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Mechanical Engineering DISCUSSION
I am working through a fluid mechanics text that uses US customary units. When I apply F = m × a to a 10 lbm object under gravity I get 322, yet the book says the object weighs 10 lbf. Some equations in the book have a factor g_c in the denominator and others do not.
What is the difference between lbm and lbf, what is g_c physically, and how do I know when to include it?
The pound-mass and the pound-force were defined independently, so that one pound-mass weighs one pound-force under standard gravity. That is convenient on a scale but it means the set of units lbm, ft, s and lbf is not coherent: multiplying mass by acceleration does not give lbf directly. 10 lbm × 32.174 ft/s² = 322 lbm·ft/s², which is a legitimate force unit (the poundal) but not the pound-force.
The conversion is 1 lbf = 32.174 lbm·ft/s². Newton's law in these units is written F = m × a / g_c with g_c = 32.174 lbm·ft/(lbf·s²), and then 10 × 32.174 / 32.174 = 10 lbf.
g_c is a unit conversion factor, not the acceleration of gravity. It has the same numerical value as standard gravity in ft/s² only because of how the pound-force was defined, and it does not change with location. On the Moon, where g is about 5.32 ft/s², the 10 lbm object still has a mass of 10 lbm and weighs 10 × 5.32 / 32.174 = 1.65 lbf.
The alternative is to use the slug as the mass unit. One slug is 32.174 lbm, and with slugs the system is coherent: F in lbf = m in slugs × a in ft/s², with no g_c. 10 lbm is 0.311 slug, and 0.311 × 32.174 = 10 lbf.
As for when to include it: whenever mass in lbm meets force in lbf in the same term. Kinetic energy is m × v² / (2 × g_c): 10 lbm at 50 ft/s has 10 × 2500 / (2 × 32.174) = 388 ft·lbf. The velocity head in the Bernoulli equation written per unit mass is v² / (2 × g_c), and the elevation term is (g / g_c) × z.
Books that leave g_c out are either working in slugs or treating it as understood. Check how density is given: water is 62.4 lbm/ft³ or 1.94 slug/ft³. If the book uses 1.94, it is a slug-based text and needs no g_c.
A reliable working method is to carry the units through every line as if they were algebra and see whether they cancel. If you are left with lbm·ft/s² where you wanted lbf, you are missing a division by g_c. The other reliable method is to convert the inputs to SI, calculate, and convert the result back: 1 lbm = 0.4536 kg and 1 lbf = 4.448 N.
The metric world has the same trap in a milder form. The kilogram-force (9.807 N) still appears on pressure gauges marked kg/cm² and in old torque figures given in kg·m, and treating those kilograms as mass gives errors of a factor of 9.81.
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