Agent Smith Posted May 4 Share Posted May 4 (edited) This is from memory of high school physics. It concerns the mathematical relationship directly proportional to between force (F), mass (m) and acceleration (a). F = ma 1. F [math]\propto[/math] m 2. F [math]\propto[/math] a Therefore, says my physics textbook, 3. F [math]\propto[/math] ma Can someone please tell me how that's derived i.e. how do we infer 3 from 1 and 2? Muchas gracias in advance! Edited May 4 by Agent Smith Link to comment Share on other sites More sharing options...

Genady Posted May 4 Share Posted May 4 1. F=xm where x does not depend on m 2. F=ya where y does not depend on a 3. F=xma/a=ya 4. y=xm/a 5. since y does not depend on a, xm/a does not depend on a 6. x/a does not depend on a 7. x=za where z does not depend on a or on m 8. F=zma @Agent Smith 1 Link to comment Share on other sites More sharing options...

MigL Posted May 4 Share Posted May 4 Directly proportional implies a linear relationship between two variables. If you double the one variable, the other doubles. If you triple it, the other variable also triples. And so on ... For F=ma,you have three variables, and the correct reading is Force and acceleration are directly proportional when mass is constant. IOW, it would not work for a rocket which changes mass as it expels fuel. Nor would it work in relativistic situations. 2 Link to comment Share on other sites More sharing options...

studiot Posted May 4 Share Posted May 4 7 hours ago, Agent Smith said: This is from memory of high school physics. It concerns the mathematical relationship directly proportional to between force (F), mass (m) and acceleration (a). F = ma 1. F [math]\propto[/math] m 2. F [math]\propto[/math] a Therefore, says my physics textbook, 3. F [math]\propto[/math] ma Can someone please tell me how that's derived i.e. how do we infer 3 from 1 and 2? Muchas gracias in advance! As MigL (+1) has noted, direct proportionality between two variables is the starting point for this. But a variable (force, F in this case) can be proportional to more than one variable at the ame time. In with your example, Force, F, is directly proportional to the mass of a body all other influences be held constant. It is also proportional to acceleratio, again all other influences be held constant. When this happens, the dependent variable (F) is given by the product of the two influencing variables as in F = ma. (Sometimes there will be three or more variables then we have a triple product. This happens in fluid mechanics and electronic signal theory) This is very good and very important mathematically because the variables are 'separated', which means we can deal with them separately or independently. Separation of variables is the method of solution for some of the most important equations in Physics. Going back to two variables, one directly proportional to the other I would add to MigL's description that a graph or plot of one variable against the other is a straight line graph through the origin. Straight lines not through the origin, are not representative of direct proportionality. 1 Link to comment Share on other sites More sharing options...

Agent Smith Posted May 5 Author Share Posted May 5 To all the above posters: Merci beaucoup! The way I understand it is as follows. F [math]\propto[/math] a Ergo, F = k_{1}a (k_{1} is the constant of proportionality) --- F [math]\propto[/math] a Ergo, F = k_{2}a (k_{2} is the constant of proportionality) Note that k_{1} contains a (it remains constant) and k_{2} contains m (it's part of the constant of proportionality) So, F = k_{3}ma (factoring out a constant from k_{1} and k_{2}) Does this make any sense 1 Link to comment Share on other sites More sharing options...

Genady Posted May 5 Share Posted May 5 4 hours ago, Agent Smith said: To all the above posters: Merci beaucoup! The way I understand it is as follows. F ∝ a Ergo, F = k_{1}a (k_{1} is the constant of proportionality) --- F ∝ a Ergo, F = k_{2}a (k_{2} is the constant of proportionality) Note that k_{1} contains a (it remains constant) and k_{2} contains m (it's part of the constant of proportionality) So, F = k_{3}ma (factoring out a constant from k_{1} and k_{2}) Does this make any sense Yes, except for the typo: you have 'a' twice and no 'm' with both k_{1 }and k_{2}. If you replace, say, F = k_{1}a with F = k_{1}m, then your k_{1} is what I've called 'x', your k_{2} is my 'y', and your k_{3} is my 'z' in the first comment above . 1 Link to comment Share on other sites More sharing options...

studiot Posted May 5 Share Posted May 5 Yes you have definitely got the idea. +1 F = k_{1}m and F = k_{2}a are simultaneous equations so we combine them as I indicated to form one equation F=k_{3}ma Having got that out of the way a couple of small points 5 hours ago, Agent Smith said: Note that k_{1} contains a (it remains constant) and k_{2} contains m (it's part of the constant of proportionality) It is not quite right to say that k_{1} 'contains' a or that k_{2} 'contains' m which brings us neatly to the subject of units. Nearly always in Physics, the constant of proportionality also 'contains' the tranformation of units. It is not just a number (as it is in pure maths) it has units of its own that are very important. F, a and m all have different units so k_{1} converts acceleration units to force units and k_{2} converts mass units to force units. Can you see why it is this way round ? Furthermore in SI units the constants are arranged so that k_{3} = 1 This is the basis for physicists saying that the force is proportional to the acceleration and the 'constant of proportionality' is mass. This is true for a particular body when mass is not changing. (again as MigL has already noted) and is the usual form of presenting Newton's second law (N2) 5 hours ago, Agent Smith said: F = k_{3}ma (factoring out a constant from k_{1} and k_{2}) This is combining the two constants k_{1} and k_{2} to form a new constant k_{3} . Factoring would be splitting a single constant k_{3} into factors (two factors k_{1} and k_{2} in this case) 1 Link to comment Share on other sites More sharing options...

Agent Smith Posted May 23 Author Share Posted May 23 On 5/5/2022 at 1:16 AM, studiot said: When this happens, the dependent variable (F) is given by the product of the two influencing variables as in F = ma. That's what I don't understand. Why is this true? F [math]\propto[/math] m F [math]\propto[/math] a [math]\therefore[/math] F [math]\propto[/math] ma ??? Link to comment Share on other sites More sharing options...

uncool Posted June 15 Share Posted June 15 I don't think that anyone really answered the question, because the argument the question is about uses assumptions that aren't that obvious. The statement "Force is proportional to mass" is somewhat imprecise; the fully precise statement is "Force is proportional to mass when acceleration is held constant." That is, if you have two masses which are accelerating in the same way, then the force on one divided by its mass will equal the force on the other divided by its mass. Alternatively, for any situation with forces and masses, F/m is a function of the acceleration. Similarly, the statement "Force is proportional to acceleration" is imprecise, with a more precise version being "Force is proportional to acceleration when mass is held constant." If you have two objects with the same mass, the force on one divided by its acceleration will equal the force on the other divided by its acceleration. F/a is a function of the mass. So we have that F/m = f(a) for some function f, and F/a = g(m) for some function g. Then F/ma = f(a)/a = g(m)/m. f(a)/a is also a function of acceleration independent of mass, and g(m)/m is also a function of mass independent of acceleration. And the only way that can happen is if f(a)/a = g(m)/m is constant - is independent of both mass and acceleration. Call that constant C. Then F = C*ma - in other words, force is proportional to the product of mass and acceleration. Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now