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 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)