Inverse Proportionality in $F=ma$ equation

If two quantities are inversely related to each other (mass and acceleration), does that mean that they do not depend on each other, i.e. changing mass does not have any effect on acceleration?

My confusion stems from this explanation to a practice problem about the mass and acceleration of a block on a surface where the only force acting on the block is force friction:

"The acceleration of the block does not depend on its mass. The net force on the block directly depends on its mass, but the acceleration is proportional to the force and inversely proportional to the mass, so the acceleration does not depend on the mass"

• I think you are contradicting yourself. If two quantities are inversely proportional, this fact establishes a relation between their values. Therefore, you cannot say that they do not depend on each other Apr 28, 2020 at 22:42
• Hmm, I understand what you are saying, but my question stems from this explanation for why acceleration does not depend on mass for a class practice question. Apr 28, 2020 at 22:44
• you're gonna have to specify the specific concept that takes into account your class practice question Apr 28, 2020 at 22:46
• "The acceleration of the block does not depend on its mass. The net force on the block directly depends on its mass, but the acceleration is proportional to the force and inversely proportional to the mass, so the acceleration does not depend on the mass" Apr 28, 2020 at 22:46
• This is the explanation that is confusing me Apr 28, 2020 at 22:47

You should be confused. Who wrote that? Anyway, we thankfully have mathematics to clear things up:

"The acceleration of the block does not depend on its mass."

"The net force on the block directly depends on its mass,"

$$F = \mu M^{\alpha}$$

where $$\mu$$ is some constant of propotionality. Presumably the adverb "directly" means "linearly", so we can set $$\alpha=1$$. Hence:

$$F = \mu M$$

"but the acceleration is proportional to the force"

$$a \propto F = \mu M$$

"and inversely proportional to the mass",

$$a \propto \frac 1 M$$

which sounds likes:

$$a \propto \frac 1 M \stackrel{?}{\propto} F \propto \mu M$$

meaning $$a \stackrel{?}\propto M^2$$,

...but that's wrong... I forgot something. They said "and", which is subtle, and means that that step should be:

$$a \propto F \times \frac 1 M = \frac F M = \frac{\mu M}M = \mu$$

"so the acceleration does not depend on the mass"

Correct.

• I believe the final character in the final string of equations should be mu, not M (it must be an expression independent of M, that's the whole point). Surely a typo Apr 29, 2020 at 6:15
• @electronpusher thx. Talk about blowing the punchline.
– JEB
May 4, 2020 at 15:07

Your confusion stems from trying to generalise the specific statement that applies only to the practice problem.

First, as noted in JEB's answer, the confusing statement:

"The acceleration of the block does not depend on its mass. The net force on the block directly depends on its mass, but the acceleration is proportional to the force and inversely proportional to the mass, so the acceleration does not depend on the mass"

can be much simply written as:

The acceleration of the block does not depend on its mass, because $$F=ma$$.

Taken by itself, the above is clearly nonsense. But the problem specifies:

...a practice problem about the mass and acceleration of a block on a surface where the only force acting on the block is force friction...

Importantly, the part in bold tells you that $$F=\mu mg$$, where $$\mu$$ is the friction coefficient. So: $$$$F=\mu mg = ma \quad\implies\quad a = \mu g\,,$$$$ which is indeed independent of the mass.