Direct Proportionality

[Agent Smith]

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, 2022 by Agent Smith

[Genady]

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

[MigL]

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.

[studiot]

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.

 

[Agent Smith]

To all the above posters:

Merci beaucoup!

The way I understand it is as follows.

F [math]\propto[/math] a

Ergo, F = k₁a (k₁ is the constant of proportionality)

---

F [math]\propto[/math] a

Ergo, F = k₂a (k₂ is the constant of proportionality)

Note that k₁ contains a (it remains constant) and k₂ contains m (it's part of the constant of proportionality)

So,

F = k₃ma (factoring out a constant from k₁ and k₂)

Does this make any sense

[Genady]

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 = k₁a (k₁ is the constant of proportionality)

---

F ∝ a

Ergo, F = k₂a (k₂ is the constant of proportionality)

Note that k₁ contains a (it remains constant) and k₂ contains m (it's part of the constant of proportionality)

So,

F = k₃ma (factoring out a constant from k₁ and k₂)

Does this make any sense

Yes, except for the typo: you have 'a' twice and no 'm' with both k₁ and k₂.

If you replace, say, F = k₁a with F = k₁m, then your k₁ is what I've called 'x', your k₂ is my 'y', and your k₃ is my 'z' in the first comment above  .  

[studiot]

 

Yes you have definitely got the idea. +1

F = k₁m

and

F = k₂a

are simultaneous equations

so we combine them as I indicated to form one equation

F=k₃ma

Having got that out of the way a couple of small points

5 hours ago, Agent Smith said:

Note that k₁ contains a (it remains constant) and k₂ contains m (it's part of the constant of proportionality)

It is not quite right to say that k₁ 'contains' a or that k₂ '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 k₁ converts acceleration units to force units and k₂ 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 k₃ = 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 = k₃ma (factoring out a constant from k₁ and k₂)

This is combining the two constants k₁ and k₂ to form a new constant k₃ .

Factoring would be splitting a single constant k₃ into factors (two factors k₁ and k₂ in this case)

[Agent Smith]

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 ???

[uncool]

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.

[Agent Smith]

On 6/15/2022 at 11:11 AM, uncool said:

And the only way that can happen is if f(a)/a = g(m)/m is constant - is independent of both mass and acceleration.

Thanks a million. I got most of what you said except the part quoted above. 

[Agent Smith]

On 5/4/2022 at 7:02 PM, Genady said:

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

Can you explain a little bit more about how the constants being independent of acceleration and mass matter to understanding this rule that if ...

[math]n \propto q ~ \& ~ n \propto r, then ~ n \propto qr[/math]

[Genady]

1.  Given n=F(r)*q

2. Given n=G(q)*r

3. From 1 and 2, F(r)*q=G(q)*r

4. From 3, G(q)/q=F(r)/r

5. To hold for all q and r the two sides of the equation in 4 have to be constant. Let's call it, c

6. From 4 and 5, G(q)=c*q

7. From 2 and 6, n=c*q*r

QED

If any line is unclear, let me know.

[Agent Smith]

8 hours ago, Genady said:

If any line is unclear, let me know

 

I would like an intuitive explanation also please, if possible. 

 

I've noticed the word "independent" appears in almost all answers to this question. What does it mean exactly and how does [math]F \propto m[/math] independent of mass and [math]F \propto a[/math] independent of acceleration lead to the conclusion that [math]F \propto ma[/math]

[Genady]

1 hour ago, Agent Smith said:

intuitive explanation

n is proportional to q. This means that making q twice as big makes n twice as big.

n is proportional to r. This means that making r 3 times as big makes n 3 times as big.

Now, what happens to n if q is made twice as big and r is made 3 times as big? n should become 6 times as big. I.e., n is proportional to q*r.

1 hour ago, Agent Smith said:

"independent"

For example, in my last derivation on step 1, F(r) is some function of r. It depends on r, but it does not depend on q, i.e., it is independent of q.

"F depends on r" means that if r varies, F might vary.

"F does not depend on q, i.e., it is independent of q" means that varying q alone does not have any effect on F.

[Reduct Blog]

On 5/4/2022 at 4:59 AM, 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!

I never liked the equation F = ma, since it confuses cause with effect in the physical context we usually refer to, i.e. bodies moving in space.

We look at a body, moving in space, and try to change its motion, i.e. effect an acceleration.

The more force we apply, the greater the resulting acceleration, a ∝ F, (all else being equal).
The more massive the body, the less the resulting acceleration, a ∝ 1/ m, (all else being equal).

So, F / m = a,  is a better description, at least to me.

(Newton's words in the Principia were ~ "The change of motion is proportional to the motive force, and is made in the direction of the straight line in which that force is impressed".)