Question about falling objects

  • Thread starter Thread starter TheGenericGeek
  • Start date Start date
  • Tags Tags
    Falling
AI Thread Summary
To calculate the trajectory of dirt or rocks dropped down a pipe, the speed of the material exiting the pipe (v_0) and the angle of the pipe (θ) are essential. The time of fall can be determined using a quadratic equation that incorporates gravitational acceleration and the vertical distance (h) from the pipe's end to the container. The horizontal distance traveled is calculated by multiplying the horizontal velocity (v_0*cosθ) by the time of fall. Additionally, v_0 can be derived from the height of the fall and the coefficient of kinetic friction between the material and the pipe. Understanding these principles of projectile motion will facilitate the creation of the desired Excel sheet.
TheGenericGeek
Messages
1
Reaction score
0
G'day Guys and Gals,

I'm working on trying to come up with a best-fit excel sheet for work.
Basically I'm trying to find out where dirt/rocks will peak if dropped down a pipe at any given angle.

As attached, my crude diagram - I'm trying to calculate the blue line.
The height of the container will vary, the angle of the pipe will vary, the length of the pipe will vary, and I'm trying to get a rough idea of what trajectory the dirt/rocks will follow, and turn this into an excel sheet (Just need assistance with the equations to calculate it all)
Currently the only assumption made is that the dirt/rocks will fall straight from wherever the pipe enters the container, straight down - Obviously this is not the case.

Any help you could offer is great appreciated.

Cheers,
TheGenericGeek
 

Attachments

  • Snapshot.jpg
    Snapshot.jpg
    5.4 KB · Views: 385
Physics news on Phys.org
Welcome to PF generic!

This is a relatively simple problem to solve if you know the speed of the material exiting the end of the pipe, ##v_0##. The direction that the material is moving is given by the angle of the pipe relative to the horizontal, ##\theta##. It is then a matter of applying principles of projectile motion.

The time of fall is the solution to the quadratic equation:

##\frac{1}{2}gt^2 + (v_0\sin\theta) t - h = 0## where h is the vertical fall distance from the end of the pipe.

The horizontal distance moved in that time is the horizontal velocity exiting the pipe: ##v_0\cos\theta## multiplied by the time of fall, t:

##\Delta x = (v_0\cos\theta) t##

##v_0## can be determined if you know the height that the material falls from an initial rest position to the end of the pipe AND the coefficient of kinetic friction between the material and the pipe, as well as the angle of the pipe. Otherwise, you have to measure it.

AM
 
Last edited:
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Hello everyone, Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed. My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds? Best regards, DaTario
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Back
Top