What Happens to 'a' When Multiplying by 2x?

[AlexKalm]

If we multiply M by 2x then what will happened to a ?

1] a
2] a/2
3]a*2

 

[Bandersnatch]

Hi, AlexKalm

What do you think will happen?

 

[AlexKalm]

1] a?

i only know that 2x f = 2x a ...

 

[CAF123]

Is that an incomplete problem statement ? If a force acts on a body of mass M producing an acceleration a, then if that same force acts on a body 2M then what is the acceleration?

 

[AlexKalm]

yes that was the question

 

[Bandersnatch]

(Note: we will not answer the question for you, nor will we tell you whether your guess is correct unless you show us your reasoning. It's against the forum rules.)

O.k., look, you've got three variables. If you change one, then one of the three things can happen:
1. the first of the remaining variables changes
2. the other one does
3. both do

If your equation is $F=ma$, and you increase F by a factor of x, then either m has to increase *x, a has to increase *x, or each of those has to change by some factors, that when combined produce *x.

However.

In most cases, when dealing with the above equation, one of the variables is treated as constant, so changing one of the two remaining ones by the factor of x, means that the second one needs to also change by the factor of x(if it's on the other side of the equation) or by 1/x(if it's on the same side).

In your example in post #3, m was treated as constant(you asked yourself what happens if you apply more force to the same mass), so increasing F *2 meant that a had to increase *2 as well.

Now it is F that is constant, and you increase m *2(you want to know what happens if the same force acts on a heavier object).
So what happens to a?

 

[AlexKalm]

nothing ! ? am i correct?

because hmm ... only when you double the F the A gets 2x, m can't affect a

and by doing maths a/2 is correct too

because say you have 1000N f , m1 = 10 , m2 = 20

1000=10 a => a=1000/10=100
1000=20 a => a=1000/20=50
so from that we can say that if we double the m we get a/2

but i believe that if we double the m nothing happends to a
maybe if we double the m the f gets m/2 or something ...
then how can i know which answer is correct?

 

[Bandersnatch]

Do you play football(or soccer)?

Imagine you kick the standard soccer ball as hard as you can(force=F₁), and you see the ball accelerating very fast(a₁) so that it can fly very far.
Now imagine you kick a much heavier ball(a baseball, for example; m₂>m₁). Again, you use all your strenght(force=F₂=F₁).
Will the ball land at the same distance, farther, or nearer?

Remember that here you're assuming the same F for both cases(you can't kick it any harder).

 

[AlexKalm]

it will be nearer i guess? but that doesn't clarify anything... :(

 

[Bandersnatch]

Doesn't it? Your maths in post #7 show clearly that the acceleration is lower, your soccer-player's intuition(basically, an experiment) tells you the same, so where's the problem?

 

[AlexKalm]

so it is a/2 ? i don't trust maths ...

 

[NascentOxygen]

because say you have 1000N f , m1 = 10 , m2 = 20

1000=10 a => a=1000/10=100
1000=20 a => a=1000/20=50
so from that we can say that if we double the m we get a/2 ✔

 

[AlexKalm]

yay thanks... i done it wrong in the exams... i will never trust myself again :D

 

[Bandersnatch]

Trust the maths, especially when doing maths exams.
(But don't trust lie algebra :tongue:)