Math Problem Statement

Pequeños paquetes que se desplazan sobre la banda transportadora caen en el carro de carga de 1 m de largo. Si la transportadora se desplaza a una rapidez constante de vC = 2 m/s, determine la distancia más corta y más larga R donde pueda colocarse el extremo A del carro con respecto a la transportadora para que los paquetes entren al carro.

Solution

The problem you've provided involves calculating the shortest and longest distances, RR, where the end AA of the cart can be placed so that the packages fall inside the cart. Here's a breakdown of how to solve it:

Given Data:

  • The conveyor belt speed: vc=2m/sv_c = 2 \, \text{m/s}
  • Angle of the conveyor belt: θ=30\theta = 30^\circ
  • Height of drop from conveyor belt to cart: h=3mh = 3 \, \text{m}
  • Length of the cart: 1m1 \, \text{m}

We need to determine the range of values for RR (distance from the conveyor belt to the cart) where the packages fall into the cart, i.e., the shortest distance RminR_{\text{min}} and longest distance RmaxR_{\text{max}}.

Step 1: Vertical Motion

The time it takes for the package to fall from the conveyor belt to the cart is determined by the vertical motion. Using the equation of motion for free fall:

h=12gt2h = \frac{1}{2} g t^2 Solving for tt, we get:

t=2hg=2×3m9.81m/s20.78secondst = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 3 \, \text{m}}{9.81 \, \text{m/s}^2}} \approx 0.78 \, \text{seconds}

Step 2: Horizontal Motion

The horizontal velocity of the packages is given by the component of the conveyor belt speed along the horizontal direction. Since the belt is inclined at 3030^\circ, the horizontal velocity is:

vx=vccos(30)=2m/s×cos(30)=1.73m/sv_x = v_c \cos(30^\circ) = 2 \, \text{m/s} \times \cos(30^\circ) = 1.73 \, \text{m/s}

The horizontal distance the package travels is:

d=vx×t=1.73m/s×0.78seconds1.35md = v_x \times t = 1.73 \, \text{m/s} \times 0.78 \, \text{seconds} \approx 1.35 \, \text{m}

Step 3: Determine Range for RR

The package must fall within the length of the cart, which is 1 meter. So, the distance RR where the cart can be placed must be such that the package falls between AA and BB.

  • The shortest distance RminR_{\text{min}} is when the package lands at point AA:

Rmin=d=1.35mR_{\text{min}} = d = 1.35 \, \text{m}

  • The longest distance RmaxR_{\text{max}} is when the package lands at point BB, which is 1 meter further than AA:

Rmax=Rmin+1=1.35m+1m=2.35mR_{\text{max}} = R_{\text{min}} + 1 = 1.35 \, \text{m} + 1 \, \text{m} = 2.35 \, \text{m}

Final Answer:

The range of distances RR where the cart can be placed is:

Rmin=1.35m,Rmax=2.35mR_{\text{min}} = 1.35 \, \text{m}, \quad R_{\text{max}} = 2.35 \, \text{m}

Would you like further details on any of the steps?

Related Questions:

  1. How does the time of fall change with different heights?
  2. What happens if the conveyor belt speed is increased to 3 m/s?
  3. How would the calculation differ if the incline angle were 45 degrees?
  4. What is the horizontal distance if the cart is moved 0.5 meters closer?
  5. How would the range of RR be affected if the cart length were increased to 1.5 meters?

Tip:

When dealing with inclined planes, always break down velocities into their horizontal and vertical components to simplify the problem.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Projectile Motion
Free Fall
Trigonometry

Formulas

Vertical motion equation: h = (1/2) * g * t^2
Horizontal motion equation: d = v_x * t
Horizontal velocity: v_x = v_c * cos(θ)

Theorems

Equations of motion for free fall
Trigonometric decomposition of velocity

Suitable Grade Level

Grades 10-12