Does Mass Affect Acceleration or Not?

[indnajns]

This question may be a bit too basic for you all, but I'm trying to help my 8th grader with his homework. I have come across the following quandry.

From his science book [Prentice Hall Science Explorer Physical Science (Virginia)]

pg 326 "... all objects in free fall accelerate at the same rate regardless of mass."
pg 322 (talking about moving a red wagon) "Another way to increase acceleration is to change the mass. According to the equation [Accel=Force/Mass], acceleration and mass change in opposite ways."

How can these two statements both be true. Are they not mutually exclusive? If F=MA, then A=F/M, which means if you change the mass it HAS to affect acceleration. I don't understand. And PLEASE give the answer in small words. While familiar with the concepts and a straight A student decades ago, I've never had physics. (and I have to be able to explain this back to an eighth grader.)

Thank you in advance.

 

[berkeman]

This question may be a bit too basic for you all, but I'm trying to help my 8th grader with his homework. I have come across the following quandry.

From his science book [Prentice Hall Science Explorer Physical Science (Virginia)]

pg 326 "... all objects in free fall accelerate at the same rate regardless of mass."
pg 322 (talking about moving a red wagon) "Another way to increase acceleration is to change the mass. According to the equation [Accel=Force/Mass], acceleration and mass change in opposite ways."

How can these two statements both be true. Are they not mutually exclusive? If F=MA, then A=F/M, which means if you change the mass it HAS to affect acceleration. I don't understand. And PLEASE give the answer in small words. While familiar with the concepts and a straight A student decades ago, I've never had physics. (and I have to be able to explain this back to an eighth grader.)

Thank you in advance.

F=ma is correct. For an object falling under the influence of gravity, the mass does not affect the acceleration -- that is a constant called "g". As the mass gets larger, it takes a greater force (supplied by the greater weight) to be able to support the same acceleration. g=F/m. The weight (the force) is proportional to mass, so the fraction is the constant g.

 

[indnajns]

Ah. So my problem is oversimplification and gravity. Bear with me, I'm almost onboard.

free fall - when the only force acting on an object is gravity.
gravity - an object's weight is a measurement of the force of gravity on that object.

A=F/M
So gravity is my F. The object's mass is my M. Oh, I get it. On Earth, those are going to be one and the same, so if my kid weighs 35kg I'd have 35kg/35kg which is 1 and no matter what object we're talking about, the ratio will always be the same.

So, 'g' is a special acceleration, unique to free fall on Earth?
A = F/M
9.8m/s2 = 35kg/35kg
9.8m/s2 = 1
g = 9.8m/s2

So, my two original statements differ because the first one is talking ONLY about free fall and the second is acceleration in general.

Am I close? (shew, I hope so!)

 

[RoyalCat]

F=ma is correct.

We're pretty much set on what m is, but F and a may prove a bit tricky.

The apparent contradiction is resolved once we understand that the gravitation force is unique in that it is directly proportional to mass.

In the most general case, let's say you apply a force F to an object of mass m
This action will result in the object accelerating with an acceleration given by Newton's Second Law: a=\frac{F}{m}

For the private case of free-fall, where the only force is that of gravity things turn a bit shifty. The gravitational force is unique in that it is a force that is directly proportional to the mass of the object it is acting on.

That means that an object with 10 times the mass, would experience 10 times the force! How does this hold up against Newton's Second Law, which we know must hold?

Well, let's give the gravitational force a name. We know that it is directly proportional to the mass, by way of some constant, \alpha

F_{gravity}=\alpha \cdot m

Plugging that into F=ma we find that the acceleration of an object of mass m is:

a=\alpha and that is an astounding result in that it is completely independent of the mass of the object in question. Whatever its mass, it will always accelerate at an acceleration equal to \alpha

And we have a special name for that proportionality constant, and for the earth, it is g\approx 9.8 [m/s^2]

And just another pitfall I'd like to comment on is that you've mistaken the terms of "Weight" and "Mass" to be one and the same!
But this is hardly the case! The colloquial use of the two has equated them in informal speech, but in the framework of physics, they are very very different.

Mass is an intrinsic property of an object. It is defined by Newton's Second Law as the constant ratio that you would find between the force you apply to an object and the acceleration it goes through. Its SI units are [kg]

Weight, on the other hand is a force. It is the force that an object experiences when under the influence of gravity. Its SI units are N=\frac{[kg][m]}{[s^2]}

Note how in your calculation you got a ratio with no units, that was equal to some quantity (g) that does have units! That is a fundamental mistake.

It is obvious that you can only equate or relate two quantities if they have the same units. You cannot say that a ratio without units is an acceleration. Just like you can't say that 2 oranges equal 3 apples.

Use the quick and simple test called dimensional analysis (Making sure that both RHS and LHS units are consistent) to screen for possible errors. This is a skill your eighth-grader will use A LOT!

Best wishes. :)