mechanical engineering - Pound-force (lbf) vs Pound-mass (lbm ...
Aug 25, 2015 · 1 pound mass is that mass that weighs one pound in 1 g gravity. For most practical cases, a pound mass and a pound weight define the same amount of stuff on the surface of the earth. To define a pound mass we rearrange Newton's law of F = mA to. m = F/A. then plug in the particulars to get pound mass: 1 pound mass = (1 pound force)/(32.174 ft/s²)
pound force and pound mass - Engineering Stack Exchange
Feb 16, 2021 · By example, an object with a mass of 5 lb $_m$ has a mass of 2.2680 kg regardless of where it sits (e.g. what planet). But this mass. exerts 5 lb $_f$ as its weight on earth; exerts 0.827 lb $_f$ as its weight on the moon with gravitation 5.32 ft/s $^2$ By inversion, an object that exerts 5 lb $_f$ as its weight on the moon has a mass of 30.2 ...
simulation - Where does weight mass conversion ratio '0.00259' …
Aug 22, 2016 · I think this is very fundamental question. I am wondering why 0.00259 is used as weight-mass conversion ratio, when inch/pound unit is used in Nastran.
fluid mechanics - Different Values for Different Calculations of ...
Apr 19, 2022 · In the Imperial system the mass unit is the slug [sl], and is derived from the pound-force by defining it as the mass that will accelerate at $1 ft/s^2$ when a $1 lb_f$ acts upon it. The density of water is $1.940 sl/ft^3$ at 39 °F (4 °C), and the specific weight in Imperial units is:
Understanding required torque for a motor lifting a weight
May 14, 2015 · 5 cm = 0.05 m. The gravitational force on your 450 g mass is 4.4 N as you say, so the torque just to keep up with gravity is (4.4 N)(0.05 m) = 0.22 Nm. However, that is the absolute minimum torque just to keep the system in steady state. It leaves nothing for actually accellerating the mass and for overcoming the inevitable friction.
Mass and weight of air in a room - Engineering Stack Exchange
A pound force is defined as the force required to accelerate a slug at 1 ft/s^2. The density of air is $\rho = 0.0724 \ lb_m/ft^3 = 0.0724/32.2 \ slugs/ft^3$. The weight of the air is $\rho V g = 0.0724/32.2 \ slugs/ft^3 \cdot32.2 ft/s^2\cdot 6000 ft^3 = 0.0724\cdot 6000 \ …
Force at end of pivot beam, based on total mass and CG?
Jun 13, 2023 · A handle lever need to be balanced so it does not move by gravity or vertical acceleration. This handle moves ±90∘ from center (top position). Supported by ball bearing in center. The unit can bare...
General Units question - Engineering Stack Exchange
Mar 24, 2022 · Either that shouldn't resolve, or it's something like length/length and you missed that simplification. Did you use "lb" for both pound-force and pound-mass, and did you use "kg" for both kilograms and 9.81 Newtons (i.e., the force exerted by a 1kg mass in one earth gravity)? Perhaps show what you actually did. $\endgroup$ –
airflow - How do I convert cubic metres/feet per minute to a …
May 17, 2016 · The remaining problem is to convert kg to pounds. Kg is a unit of mass, and pounds is a unit of weight. The conversion depends on gravity. You didn't say if whether this is on the surface of the earth, on the internal space station, or somewhere else. I'll assume the surface of the earth where 1 kg of mass weighs about 2.2 pounds.
Converting air flow rate between kg/s and m^3/s
Apr 21, 2015 · Constant Mass Flow: $$ \dot{m} = \rho A V $$ (Hence mass flow is equal to density times area times velocity. So assuming you have an area of $1$ m^2 with a velocity of $1$ m/s, air with a density of $1.225$ kg/m^3 will equate to a mass flow of $1.225$ kg/s. For incompressible flows, the mass flow rate is constant.