Pressure on face of object when dropped

Victor

If I vertically drop a flat-faced object of 55mm diameter from various heights on the floor, surely as the height increases, the pressure exerted on the face also increases?

I presume I am using the wrong formula here somewhere.

y (height) m 1.000 2.000 3.000 4.000 5.000 6.000 7.000 8.000

v = √(2gy)
v (velocity) m/s = 4.429 6.264 7.672 8.859 9.905 10.850 11.719 12.528
g (gravity) m/s^2 9.810 9.810 9.810 9.810 9.810 9.810 9.810 9.810
y (height) m 1.000 2.000 3.000 4.000 5.000 6.000 7.000 8.000

F=ma kg.m/s^2
F (force) kg.m/s^2 = 9.810 9.810 9.810 9.810 9.810 9.810 9.810 9.810
m (mass) kg = 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000
a (acceleration) m/s^2 = 9.810 9.810 9.810 9.810 9.810 9.810 9.810 9.810

p=F/A
p (pressure) kg.m/s^2/m^2 = 4,129 4,129 4,129 4,129 4,129 4,129 4,129 4,129

A= πr^2
r m = 0.028 0.028 0.028 0.028 0.028 0.028 0.028 0.028
A (area) m^2 = 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002

dlouth15

You are using F/A which would work if the object were resting on a solid surface since the solid surface will exert a force equal and opposite to that of gravity. But what happens with air is that it flows around the body and the slower it is moving the better able it is to do this. As the speed builds up the force of air (drag) acting against the object also build up until it matches the force due to gravity and the terminal velocity is reached.

Have a look at the article for the drag equation. The formula is given near the top. You can see that the force is proportional to the square of the velocity. The average pressure due to drag on the bottom of the object would then be [latex]F_D / A[/latex] (in addition to the normal atmospheric pressure).

You will have to take into account the force due to drag in addition to the force due to gravity in your calculations of velocity as the object falls from various heights.

Mellor

Not the wrong formula, just a wrong number.

The issue that I can see is that in the F=ma section, you have the acceleration as 9.81, ie gravity. But that's not correct, is also why all the answers are the same, as force of gravity is constant. We are looking for the force when the object hits the table, so the acceleration value should be the rate of acceleration as the object slows to zero.
Higher drop, great velocity, greater acceleration, greater force.

If you don't have a value for that, you'll need to estimate it. If its almost instant, just use a very small unit of time (as its never instant)

dlouth15 wrote: »

You are using F/A which would work if the object were resting on a solid surface since the solid surface will exert a force equal and opposite to that of gravity. But what happens with air is that it flows around the body and the slower it is moving the better able it is to do this. As the speed builds up the force of air (drag) acting against the object also build up until it matches the force due to gravity and the terminal velocity is reached.

Drag (or lack of it) isn't the issue with the OP.
Drag will affect the speed of the object by the time it hits the floor, and by extension the force of it hitting the floor. But it the rapid deceleration removes the drag, so adding it on top isn't right.

The speeds here are so far from terminal velocity, that its not worth bothering with drag. And assume its negligible, or in a vacuum.

dlouth15

Mellor wrote: »

Drag (or lack of it) isn't the issue with the OP.
Drag will affect the speed of the object by the time it hits the floor, and by extension the force of it hitting the floor. But it the rapid deceleration removes the drag, so adding it on top isn't right.

The speeds here are so far from terminal velocity, that its not worth bothering with drag. And assume its negligible, or in a vacuum.

Doesn't he want the air pressure on the bottom surface as the object falls? I may have misread the question. When I think of pressure I tend to think of fluids etc.

I can see where you are coming from with your answer. The pressure you are giving is due to the rapid deceleration as the object hits the floor. I agree this is what Victor wants.

Victor

For the purposes of this exercise, we can ignore air resistance / treat it as being in a vacuum.

I'm talking about the force and pressure values that the object and the floor exert on each other.

So, the object makes a 30mm dent in the metal floor. Let us ignore that a greater height may mean a greater dent.

Do I have it right (let us exclude bounce, compression of the object, etc.) that deceleration time is stop distance divided by velocity?

So force and pressure are proportional to height dropped.

h (height) m 1.000 2.000 3.000 4.000 5.000 6.000 7.000 8.000 9.000 10.000
g (gravity) m/s^2 9.810 9.810 9.810 9.810 9.810 9.810 9.810 9.810 9.810 9.810

v = √(2gh)
v (velocity) m/s = 4.429 6.264 7.672 8.859 9.905 10.850 11.719 12.528 13.288 14.007

t=s/v ???
t (deceleration time) seconds = 0.007 0.005 0.004 0.003 0.003 0.003 0.003 0.002 0.002 0.002
s (stop distance) m = 0.030 0.030 0.030 0.030 0.030 0.030 0.030 0.030 0.030 0.030

s = vt + (1/2)dt^2
'=>d= 2(s/t^2 -vt)
d (deceleration) m/s^2 = 1,308 2,616 3,924 5,232 6,540 7,848 9,156 10,464 11,772 13,080

F=md kg.m/s^2
F (force) kg.m/s^2 = 1,308 2,616 3,924 5,232 6,540 7,848 9,156 10,464 11,772 13,080
m (mass) kg = 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000

A= πr^2
r m = 0.028 0.028 0.028 0.028 0.028 0.028 0.028 0.028 0.028 0.028
A (area) m^2 = 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002

p=F/A
p (pressure) kg.m/s^2/m^2 = 550,536 1,101,096 1,651,657 2,202,218 2,752,779 3,303,340 3,853,900 4,404,461 4,955,022 5,505,583

Lbeard

Victor wrote: »

So force and pressure are proportional to height dropped.

Pressure is defined as force per unit area.

The impact force/ impact pressure will be proportional to the height dropped.

What you doing? What are you building in there?