Calculating pressure on a surface after reaching final velocity

AI Thread Summary
To calculate the pressure at the point of impact for a drop test, it's essential to first determine the final velocity of the object, which in this case is approximately 3 m/s upon hitting the ground. The pressure can then be calculated using the formula Pressure = Force/Area, where Force is derived from the change in momentum over time. The time taken for the ball to come to rest significantly affects the force calculation; for a soft ball, this could be around 1 second, while a hard ball might take only 1/100 of a second. Without precise measurements of impact time, an educated guess is necessary to estimate the force. Understanding these variables is crucial for accurately determining the pressure exerted on the surface during impact.
tegefe
Messages
1
Reaction score
0
I have a question. I'm conducting a drop test for packaging material and I need to find the pressure at the point of impact.
What I did was drop a 18 lb ball 18".
I converted the mass to 3.629g and the distance to 0.4572m to work in S.I. units
then used the equation 2ad=Vf^2-V0^2. Afetr doing the math I got my Vf=2.993 m/s^2, but now I have to convert that final velocity to pressure and I have no clue as to how to go about it. Can you assist?
 
Physics news on Phys.org
Some things wrong with your information! an 18 lb ball is more like 9kg.
Its velocity when it hits the ground is 3m/s (not m/s^2).
To find pressure you need Force/area and to calculate the force due to an impact you need to have some idea of the time taken for the ball to come to rest.
(Force = change in momentum per second) If you do not know the time for the impact and you have no way of measuring it then you must make an educated guess!
If it is a 'soft' ball it may take up to 1 second to come to rest.
If it is a 'hard' ball (like a bowling ball) it may come to rest in about 1/100 of a second.
This will give a wide range of force during impact but if your investigation is a practical exercise I hope this will be of some help.
 
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