Chapter 14.3, Problem 14.68P

Coal is being discharged from a first conveyor belt at the rate of 120 kg/s. It is received at A by a second belt discharges it again at B. Knowing that $v_1=3$ m/s and $v_2=4.25$ m/s and that the second belt assembly and the coal it supports have a total mass of 472 kg, determine the components of the reactions at C and D.

    

To determine

The components of reaction at C and D.

Answer to Problem 14.68P

The $y$ component of reaction at D is $2900\text{N}$ and the $x$ component of reaction at D is $0\text{N}$, the $y$ component of reaction at C is $2362.7\text{N}$ and the $x$ component of reaction at C is $90\text{N}$.

Explanation of Solution

Given:

Rate of discharge of coal is $120\text{ kg/s}$.

Velocity $v_1$ is $3\text{ m/s}$.

Velocity $v_2$ is $4.25\text{ m/s}$.

Total mass is $472\text{kg}$.

Concept used:

Draw the diagram to illustrate the path travelled by coal on the belt as shown below:

The path followed by the coal over belt can illustrated as:

Write the expression for the velocity component in $y$ direction.

${\left(v_1\right)}_y=\sqrt{2gh}$ ...... (1)

Here, ${\left(v_1\right)}_y$ is the velocity component in $y$ direction, $g$ is acceleration due to gravity and $h$ is the height from which coal fall.

Write the expression for angle $\theta$ in triangle EBF.

$\theta ={\tan }^{-1}\left(\frac{BF}{EF}\right)$ ...... (2)

Here, $\theta$ is the angle of inclination of conveyor.

Write the expression for the moment about point C.

$\left[\begin{array}{l}\left(\Delta m\right){\left(v_1\right)}_x\left(1.2\text{}\text{m}\right)+\left(\Delta m\right){\left(v_1\right)}_y\left(0.75\text{m}\right)\\ +\left(W\right)1.8-\left(D\right)\left(3\text{m}\right)\end{array}\right]=\left[\begin{array}{l}\left(\Delta m\right)\left(v_2\cos \theta \right)\left(2.4\text{ m}\right)\\ -\left(\Delta m\right)\left(v_2\sin \theta \right)\left(3\text{ m}\right)\end{array}\right]$

Simplify the above expression.

$\left(D\right)\left(3\text{m}\right)=\left[\begin{array}{l}1.2\left(\Delta m\right){\left(v_1\right)}_x+0.75\left(\Delta m\right){\left(v_1\right)}_y+1.8\left(W\right)\\ -2.4\left(\Delta m\right)\left(v_2\cos \theta \right)+3\left(\Delta m\right)\left(v_2\sin \theta \right)\end{array}\right]$ ...... (3)

Here, $D$ is the reaction at point D, $\Delta m$ is the mass of coal entering and leaving the system, $W$ is the rate of discharge of coal, ${\left(v_1\right)}_x$ is the velocity component in $x$ direction and $v_2$ is velocity.

Write the expression for the $x$ component of reaction at C.

$C_x\Delta t=\left(\Delta m\right)\Delta t{v}_2\cos \theta -\Delta m\Delta t{\left(v_1\right)}_x$ ...... (4)

Here, $C_x$ is the $x$ component of reaction at C.

Write the expression for $y$ component of C.

$C_y\Delta t=\Delta m{v}_2\sin \theta \Delta t+\Delta m{\left(v_1\right)}_y\Delta t-D\Delta t+W\Delta t$ ...... (5)

Here, $C_y$ is the $y$ component of reaction at C.

Calculation:

Substitute $9.81{\text{ m/s}}^2$ for $g$ and $0.545\text{ m}$ for $h$ in equation (1).

$\begin{array}{c}{\left(v_1\right)}_y=\sqrt{2\left(9.80\right)\left(0.545\right)}\\ =3.27\text{m/s}\end{array}$

Substitute 1.2 for BF and 2.25 for EF in equation (2).

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{1.2}{2.25}\right)\\ \theta =28.07°\end{array}$

Substitute $472\text{kg}$ for $W$, $120\text{ kg}$ for $\Delta m$, $28.07°$ for $\theta$, $3.27\text{ m/s}$ for ${\left(v_1\right)}_y$, $3$ for ${\left(v_1\right)}_x$ and $4.25\text{ m/s}$ for $v_2$ in equation (3).

$\begin{array}{c}3D=\left[\begin{array}{l}\left(472\right)\left(9.81\right)\left(1.8\right)+120\left(3\right)\left(1.2\right)+120\left(3.27\right)\left(0.75\right)\\ -120\left(4.25\right)\left(2.4\right)\left(\cos 28.07°\right)+120\left(4.25\right)\left(3\right)\left(\sin 28.07°\right)\end{array}\right]\\ 3D=8700\\ D=2900\text{ N}\end{array}$

Substitute $120\text{kg/s}$ for $\Delta m$, $28.07°$ for $\theta$, $4.25\text{ m/s}$ for $v_2$ and $3\text{ m/s}$ for ${\left(v_1\right)}_x$ in equation (4).

$\begin{array}{c}{C}_x\Delta t=120\left(4.25\right)\Delta t\cos 28.07°-120\Delta t\left(3\right)\\ C_x=90\text{ N}\end{array}$

Substitute $120\text{kg/s}$ for $\Delta m$, $28.07°$ for $\theta$, $4.25\text{ m/s}$ for $v_2$, $472\text{}\text{kg}$ for $W$ and $2900\text{N}$ for $D$ in equation (5).

$\begin{array}{c}{C}_y\Delta t=120\Delta t\left(4.25\right)\sin 28.07°+120\Delta t\left(3.27\right)-2900\Delta t+472\Delta t\left(9.81\right)\\ C_y=239.98+392.4-2900+4630.32\\ =2362.7\text{N}\end{array}$

The $y$ component of reaction at D is $2900\text{N}$ and the $x$ component of reaction at D is $0\text{N}$, the $y$ component of reaction at C is $2362.7\text{N}$ and the $x$ component of reaction at C is $90\text{N}$.

Conclusion:

Thus, the $y$ component of reaction at D is $2900\text{N}$ and the $x$ component of reaction at D is $0\text{N}$, the $y$ component of reaction at C is $2362.7\text{N}$ and the $x$ component of reaction at C is $90\text{N}$.