How Do You Calculate Pressure of a Falling Object at Impact?

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To calculate the pressure of a falling object at impact, one must consider both the weight of the object and its velocity upon reaching the ground. The basic formula for pressure is P = F/A, where F is the force and A is the area of contact. At impact, the force can be expressed as F = mg + mv/Δt, incorporating the object's mass, gravitational force, and the change in momentum over time. The velocity at impact can be determined using kinematic equations, and it is important to note that this calculation assumes no bouncing occurs. Understanding these mechanics allows for a more accurate assessment of the pressure generated by a falling object.
LasTSurvivoR
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Ok guys everyone knows teh solid Pressure = P / S but i have a question..
I though about an object that falls from 20 meters.(Its mass is 2 kg, 2x2x2 meters cube)
When it reaches the ground how much pressure does it make , in normal case
20 . 10 / 2² = 50 Pascal it makes.However how can I calculate the pressure made by a falling object that has a downwards velocity ?

THankkss , that can be a foolish question but I couldn't understand this.
 
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LasTSurvivoR said:
Ok guys everyone knows teh solid Pressure = P / S but i have a question..
I though about an object that falls from 20 meters.(Its mass is 2 kg, 2x2x2 meters cube)
When it reaches the ground how much pressure does it make , in normal case
20 . 10 / 2² = 50 Pascal it makes.However how can I calculate the pressure made by a falling object that has a downwards velocity ?

THankkss , that can be a foolish question but I couldn't understand this.
It depends on how long it takes to stop. It is just a mechanics question involving change of momentum.

[Corrected:]

F = mg + mv/\Delta t (assuming constant stopping force)

P = F/A = \frac{mg + mv/\Delta t}{A}

AM
 
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Thanks that is a very logical thought , and probably the correct one I had never thought this :) I will Try :=)
 
Those equations are alittle messed up andrew, they say that the force decreses the faster you stop :)

It should be:

F = mg + \frac{mv}{\Delta t}

P = F/A = \frac{mg + \frac{mv}{\Delta t}}{A}

v is the velocity at the time of impact, which you can find from kinematics.

For the record, this equation assumes no bouncing.
 
Crosson said:
Those equations are alittle messed up andrew, they say that the force decreses the faster you stop :)

It should be:

F = mg + \frac{mv}{\Delta t}

P = F/A = \frac{mg + \frac{mv}{\Delta t}}{A}

v is the velocity at the time of impact, which you can find from kinematics.

For the record, this equation assumes no bouncing.
Right you are! F\Delta t = mv

AM
 
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