_x$, $\left(-3.2139\right)i+\left(-16.501\right)j$ for $v_C$ and $-0.985v_{B}i+0.1736v_{B}j$ for $v_B$ in Equation (1).

$\begin{array}{c}0i=\left(-3.2139\right)i+\left(-16.501\right)j-\left(-0.985v_{B}i+0.1736v_{B}j\right)\\ 0i=\left(-3.2139\right)i+0.985v_{B}i\end{array}$

Solve the above equation for velocity $\left(v_B\right)$ of block B.

$\begin{array}{c}{v}_B=\frac{3.2139}{0.985}\\ =3.26\text{ft/s}\end{array}$

Therefore, the required velocity $\left(v_B\right)$ of the belt if the relative velocity with which the coal hit the belt is to be vertical is $\underset{\_}{3.26\text{ft/s}}$ at an angle of $\underset{\_}{10°}$.

To determine

The required velocity $\left(v_B\right)$ of the belt if the relative velocity with which the coal hit the belt is to be as small as possible.

Answer to Problem 11.127P

The required velocity $\left(v_B\right)$ of the belt if the relative velocity with which the coal hit the belt is to be as small as possible is $\underset{\_}{0.3\text{ft/s}}$ at an angle of $\underset{\_}{10°}$.

Explanation of Solution

Given Information:

The initial ${\left(v_C\right)}_0$ velocity of coal $5\text{ft/s}$ at an angle $\theta$ of $50°$.

The angle $\left(\alpha \right)$ between the horizontal surface and the belt is $10°$.

Calculation:

Write vector for relative velocity $\left(v_{C/B}\right)$ of C with respect to B:

Substitute $\left(-3.2139\right)i+\left(-16.501\right)j$ for $v_C$ and $-0.985v_{B}i+0.1736v_{B}j$ for $v_B$ in Equation (1).

${\left(v_{\text{C/B}}\right)}_{x}i+{\left(v_{\text{C/B}}\right)}_{y}j=\left(-3.2139\right)i+\left(-16.501\right)j-\left(-0.985v_{B}i+0.1736v_{B}j\right)$

Separate the i and j component.

Consider i component.

${\left(v_{\text{C/B}}\right)}_{x}i=\left(-3.2139\right)i+0.985v_{B}i$

Consider j component.

${\left(v_{\text{C/B}}\right)}_{y}j=\left(-16.501\right)j-\left(0.1736v_{B}j\right)$

Find the expression for ${\left(v_{C/B}\right)}^2$ using the relation:

${\left(v_{C/B}\right)}^2={\left(v_{C/B}\right)}_x^2+{\left(v_{C/B}\right)}_y^2$

Substitute $\left(-3.2139\right)i+0.985v_{B}i$ for ${\left(v_{C/B}\right)}_x$ and $\left(-16.501\right)j-\left(0.1736v_{B}j\right)$ for ${\left(v_{\text{C/B}}\right)}_y$.

$\begin{array}{c}{\left(v_{C/B}\right)}^2={\left(\left(-3.2139\right)+0.985v_B\right)}^2+{\left(\left(-16.501\right)-\left(0.1736v_B\right)\right)}^2\\ =\left\{\begin{array}{l}10.329+0.970v_B^2+2\left(-3.2139\times 0.985v_B\right)+272.283+0.0301v_B^2\\ -2\left(-16.501\times 0.1736v_B\right)\end{array}\right\}\\ =10.329+0.970v_B^2-6.331v_B+272.283+0.0301v_B^2+5.729v_B\\ =282.612-0.602v_B+v_B^2\end{array}$ (2)

Calculate the required velocity $\left(v_B\right)$ of the belt if the relative velocity with which the coal hit the belt is to be as small as possible:

Differentiate the equation 2 with respective to velocity $\left(v_B\right)$

$\begin{array}{c}\frac{d{\left(v_{C/B}\right)}^2}{d{v}_B}=282.612-0.602v_B+v_B^2\\ =-0.602+2v_B\end{array}$

Solve the above equation for $\left(v_B\right)$:

$\begin{array}{c}{v}_B=\frac{0.602}{2}\\ =0.3\text{ft/s}\end{array}$

Therefore, the required velocity $\left(v_B\right)$ of the belt if the relative velocity with which the coal hit the belt is to be as small as possible is $\underset{\_}{0.3\text{ft/s}}$ at an angle of $\underset{\_}{10°}$.