# Converting to lbs of force

Im trying to find out how much i would weigh (in pounds) if I were to land on a scale from a 10 foot drop, in a vacuum of course. Does anyone know the equation that would solve this? I know the scale would level back out to my original weight, but I want the weight it would initially display.

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russ_watters
Mentor
It depends on how fast you decelerate when you hit the ground.

russ_watters said:
It depends on how fast you decelerate when you hit the ground.
I would round it off to instantly, no? Just imagine my legs being perfectly straight as i hit the ground..

Mmats said:
I would round it off to instantly, no? Just imagine my legs being perfectly straight as i hit the ground..
Thing is, when you lose momentum like this, you exert what is called an impulse, or force * time. If you decelerate instantly, that means you exert an infinitely large force for an infinitely small period of time.

3.048m/s * 68.04kg = 207.38 kg m/s

so the scale would read 207.38kg?

russ_watters
Mentor
No. Notice that your units don't match...

jtbell
Mentor
Mmats said:
3.048m/s * 68.04kg = 207.38 kg m/s
If it takes you one second to slow to a stop, the average force that the ground exerts on you is 207.38 N. If it takes 0.1 second, the average force is 2073.8 N. If it takes 0.01 second, the average force is 20738 N. And so forth.

i guess its alot more complicated than i thought it would be, actually it seems somewhat impossible to even come up with even a good estimate. thanks for all the help anyways

A physics book could present such a problem and include the number of springs in the scale and the spring constants of each spring....

Andrew Mason
Homework Helper
Mmats said:
Im trying to find out how much i would weigh (in pounds) if I were to land on a scale from a 10 foot drop, in a vacuum of course. Does anyone know the equation that would solve this? I know the scale would level back out to my original weight, but I want the weight it would initially display.
You would need to know the spring constant of the scale, then use:

$$mgh = \frac{1}{2}k(\Delta x)^2$$

where $\Delta x$ is the displacement of the spring from its equilibrium position when you are just standing on it. This gives:

$$\Delta x = \sqrt{2mgh/k}$$

You would then add mg/k (the spring extension due to your weight when just standing on it) to that to find the maximum displacement.

So:

$$F = k\Delta x + mg = \sqrt{2mghk} + mg$$

The higher the spring k, the higher the force.

AM