Question

Sand moves without slipping at 6.0 m/s down a conveyer that is tilted at 15°. The sand enters a pipe 3.0 m below the end of the conveyer belt, as shown in FIGURE P4.56. What is the horizontal distance d between the conveyer belt and the pipe?

Short answer

The horizontal distance d between the conveyer belt and the pipe is$2.72\text{m}$.

Step-by-step solution

Step 1. Given information

The diagram is as follows

The initial speed of the sand is $6.0\text{m/s}$, the inclination angle is$15°$and the vertical height travelled by the sand is$3.0\text{m}$.

Step2. Calculation of the time taken by the sand to travel the vertical height

The formula to calculate the vertical height travelled by the sand is given by

$h=v_{iy}t+\frac{1}{2}g{t}^2.............\left(1\right)$

Here, $h,v_{iy},t,g$are the vertical distance travelled by the sand, the vertical component of its initial speed, time taken by the sand and the acceleration due to gravity respectively.

Simplify equation (1) to obtain an expression for the time taken by the sand to travel the vertical distance.

role="math" localid="1647580699932" $\begin{array}{rcl}g{t}^2+v_{iy}t-h& =& 0\\ g{t}^2+v_i\sin \theta t-h& =& 0\\ t& =& \frac{-v_i\sin \theta \pm \sqrt{{\left(-v_i\sin \theta \right)}^2+4gh}}{2g}.....................\left(2\right)\end{array}$

Here, $\theta$is the inclination angle.

Substitute the values of the parameters from the given information into equation (2) and solve to calculate the required time taken by the sand.

role="math" localid="1647580830758" $\begin{array}{rcl}t& =& \frac{-6\sin 15°\pm \sqrt{{\left(-6\sin 15°\right)}^2+4\times 9.8\times 3}}{2\times 9.8}\text{s}\\ & \approx & 0.47\text{s}\text{and}-0.63\text{s}\end{array}$

As time cannot be negative, the allowed value of the time is$0.47\text{s}$.

Step 3. Calculation of the horizontal distance

The formula to calculate the required horizontal distance$\left(d\right)$ is given by

role="math" $d=v_i\cos \theta t..............\left(3\right)$

Substitute the values of the parameters into equation (3) to calculate the required horizontal distance.

$\begin{array}{rcl}d& =& 6\times \cos 15°\times 0.47\text{m}\\ & \approx & 2.72\text{m}\end{array}$