Two block friction problem

In summary, when a force of 10N is applied to a 4kg block, the force of friction between the two blocks will also equal 10N. However, this is not a maximum value and cannot be assumed. The forces of friction are equal and opposite, and the blocks will move together with the same acceleration only when the force applied is less than or equal to the maximum friction force. It is important to calculate the forces on both blocks properly, rather than speculating, in order to determine the correct acceleration of each block.
  • #1
Hamiltonian
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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
friction.jpeg
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
 
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  • #2
What would happen if the blocks were stuck together?
 
  • #3
PeroK said:
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##
 
  • #4
Hamiltonian299792458 said:
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?
 
  • #5
PeroK said:
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
 
  • #6
Hamiltonian299792458 said:
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.
 
  • #7
Hamiltonian299792458 said:
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.
 
  • #8
Hamiltonian299792458 said:
I want to know what is wrong with the steps I have mentioned above.
You can't assume the maximum frictional force.
 
  • #9
PeroK said:
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.
 
  • #10
Hamiltonian299792458 said:
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.
 
  • #11
PeroK said:
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
 
  • #12
Hamiltonian299792458 said:
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.

Hamiltonian299792458 said:
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.
 
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  • #13
PeroK said:
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 ...
 
  • #14
Hamiltonian299792458 said:
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.
 
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  • #15
PeroK said:
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.
 
  • #16
Hamiltonian299792458 said:
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.
 

What is a two block friction problem?

A two block friction problem is a physics problem that involves two blocks in contact with each other, where one block is on top of the other and there is friction between the two blocks. The problem usually requires finding the acceleration, force, or coefficient of friction between the blocks.

What are the key principles involved in solving a two block friction problem?

The key principles involved in solving a two block friction problem are Newton's laws of motion, specifically the second law (F=ma), and the concept of friction. The normal force and the coefficient of friction are also important factors to consider.

How do I set up the equations for a two block friction problem?

To set up the equations for a two block friction problem, you will need to draw a free body diagram for each block, showing all the forces acting on them. Then, use Newton's second law to write equations for the forces in the x and y directions. Finally, use the friction equation (F=μN) to solve for the unknown variables.

What are the common assumptions made in a two block friction problem?

The common assumptions made in a two block friction problem are that the surfaces of the blocks are rough enough to create friction, but not so rough that the blocks do not slide at all. Other assumptions may include neglecting air resistance and assuming that the blocks are in contact with each other at a single point.

What are some strategies for solving a two block friction problem?

Some strategies for solving a two block friction problem include drawing accurate free body diagrams, identifying the forces acting on each block, setting up and solving the equations for each block, and checking the solution for reasonableness. It can also be helpful to break the problem down into smaller parts and solve them separately before combining the solutions.

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