# Two boxes; one pushing the other

• Karajovic
In summary: F1 and F2?There is only one external force on M+m (F2 is an internal force … they don't count).yes, but f1 does equal Ma. but for c) doesn't the F2 force sort of act like a frictional force to the F1 force? Like its a reaction force, but it pushes against the box with f1, so the net is f1 - f2, or not?(and what do you mean by "net acceleration" … there's no such thing :confused:)oh, sorry then :/ what i meant is the acceleration of the system... so a = F1/(M + m) ? then... is there any way
Karajovic

## Homework Statement

Two blocks, of mass M and mass m, are in contact on a horizontal frictionless table (with the block of mass M on the left and the block of mass m on the right). A force F1 is applied to the block of mass M and the two blocks move together to the right.

a. draw a free body diagram

b. Suppose the larger block M exerts a force F2 on the smaller mass m. By Newton’s third law, the smaller block m exerts a force F2 on the larger block M. Argue whether F1 = F2 or not. Justify your reasoning.

c. Derive an expression for the acceleration of the system.

d. Derive an expression for the magnitude of the force F2 that the larger block exerts on the smaller block.

e. Choose different values of M and m (e.g. M = 2m, M = 5m, including the case M = m) and compare the magnitudes of F1 and F2.

## The Attempt at a Solution

a. For box of mass M:
FN
|
F2<--o--->F1
|
Fg

For box of mass m:
FN
|
o--->F2
|
Fg
b. Using Newton's Third Law, for every force there is an equal reaction force. Since the box with mass M is being pushed by the force F1, doesn't that mean that it is pushing the smaller box of mass m with the same force? But this doesn't seem right because they don't have the same masses... and if F1=F2 that means the boxes wouldn't be moving right?

c. a = Fnet/m
a = (mnet*anet)/(M+m)?

d. F2 = Ma

e. F1 = M*a, while F2 = 2m*a (I don't really understand this part)...

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Welcome to PF!

Hi Karajovic! Welcome to PF!

The important point is that both boxes have the same acceleration, a.

Write out the F = ma equation for:
i] both boxes
ii] the left box
iii] the right box.

That will give you all the equations you need relating m M F1 and F2

what do you get?

tiny-tim said:
Write out the F = ma equation for:
i] both boxes
ii] the left box
iii] the right box.

Okay.. so I would get:

i] F = F1 + F2 (?)
ii] F1 = Ma
iii] F2 = ma

I don't know if that's it

Hi Karajovic!
Karajovic said:
i] F = F1 + F2 (?)
ii] F1 = Ma
iii] F2 = ma

iii] is correct.

for ii], you need to use both forces on the M block (there was only one force on the m block, so iii] was ok).

i] isn't an F = ma equation, is it?

tiny-tim said:
Hi Karajovic!

iii] is correct.

for ii], you need to use both forces on the M block (there was only one force on the m block, so iii] was ok).

i] isn't an F = ma equation, is it?

Haha, you're right Then how would I use two forces in one equation? Because for ii] you have F1 = Ma (acting on the M box) and you have F2 = ma (also acting on the M box)...

Karajovic said:
Because for ii] you have F1 = Ma (acting on the M box) and you have F2 = ma (also acting on the M box)...

Nope … ii] is for the M box.

On the LHS, you have all the forces on the M box (ie, the net force).

On the RHS, you have only the mass you're looking at.

(and I'm off to bed :zzz: … see you tomorrow!)

tiny-tim said:
Nope … ii] is for the M box.

On the LHS, you have all the forces on the M box (ie, the net force).

On the RHS, you have only the mass you're looking at.

(and I'm off to bed :zzz: … see you tomorrow!)

Okay so my conclusion:
b) F1 does not equal F2. First of all F1 = Ma while F2 = ma, also, if the forces were equal, the boxes would not be moving

c) anet = (f1-f2)/(M+m)

d) F2 = ma

now for e) I'm not sure.. F1 = Ma while F2 = ma... Now how do I use different values of M and m? I don't seem to understand what it is asking me to find...

Last edited:
No, F1 is not equal to Ma … there are two forces on M.

And c) is wrong … there is only one force on M+m.

Try again.

tiny-tim said:
No, F1 is not equal to Ma … there are two forces on M.

And c) is wrong … there is only one force on M+m.

Try again.

F1 has to equal Ma ? That is the force exerted on the big box, of mass M.

and for c) its asking for the net acceleration, right? So isn't that equal to Fnet/mnet, Fnet equals (F1 - F2 (because F2 is the force hitting the little box by the big, but then an equal force hits the big box)) while mnet is (M+m)

I attached a FBD diagram.

Thanks for the help!

#### Attachments

• M and m boxes.jpg
6.7 KB · Views: 1,220
There is only one external force on M+m (F2 is an internal force … they don't count).

(and what do you mean by "net acceleration" … there's no such thing )

tiny-tim said:
There is only one external force on M+m (F2 is an internal force … they don't count).

yes, but f1 does equal Ma. but for c) doesn't the F2 force sort of act like a frictional force to the F1 force? Like its a reaction force, but it pushes against the box with f1, so the net is f1 - f2, or not?

(and what do you mean by "net acceleration" … there's no such thing )

oh, sorry then :/ what i meant is the acceleration of the system... so a = F1/(M + m) ? then my above theory is wrong :(

Karajovic said:
yes, but f1 does equal Ma. but for c) doesn't the F2 force sort of act like a frictional force to the F1 force? Like its a reaction force, but it pushes against the box with f1, so the net is f1 - f2, or not?

c) is the whole system. Internal forces don't count.
... so a = F1/(M + m) ?

Yes!

tiny-tim said:
c) is the whole system. Internal forces don't count.

Okay, thank you! But may I ask, the net force is just f1 then? You do not subtract f2 from it and is my b) statement still okay?

Karajovic said:
Okay, thank you! But may I ask, the net force is just f1 then? You do not subtract f2 from it

The net force on M+m is F1.

The net force on m is F2.

The net force on M is F1 - F2.

Write out the three F = ma equations (for each of the three systems), to see how they work together …

you should find that 2 of them add to make the third one.
… and is my b) statement still okay?

which b)?

do you mean
b) F1 does not equal F2. First of all F1 = Ma while F2 = ma, also, if the forces were equal, the boxes would not be moving

that's ok except for F1 = Ma.

tiny-tim said:
The net force on M+m is F1.

The net force on m is F2.

The net force on M is F1 - F2.

Write out the three F = ma equations (for each of the three systems), to see how they work together …

you should find that 2 of them add to make the third one.

okay, well i get f1 = (M+m)a while F2 = ma and F1-F2 = Ma + ma - ma = Ma

and f1 does not equal f2 because f1 = (M+m)a while F2 = ma, if they equal, no movement would occur!

How's that it makes sense now!

The total net force of the system is F1, therefore the acceleration of the system is a = f1/(M+m) ??

and what do they mean for e?

thanks again hey, i asked my teach and he said that f1 does equal f2 ? Is he wrong?

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## What is the concept of "Two boxes; one pushing the other"?

The concept of "Two boxes; one pushing the other" refers to a scenario in physics where two boxes are placed on a surface and one box is used to push the other. This is an example of Newton's third law of motion, which states that for every action, there is an equal and opposite reaction.

## What is the significance of this concept?

This concept is significant because it illustrates the principles of force and motion, as well as the application of Newton's third law. It also demonstrates how two objects interacting with each other can affect their motion and the forces involved.

## How does the mass of the boxes affect their motion in this scenario?

The mass of the boxes does not affect their motion in this scenario, as long as the surface they are on is frictionless. According to Newton's second law, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. In this scenario, the net force and acceleration are the same for both boxes, regardless of their mass.

## What would happen if the surface was not frictionless?

If the surface was not frictionless, the friction between the two boxes and the surface would affect their motion. This would result in a decrease in the acceleration of the boxes and an increase in the amount of force needed to push them.

## Can this concept be applied in real-life situations?

Yes, this concept can be applied in real-life situations. For example, when pushing a shopping cart, the force exerted on the cart causes it to move forward, while an equal and opposite force is exerted on the person pushing the cart. This also applies to scenarios such as pushing a car or riding a skateboard.

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