Chapter 14.3, Problem 14.68P

To determine

The components of the reactions at C and D of the conveyor belt system.

Answer to Problem 14.68P

The reaction at D is $\underset{\_}{3,120\text{N}}$.

The reaction component at C along x direction is $\underset{\_}{C_x=112.5\text{N}}$.

The reaction component at C along y direction is $\underset{\_}{C_y=2,660\text{N}}$.

Explanation of Solution

Given information:

The discharge of coal from first conveyor belt at the rate $Q=150\text{kg}/\text{s}$.

The magnitude of velocity $v_1=3\text{m}/\text{s}$ and $v_2=4.25\text{m}/\text{s}$

The total mass is $m=500\text{kg}$.

Calculation:

Consider the acceleration due to gravity as $g=9.81\text{m}/{\text{s}}^2$.

Sketch the Free Body Diagram of coal discharge from first conveyor belt as shown in Figure 1.

Refer to Figure 1.

Calculate the mass of coal $\left(\Delta m\right)$ entering and leaving the system in time interval $\Delta t$ as shown below.

$\Delta m=Q\Delta t$

Substitute $150\text{kg}/\text{s}$ for Q.

$\Delta m=150\left(\Delta t\right)$

Calculate the velocity of the coal hits the second belt $\left(v_A\right)$ as shown below.

Velocity along x direction.

${\left(v_A\right)}_x=3\text{m}/\text{s}$

Velocity along y direction.

${\left(v_A\right)}_y=\sqrt{2gh}$

Substitute $9.81\text{m}/{\text{s}}^2$ for g and $0.75\text{m}$ for h.

$\begin{array}{c}{\left(v_A\right)}_y=\sqrt{2\times 9.81\times 0.75}\\ =\sqrt{14.715}\\ =3.836\text{m}/\text{s}\end{array}$

Sketch the Free Body Diagram of the second conveyor belt assembly as shown in Figure 2.

Refer to Figure 2.

Calculate the velocity $\left(v_B\right)$ as shown below.

Velocity along x direction.

${\left(v_B\right)}_x=\frac{3}{3.4}{v}_B$

Velocity along y direction.

${\left(v_B\right)}_y=\frac{1.6}{3.4}{v}_B$

Apply angular momentum about C as shown below.

$\begin{array}{l}1.6\left(\Delta m\right){\left(v_A\right)}_x+1\left(\Delta m\right){\left(v_A\right)}_y+2.4W\Delta t-4D\Delta t=3.2\left(\Delta m\right){\left(v_B\right)}_x-4\left(\Delta m\right){\left(v_B\right)}_y\\ 1.6\frac{\Delta m}{\Delta t}{\left(v_A\right)}_x+\frac{\Delta m}{\Delta t}{\left(v_A\right)}_y+2.4W-4D=3.2\frac{\Delta m}{\Delta t}{\left(v_B\right)}_x-4\frac{\Delta m}{\Delta t}{\left(v_B\right)}_y\\ 2.4W-4D=\frac{\Delta m}{\Delta t}\left[3.2{\left(v_B\right)}_x-4{\left(v_B\right)}_y-1.6{\left(v_A\right)}_x-{\left(v_A\right)}_y\right]\\ 4D=2.4W-\frac{\Delta m}{\Delta t}\left[3.2{\left(v_B\right)}_x-4{\left(v_B\right)}_y-1.6{\left(v_A\right)}_x-{\left(v_A\right)}_y\right]\end{array}$

Substitute $4,905\text{N}$ for W, $150\text{kg}/\text{s}$ for $\frac{\Delta m}{\Delta t}$, $3\text{m}/\text{s}$ for ${\left(v_A\right)}_x$, $3.836\text{m}/\text{s}$ for ${\left(v_A\right)}_y$, $\frac{3}{3.4}{v}_B$ for ${\left(v_B\right)}_x$, and $\frac{1.6}{3.4}{v}_B$ for ${\left(v_B\right)}_y$.

$\begin{array}{c}4D=2.4\times W-150\left[3.2\times \frac{3}{3.4}{v}_B-4\times \frac{1.6}{3.4}{v}_B-1.6\times 3-3.836\right]\\ =2.4W-150\left[0.9412v_B-8.636\right]\\ =2.4W-141.18v_B+1,295.4\end{array}$

$D=0.6W-35.295v_B+323.85$ (1)

Calculate the weight of the coal $\left(W\right)$ as shown below.

$W=mg$

Substitute $500\text{kg}$ for m and $9.81\text{m}/{\text{s}}^2$ for g.

$\begin{array}{c}W=500\times 9.81\\ =4,905\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\times \frac{\text{1}\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\\ =4,905\text{N}\end{array}$

Calculate the component of the reaction at D as shown below.

Substitute $4,905\text{N}$ for W and $4.25\text{m}/\text{s}$ for $v_B$.

$\begin{array}{c}D=0.6\times 4,905-35.295\times 4.25+323.85\\ =3,116.8\text{N}\\ =3,117\text{N}\\ \approx \text{3120 N}\end{array}$

Hence, the reaction at D is $\underset{\_}{3,120\text{N}}$.

Apply the Equations of Equilibrium along x direction as shown below.

$\begin{array}{l}\Delta m{\left(v_A\right)}_x+C_x\Delta t=\Delta m{\left(v_B\right)}_x\\ C_x\Delta t=\Delta m\left[{\left(v_B\right)}_x-{\left(v_A\right)}_x\right]\\ C_x=\frac{\Delta m}{\Delta t}\left[{\left(v_B\right)}_x-{\left(v_A\right)}_x\right]\end{array}$

Substitute $150\text{kg}/\text{s}$ for $\frac{\Delta m}{\Delta t}$, $\frac{3}{3.4}{v}_B$ for ${\left(v_B\right)}_x$, and $3\text{m}/\text{s}$ for ${\left(v_A\right)}_x$.

$\begin{array}{c}{C}_x=150\times \left(\frac{3}{3.4}{v}_B-3\right)\\ =132.35v_B-450\end{array}$

Substitute $4.25\text{m}/\text{s}$ for $v_B$.

$\begin{array}{c}{C}_x=132.35\times 4.25-450\\ =112.5\text{N}\end{array}$

Hence, the reaction component at C along x direction is $\underset{\_}{C_x=112.5\text{N}}$.

Apply the Equations of Equilibrium along y direction as shown below.

$\begin{array}{l}-\Delta m{\left(v_A\right)}_y+C_y\Delta t-W\Delta t+D\Delta t=\Delta m{\left(v_B\right)}_y\\ C_y\Delta t=W\Delta t-D\Delta t+\Delta m\left[{\left(v_B\right)}_y+{\left(v_A\right)}_y\right]\\ C_y=W-D+\frac{\Delta m}{\Delta t}\left[{\left(v_B\right)}_y+{\left(v_A\right)}_y\right]\end{array}$

Substitute $150\text{kg}/\text{s}$ for $\frac{\Delta m}{\Delta t}$, $\frac{1.6}{3.4}{v}_B$ for ${\left(v_B\right)}_y$, $3,117\text{N}$ for D, $4,905\text{N}$ for D, and $3.836\text{m}/\text{s}$ for ${\left(v_A\right)}_y$.

$\begin{array}{c}{C}_y=4,905-3,117+150\left[\frac{1.6}{3.4}{v}_B+3.836\right]\\ =2,363.4+70.588v_B\end{array}$

Substitute $4.25\text{m}/\text{s}$ for $v_B$.

$\begin{array}{c}{C}_y=2,363.4+70.588\times 4.25\\ =2,663\text{N}\\ \approx \text{2,660 N}\end{array}$

Therefore, the reaction component at C along y direction is $\underset{\_}{C_y=2,660\text{N}}$.