Why Does the Friction Force Not Match in the Two Block Problem?

[Hamiltonian]

Homework Statement

a block of mass 4kg is kept on top of another block of mass 1kg. the coefficient of friction between the two blocks is 0.5. The surface on which the 1kg block is kept is smooth. does the 4kg block accelerate if a force of 10N is applied on it?

Relevant Equations

F = ma

when a force of 10N is applied to the 4kg block the force of friction between the two blocks will also equal 10N as the maximum value of friction between the two surfaces is f = N$\mu$ = 20N. if you look at this free body diagram f = F = 10N so the net force acting on the top(4kg) block will be 0N and the net force acting on the bottom block will be 10N hence from this you can conclude that the bottom block will have an acceleration of 10m/$s^2$ and the top block will have an acceleration of 0m/$s^2$ which is physically impossible! I don't understand in which step I am making a mistake. I realize that both the blocks should move together with the same acceleration till the time
F $\leq$ $\mu$N

 

[PeroK]

What would happen if the blocks were stuck together?

 

[Hamiltonian]

What would happen if the blocks were stuck together?

if they were to be "stuck" together and a force of 10N is applied then they would move with the same acceleration = 10/5 = 2m/$s^2$

 

[PeroK]

if they were to be "stuck" together and a force of 10N is applied then they would move with the same acceleration = 10/5 = 2m/$s^2$

What force (of friction) do you need between the blocks for them to move like this?

 

[Hamiltonian]

What force (of friction) do you need between the blocks for them to move like this?

well the force of friction between them should > 0, and for them to move with the same acceleration it depends on the force applied F, they move together only when the force applied F $\leq$ $\mu$N

 

[PeroK]

well the force of friction between them should > 0, and for them to move with the same acceleration it depends on the force applied F, they move together only when the force applied F $\leq$ $\mu$N

You should be able to work it out more precisely.

 

[Hamiltonian]

when a force of 10N is applied to the 4kg block the force of friction between the two blocks will also equal 10N as the maximum value of friction between the two surfaces is f = N$\mu$ = 20N. if you look at this free body diagram f = F = 10N so the net force acting on the top(4kg) block will be 0N and the net force acting on the bottom block will be 10N hence from this you can conclude that the bottom block will have an acceleration of 10m/$s^2$ and the top block will have an acceleration of 0m/$s^2$ which is physically impossible! I don't understand in which step I am making a mistake. I realize that both the blocks should move together with the same acceleration till the time
F $\leq$ $\mu$N

I want to know what is wrong with the steps I have mentioned above.

 

[PeroK]

I want to know what is wrong with the steps I have mentioned above.

You can't assume the maximum frictional force.

 

[Hamiltonian]

You can't assume the maximum frictional force.

i am not assuming the maximum friction force. the max friction force is given by f = $\mu$N, where N is the normal reaction.

 

[PeroK]

i am not assuming the maximum friction force. the max friction force is given by f = $\mu$N, where N is the normal reaction.

No it isn't. That's the maximum friction force possible. Friction is a reactive force and needs to be calculated.

Also, think about Newton's third law between the blocks.

 

[Hamiltonian]

No it isn't. That's the maximum friction force possible. Friction is a reactive force and needs to be calculated.

yes I know the frictional force is self-adjusting up to a maximum value and hence I can conclude that in this case, the force of friction is equal to 10N.

Also, think about Newton's third law between the blocks.

are you suggesting that the forces of friction get canceled, if so then none of the blocks will accelerate

 

[PeroK]

yes I know the frictional force is self-adjusting up to a maximum value and hence I can conclude that in this case, the force of friction is equal to 10N.

There is absolutely no reason to assume that.

You have to start calculating and stop speculating.

are you suggesting that the forces of friction get canceled, if so then none of the blocks will accelerate

The forces of friction are equal and opposite. One block on the other.

 

[Hamiltonian]

There is absolutely no reason to assume that.

You have to start calculating and stop speculating.

I have calculated the value of acceleration to be 0m/$s^2$ for the 4kg block
but I know that is wrong ...

 

[PeroK]

I have calculated the value of acceleration to be 0m/$s^2$ for the 4kg block
but I know that is wrong ...

We know that already. Calculate the forces on both blocks properly.

 

[Hamiltonian]

We know that already. Calculate the forces on both blocks properly.

10 - f = 4a
f = 1a
'a' is the acceleration of both the blocks( i have henced assumed they move together)

solving this I get f = 2N and not 10N.

 

[PeroK]

10 - f = 4a
f = 1a
'a' is the acceleration of both the blocks( i have henced assumed they move together)

solving this I get f = 2N and not 10N.

Looks okay.