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Pound into Newton

PostPosted: Mon Sep 24, 2007 10:41 am
by Newb
This is what I was given:

One inch= 25.4 mm
Mass of one cubic meter of water= 1000 kg
g=9.81 m/s^2
Weight of one cubic meter of water= 62.4 lb

I have to show from the above data how many Newtons is equal to one pound...I know its 4.448 N, but I cannot seem to show it. SO CONFUSING!!!

Any help would be appreicated. Thanks.

N

Re: Pound into Newton

PostPosted: Mon Sep 24, 2007 6:25 pm
by Guest
Newb wrote:This is what I was given:

One inch= 25.4 mm
Mass of one cubic meter of water= 1000 kg
g=9.81 m/s^2
Weight of one cubic meter of water= 62.4 lb

I have to show from the above data how many Newtons is equal to one pound...I know its 4.448 N, but I cannot seem to show it. SO CONFUSING!!!

Any help would be appreicated. Thanks.

N


Uhh, one cubic foot of water weighs 62.4 pounds. But I don't know what theat has to do with pounds per newton.

The "english" system has both pound-mass (lbm) and pound-force (lbf). At "normal" gravity, 1 lbm exerts a 1 lbf force on whatever holds it up. This certainly isn't true on the moon, or for that matter at the poles or the equator, or on mountain tops, but it is remarkably true at 45 degrees latitude and sea level. "Normal" gravity is taken as 9.80665 m/s^2 exactly, which rounds to your value.

What you really need to know is that 1 lbm is defined as 0.453 592 37 kg in the US (and UK). The force ( f = ma) exerted at standard gravity is 0.453 592 37 kg x 9.80665 m/s^2 = 4.448 222 N, but at standard gravity, 1 lbm exerts force of 1 lbf. Therefore 1 lbf = 4.448 222 N. (Both constants above are exact by declaration, but I rounded the product to 7 figures)

I suppose you are expected to manipulate the water data to reconstruct that 1 lbm = 0.453 592 37 kg, but it is so defined by NIST which is charged by Congress to define weights and measures for the United States.

I hope the teacher has led you through this discussion of pounds-mass and pounds-force and standard gravity. Otherwise, the problem is horribly framed and mostly demonstrates the teacher's ignorance of the SI.