Force Applied to Two Blocks: Which is Bigger?

[thonwer]

Homework Statement

Two blocks with masses m1 and m2, m2>m1, are attached, I push from the left and from the right side, which force is bigger? The surface is frictionless.

http://imagizer.imageshack.us/v2/800x600q90/194/wkft.png

2. The attempt at a solution

I think it's bigger when applied from the left because of the reaction of m2 is bigger than the reaction of m1 because m2>m1, but I don't know any equations to corroborate it.

 

[tiny-tim]

hi thonwer! welcome to pf!

Two blocks with masses m1 and m2, m2>m1, are attached, I push from the left and from the right side, which force is bigger? The surface is frictionless.

i don't understand the question, there seems to be something missing

why can't the forces be whatever you want?

 

[thonwer]

The force I have to consider is the arrow drawn in each case because it is what they ask me, the picture is part of the formulation.
What I mean on the question is, in one case I push from the left, in other case I push from the right. In which case the force is bigger so as to move the blocks?

 

[tiny-tim]

hi thonwer!

The force I have to consider is the arrow drawn in each case because it is what they ask me, the picture is part of the formulation.

but the picture shows the same letter, F, for both forces, which presumably means that they are equal

do you mean which acceleration is bigger?

 

[thonwer]

Ok sorry, I did not explain myself quite well, the Force applied is the same in both cases.

The correct question would be, which reaction force is bigger?

 

[tiny-tim]

The correct question would be, which reaction force is bigger?

ahhh!

well, as you probably know, the acceleration is the same in both cases

so do good ol' Newton's second law for the block that the force F isn't on in each case …

what do you get? ​

 

[thonwer]

I didn't understand you sorry, English is not my first language,
what did you want to say with "do good ol' Newton's second law for the block that the force F isn't on in each case" ?

 

[tiny-tim]

I didn't understand you sorry, English is not my first language,
what did you want to say with "do good ol' Newton's second law for the block that the force F isn't on in each case" ?

sorry!

in the first picture, the force F is on the left, pushing on the smaller block

so do the equation "force = mass times acceleration" (ie Newton's second law) for the larger block

(and then do the same for the smaller block in the second picture)

 

[thonwer]

For 1: F-R=m1*a -> R=F-m1*a

For 2: F-R=m2*a -> R=F-m2*a

F is the same, a is the same => m1*a<m2*a so R in 1 is bigger

Is it correct?

 

[tiny-tim]

yes!

(though it would have been slightly quicker to do the other block in each case:

For 1: R=m2*a

For 2: R=m1*a )​