# Force (in lbs) to speed?

• B
Hello,

I have been looking at physics calculators, but cannot seem to find precisely what I need. I don't need a very precise answer, merely a realistic ballpark one.

I need to know how we find the rough speed of an object's movement when a certain amount of force is applied to it.

For example, if I apply a steady 5 lbs of force to a rock weighing 1 lb for 6 seconds, how fast would it move (in mph, or if kph is much easier, that answer instead). Does the speed increase over those 6 seconds? (Assume 0 wind speed.)

Similarly, what happens if a disk on the ground weighs 10 lbs, and I apply 5 lbs of pressure pushing/pulling it on the ground for 6 seconds. How in these instances would I discover the speed of movement? I'm sure the surface friction is relevant to some degree, but how?

Again, I don't need these answers to be exacting, just a general formula for educated guesses.

Daniel

Related Other Physics Topics News on Phys.org
mfb
Mentor
F=ma or equivalently a=F/m - acceleration is force divided by mass. If the force doesn’t change, the final speed is the acceleration multiplied by the time. If the force changes (e.g. with drag) it can get more complicated.

Again, I don't need these answers to be exacting, just a general formula for educated guesses.
Newton's second law relates the force on an object to its acceleration.
$$\mathbf{F}=m\mathbf{a}$$
For realistic cases, friction will create a force which opposes motion. But to calculate this you would generally need to know the coefficient of friction for a particular situation.
I need to know how we find the rough speed of an object's movement when a certain amount of force is applied to it.
Once you know the acceleration of an object, the velocity ##\mathbf{v}## is given as a function of time ##t##,
$$\mathbf{v}=\mathbf{v}_{0}+\mathbf{a}t$$
where ##\mathbf{v}_{0}## is the initial velocity.

PhanthomJay
Homework Helper
Gold Member
In the USA system of measure, do not forget that when you find the acceleration a=F/m, that you must convert the weight of the object (in pounds) into mass units ( in slugs) by dividing weight by acceleration of gravity (m = W/g, where g on planet earth is about 32 ft/sec/sec). The result for 'a' will be in units of ft/sec/sec, and when you find v from v = at, the speed v will be in units of ft/sec.
It's easy ?!
In your first example, a = F/m = 5/(1/32) = 160, and the velocity is v =at = 160(6) = 960 ft/sec, or about 650 miles/hour, faster than a speeding jet!
Nothing is easy in this crazy wonderful mixed up world of Physics.

jbriggs444
Homework Helper
2019 Award
Newton's second law relates the force on an object to its acceleration.
$$\mathbf{F}=m\mathbf{a}$$
[ @PhanthomJay beat me to it]
That formula is valid for coherent systems of units -- i.e. those in which the unit of force is equal to the unit of mass multiplied by the unit of acceleration. SI is a coherent system of units. A force of one newton applied to a mass of one kilogram gives rise to an acceleration of one meter per second squared.

The generalized formula (valid for all systems of units) is
$$\mathbf{F}=km\mathbf{a}$$Where k is a constant that depends on the system of units. In U.S. customary system a force of one pound (force) applied to a mass of one pound (mass) gives rise to an acceleration of one gee -- approximately 32 feet per second per second. The corresponding value for ##k## is approximately ##\frac{1}{32}##

CWatters
Homework Helper
Gold Member
For example, if I apply a steady 5 lbs of force to a rock weighing 1 lb for 6 seconds, how fast would it move (in mph, or if kph is much easier, that answer instead). Does the speed increase over those 6 seconds? (Assume 0 wind speed.)

Similarly, what happens if a disk on the ground weighs 10 lbs, and I apply 5 lbs of pressure pushing/pulling it on the ground for 6 seconds. How in these instances would I discover the speed of movement? I'm sure the surface friction is relevant to some degree, but how?
As others have said.... Use F=ma to work out the acceleration. Note that F is the total or net force so if there is any friction you would typically subtract that from the applied force to give F.

Then use the equations of motion, for example

V = U + at
where
U is the initial velocity (perhaps zero)
V is the final velocity
a is the acceleration
t is the time for which the force is applied.

This works if the acceleration is constant. It's a fair bit more complicated if it's not constant.

The generalized formula (valid for all systems of units) is
F=kma​
\mathbf{F}=km\mathbf{a}Where k is a constant that depends on the system of units. In U.S. customary system a force of one pound (force) applied to a mass of one pound (mass) gives rise to an acceleration of one gee -- approximately 32 feet per second per second. The corresponding value for kk is approximately 132\frac{1}{32}
There is no "k" factor in Newton's Second Law, and to insert on in this fashion is not a good idea at all.

There are actually two unit systems in common use in the USA. One is the ft-lb-s system, and the other is the in-lb-s system. Both are force based, rather than mass based. In the ft-lb-s system, the unit of mass is the slug = 1 lb-s^2/ft. In the in-lb-s system, the unit of mass is the lb-s^2/in. Both work quite well, and Newton's Second law holds with no need for a factor "k."

The pound mass only properly exists in the archaic units system for which the force unit is the poundal.

• mfb and NFuller
jbriggs444
Homework Helper
2019 Award
There are actually two unit systems in common use in the USA.
Those two systems of units are engineering systems. They are rarely encountered in day to day life. The customary system of units has the pound-force as the unit of force and the pound-mass as the unit of mass. It is entirely usual for someone in the U.S. to speak of a 5 pound force applied to a 1 pound mass. One need not convert to engineering units to solve a problem posed this way with customary units.

For purposes of commerce, the pound is legally and most commonly used in the sense of a unit of mass equal to 0.45359237 kg.

Edit: Please do not take this as a defense of the U.S. customary system. Introductory physics curriculum bends over backwards trying to expunge it from students minds -- for good reason.

Those two systems of units are engineering systems. They are rarely encountered in day to day life.
I suggest to you that these unit systems are quite common among people who talk in terms of F = m*a. Your pound force/pound mass confusion is common among folk who never give Newton a thought. Why would you bring that confusion into a technical discussion?

jbriggs444
Homework Helper
2019 Award
I suggest to you that these unit systems are quite common among people who talk in terms of F = m*a. Your pound force/pound mass confusion is common among folk who never give Newton a thought. Why would you bring that confusion into a technical discussion?
I am not confused. Nor are gravitational systems of units invalid.

I will agree with you that coherent systems constructed so that f=ma are superior for engineering use and avoid the confusing feature where the unit of force and of mass share the same name. The original poster's question seems to have been satisfactorily answered. Perhaps we can leave it at that.

Last edited:
Redbelly98
Staff Emeritus
Homework Helper
In the USA system of measure, do not forget that when you find the acceleration a=F/m, that you must convert the weight of the object (in pounds) into mass units ( in slugs) by dividing weight by acceleration of gravity (m = W/g, where g on planet earth is about 32 ft/sec/sec). The result for 'a' will be in units of ft/sec/sec, and when you find v from v = at, the speed v will be in units of ft/sec.
That sounds overly complicated for the OP's scenario. It's 5 lbs of force on a 1 lb object. Clearly the acceleration is 5g. No need to bring slugs into it.

PhanthomJay
Homework Helper
Gold Member
That sounds overly complicated for the OP's scenario. It's 5 lbs of force on a 1 lb object. Clearly the acceleration is 5g. No need to bring slugs into it.
I guess you are saying that a 5 newton force on an object that weighs 1 newton clearly produces an acceleration of 5g, no need to bring kg into it? Now that is confusing. It's F = ma, where m is in slugs in USA system and in kg in the SI system. I think It is unclear to a newcomer that a 1 lbf acting on an archaic 1 lbm produced an acceleration of g. What formula did you use to arrive at that conclusion? Looks like you used F = kma where k is 1/g and m is in lbm, or F = ma, where , m is W/g ( unit is slug). Incidentally, I sure hope text books are not moving away from US units, which are here to stay in the US for many many years. About 50 years have passed since metric was first introduced here, with little progress.
I think the OP must be more confused than before.

I sure hope text books are not moving away from US units,
I have to disagree with you there .

I once had a chemistry professor who would chant to students the phrase "Remember the Mars Climate Orbiter!"
https://en.wikipedia.org/wiki/Mars_Climate_Orbiter

jbriggs444
Homework Helper
2019 Award
1 lbf acting on an archaic 1 lbm produced an acceleration of g. What formula did you use to arrive at that conclusion?
The fact that 1 lbf is the force produced by the Earth's gravity (i.e. one g) acting on 1 lbm. It's a built in feature of the system of units.

I once had a chemistry professor who would chant to students the phrase "Remember the Mars Climate Orbiter!"
This says nothing more than that somebody (I wonder who?) failed to read and follow the contract documents. Has this ever happened previously?

Khashishi
About 50 years have passed since metric was first introduced here, with little progress.
Well, when US loses its dominance, it will be forced to adapt to the global marketplace.

Well, when US loses its dominance, it will be forced to adapt to the global marketplace.
We'll be happy to wait.

mfb
Mentor
Incidentally, I sure hope text books are not moving away from US units, which are here to stay in the US for many many years.
And why does it stay? Textbooks keeping archaic units (your words!) are certainly contributing to the continuous use of such a confusing, incoherent system used by just 5% of the world population, while the remaining 95% use a single coherent system of units.

Meanwhile physics is done exclusively in the metric system even in the US, and engineering is moving in that direction as well. NASA has certainly learned from the past and uses metric where possible.

And why does it stay? Textbooks keeping archaic units (your words!) are certainly contributing to the continuous use of such a confusing, incoherent system used by just 5% of the world population, while the remaining 95% use a single coherent system of units.
There is nothing even faintly incoherent about US Customary units if used properly.

Why is it that in the sol-called SI countries, people routinely give their weight in kilograms, a mass unit? Isn't that supposed to be impossible with the perfect SI system? No, its not, because people can, and will, be confused in either USC or SI.

In one of my earlier jobs (a long time ago), I recall looking at some experimental data done before my time where a series of torque measurements were recorded in gram-inch units. This, of course, referred to gram-force units, with a distance factor in inches. People can and will make mistakes, particularly when academics misinform them about one system versus another (and yes, I say this as a retired Professor).

mfb
Mentor
There is nothing even faintly incoherent about US Customary units if used properly.
There are weird and arbitrary conversion factors everywhere. You need a calculator whenever you want to convert anything. Even if you want to convert something volumes to volumes because it is obvious that 231 cubic inch is the best value for a gallon and 1.8046875 cubic inch (that value is exact) is a natural definition for a fluid ounce (which, of course, has no relation to the ounce as mass unit, because it is not linked via the density of water any more).

Why is it that in the sol-called SI countries, people routinely give their weight in kilograms, a mass unit?
They give their mass and call it weight, but what they are actually interested in is the mass anyway. You can plug that into formulas without having to worry about conversion factors.

• jerromyjon and NFuller
PhanthomJay
Homework Helper
Gold Member
And why does it stay?
Enormous cost of conversion is one reason, but that is secondary to the major reason: Familiarity with the USA units. Every layperson, engineer, construction worker, etc. knows how much an inch or a foot is( About the length of an adult male's shod foot) . But how much is a meter? Not many know. So the construction super calls me the engineer on the phone and asks 'say there, Phantom, how much should I stick this foundation above the ground? It doesn't say on the drawings.' And I answer 'oh, sorry, stick it up about 6 to 12 inches" and the super says thanks and that's it. Now if using SI, and the same question is asked, I'd have to mentally or with calculator respond after several minutes 'about .15 to .3 meters' , hoping I did the conversion right without slipping a decimal point, and the duper would respond' huh? How the heck much is that?' And I would respond 'about 6 to 12 inches' and he is very happy and so am I, relieved that I didn't have to recheck my metric calc. Same goes for building materials be it steel or wood or concrete ; we all as engineers or fabricators know instantaneously that A36 steel has a yield strength of 36,000 pounds per square inch....I can't imagine specifying it as 200 MPa or whatever it is, hoping I didn't mean 200 Gpa, both of which are wrong anyway.
....engineering is moving in that direction as well
On the contrary, in civil engineering, its moving backwards. 20 years ago, the government mandated SI units on state and federal highway plans, expressing contour elevations, distances, road elevations, etc. in meters. And assuming the engineer specified it correctly ,the construction folks were all messed up, and elevations etc came out wrong, and in a short time the drawings went back to good old fashioned feet and inches and SI was gone.

It would be nice to have one global system of measure, but in reality, it will not happen in the foreseeable future.

.

Last edited:
mfb
Mentor
But how much is a meter? Not many know.
School books introducing the SI would change that. That's the point.
Now if using SI, and the same question is asked, I'd have to mentally or with calculator respond after several minutes 'about .15 to .3 meters'
Oh come on. I have used SI exclusively my whole life and only seen US units from internet forums, and I can still do approximate conversions in seconds.

lekh2003
Gold Member
I have to disagree with you there .

I once had a chemistry professor who would chant to students the phrase "Remember the Mars Climate Orbiter!"
https://en.wikipedia.org/wiki/Mars_Climate_Orbiter
I agree with you and find the need for a separate US system unnecessary, besides creating chaos. A universal measuring system was created, but yet I find the US system being used for physics problems. Over complications of unit conversions lead to disasters. Long live the SI system.

In mechanics, the fundamental requirement for consistency is that Newton's Second Law be satisfied with no fudge factors included.

For those who are familiar with Greek and Latin prefixes, the whole SI system is inconsistent if you think about it. The "kilogram" is essentially (by the meaning of the kilo-prefix) 1000 grams. Thus, like it or not, the gram is the fundamental unit of mass for SI, not the kilogram. With that understanding of what is really fundamental, Newton's 2nd Law is again not satisfied in the SI system. Look at the units in Newton's Second Law:

1 lb = 1 slug * 1 ft/s^2 --> FPS system is consistent
1 lb = (1 lb-s^2/in) * 1 in/s^2 --> IPS system is consistent
1 N =/ 1 gram * 1 m/s^2 --> SI system is inconsistent

Prefixes such as milli-, kilo-, centi-, mega-, etc are essentially fudge factors inserted into the equations. The fact that they are powers of 10 makes them somewhat easier to work with, but it does not change their basic nature as fudge factors.

Regarding "weight" expressed in kilograms, it is NOT TRUE that 1 kilogram is 1 kilogram * 1 meter/s^2, so this is an inconsistent usage. It is essentially the same mistake made by expressing both force and mass in pounds, only reversed. Who is to say they are more interested in their mass than in their weight? Are they more interested in the pull of the earth on their bodies and the resultant ache in their ankles and the soles of their feet? Or are they really thinking about how their momentum will be affected in a collision?

Gallons are not standard engineering units in the US Customary system; volume is either in ft^3 or in^3. They are a hold over from earlier time, much like the use of pints for beer or stones for personal weight in the fully metricated UK. In any forced conversion, such as the UK has undergone, there will certainly be holdovers in trade. But we are talking about technical usage, not trade usage.

jbriggs444