Case when there is no normal force in spite of contact

[gracy]

Homework Statement

"There is a pulley with 2 blocks on the table.Force "F" is acting on it ,as it can be seen here

A=10 kg and B=20 kg
F=294 N
Now,my exact problem is my teacher says "there is no contact force between A and the table.."

Why block A is not pushing against table while block B is?

Homework Equations

:[/B]

F=2T
T=147 N

The Attempt at a Solution

As it is not a numerical problem instead conceptual doubt,I can not attempt anything.But I will try some in order to obey the guidelines of filling the template.Condition for normal force is there should be deformation.The objects should press against each other.I don't know why there is no deformation.Evertytime when something is on the table
there is normal force equal to weight.
Why not in this case?Why block A is not pushing against table while block B is?

 

[berkeman]

Homework Statement

"There is a pulley with 2 blocks on the table.Force "F" is acting on it ,as it can be seen here
View attachment 81996
A=10 kg and B=20 kg
F=294 N
Now,my exact problem is my teacher says "there is no contact force between A and the table.."

Why block A is not pushing against table while block B is?

Homework Equations

:[/B]

F=2T

The Attempt at a Solution

As it is not a numerical problem instead conceptual doubt,I can not attempt anything.But I will try some in order to obey the guidelines of filling the template.Condition for normal force is there should be deformation.The objects should pess each other.I don't know why there is no deformation.Evert time when something is on the table
there is normal force equal to weight.
Why not in this case?Why block is not pushing against table while block B is?

What is the normal force down on the table for the larger block?

If you moved the table 1cm down and released the blocks, what would happen?

 

[gracy]

What is the normal force down on the table for the larger block?

200N (Taking g=10 m/s^2)

 

[ehild]

Touch the table with your hand and move your hand on the surface. You feel the surface of the table, but your hand does not push it. there is contact, but the weight of your hand is opposed by your muscles, the normal force between the table and your hand is zero.

The situation is the same here. What should be the tension so as it keeps block A in rest just touching the table?
What should be the normal force acting on block B?

 

[SammyS]

Homework Statement

"There is a pulley with 2 blocks on the table.Force "F" is acting on it ,as it can be seen here
View attachment 81996
A=10 kg and B=20 kg
F=294 N
Now,my exact problem is my teacher says "there is no contact force between A and the table.."

Why block A is not pushing against table while block B is?

Homework Equations

:[/B]

F=2T
T=147 N

The Attempt at a Solution

As it is not a numerical problem instead conceptual doubt,I can not attempt anything.But I will try some in order to obey the guidelines of filling the template.Condition for normal force is there should be deformation.The objects should press against each other.I don't know why there is no deformation.Evertytime when something is on the table
there is normal force equal to weight.
Why not in this case?Why block A is not pushing against table while block B is?

It seems to me that there is a problem here with the problem as it's stated.

In order that block A not accelerate upward, the most that T can be is 10g = 98 N ,(assuming g = 9.8 m/s², which seems to be what's assumed here.)

Therefore, the upper limit for F is 196 N.

In that case, (F = 196 N) we have T =98 N so the net force on block A is:

T - m_Ag = 98 - 10⋅9.8 = 0​

So, block A is resting on the table. Does the table exert any force on the block?

 

[Chestermiller]

Hi Gracy,

Let's consider a free body diagram on mass A, and the forces acting on mass A. If mass A is not glued to the table, what is the least that the normal force can be (and still remain in contact with the table)? With this value of the normal force, from a force balance on mass A, will it be accelerating upwards or not? (Note that, if it is accelerating upwards, it is not in contact with the table).

Chet

 

[ehild]

@gracy, what is the exact wording of the problem? Are the blocks in rest? Is the force F given as F=294 N? Do you need to find the forces between the blocks and the ground? Is the pulley massless?Is the rope taut?
If the blocks are in rest, you can find the tension in the rope and the normal forces by considering the free body diagrams at each block.

 

[gracy]

Is the pulley massless?

Yes.

 

[gracy]

Is the force F given as F=294 N?

Yes.

 

[ehild]

And what is the question?

 

[gracy]

There is a massless pulley on which force "F" is applied.The pulley is connected to block A and block B.Both of which were resting on a ground when force "F" was being applied.A=10 kg and B=20 kg
Question was find acceleration of A and B.
But I have a solution for that.My teacher has given me.But in that solution it is said that "there is no contact force between A and the table"
That's my exact problem...

 

[haruspex]

Condition for normal force is there should be deformation.

Quite so. So consider starting with each block resting happily on the table, spring slack. Each weight deforms the table slightly. Now gradually raise the pulley. The tension increases in the string as the deformations decrease. At some point, the deformation from the lighter block becomes zero. What is the normal force there now?

 

[gracy]

At some point, the deformation from the lighter block becomes zero. What is the normal force there now?

Zero .(Normal force on lighter block.)

 

[haruspex]

Zero .(Normal force on lighter block.)

Right. Does that answer your question?

 

[gracy]

Does that answer your question?

Yes.But one thing I would like to ask is

Now gradually raise the pulley

Where is it mentioned in my problem as stated in #11 post.

 

[haruspex]

Yes.But one thing I would like to ask is

Where is it mentioned in my problem as stated in #11 post.

It doesn't need to be. You are given that there is no normal force. How that came to be is irrelevant. You asked how it is possible, and now you know how it is possible.

 

[ehild]

Yes.But one thing I would like to ask is

Where is it mentioned in my problem as stated in #11 post.

There is a massless pulley on which force "F" is applied.The pulley is connected to block A and block B.Both of which were resting on a ground when force "F" was being applied.
Initially the blocks were in rest on the table. Then the force was applied. That force accelerated the pulley. The raising of the pulley made the rope taut. Block A lost contact with the ground.
The net force acting on the whole system became F-W_A-W_B+N_B=N_B. This is still an upward net force, so the CM accelerates upward. At the end, B also loses contact with the ground. With no interaction between the blocks and the ground, the net force is just zero. The CM does not accelerate any more, but the pulley does, and so are the blocks.

The blocks accelerate with respect to the accelerating pulley, and the relative accelerations have the same magnitude.

 

[gracy]

NB.

What does it denote?

 

[ehild]

What does it denote?

N_B is the normal force on B from the ground.

 

[Chestermiller]

Gracy,
Are you saying that the force F is unspecified, or are you saying that it is 294 N?

Chet

 

[gracy]

or are you saying that it is 294 N?

It is 294 N.

 

[gracy]

There is a massless pulley on which force 294 N is applied.The pulley is connected to block A and block B.Both of which were resting on a ground when force "F" was being applied.A=10 kg and B=20 kg
Question was find acceleration of A and B.
But I have a solution for that.My teacher has given me.But in that solution it is said that "there is no contact force between A and the table"
That's my exact problem.

 

[Chestermiller]

There is a massless pulley on which force 294 N is applied.The pulley is connected to block A and block B.Both of which were resting on a ground when force "F" was being applied.A=10 kg and B=20 kg
Question was find acceleration of A and B.
But I have a solution for that.My teacher has given me.But in that solution it is said that "there is no contact force between A and the table"
That's my exact problem.

OK. I'm going to help you resolve this dilemma on your own.

Can we agree that the maximum contact force that the table can exert on mass A in this system is 9.8 N (it's weight), and the minimum that it can exert on mass A is zero (assuming that mass A is not glued to the table)?

Chet

 

[gracy]

Can we agree that the maximum contact force that the table can exert on mass A in this system is 9.8 N (it's weight), and the minimum that it can exert on mass A is zero (assuming that mass A is not glued to the table)?

Yes.

 

[Chestermiller]

Yes.

Excellent. OK, for each of these two extremes, (using a free body diagram on mass A) write down the force balance equation on mass A (including the ma term) that would apply.

Chet

 

[gracy]

Are we including tension in a string?For the first case i.e normal force=weight of A

Initially the blocks were in rest on the table. Then the force was applied. That force accelerated the pulley. The raising of the pulley made the rope taut. Block A lost contact with the ground.

Keeping this in mind ,I don't think so.

 

[gracy]

In case of maximum contact force
Normal contact force=weight
This case occurs when force 294 N is not applied,therefore tension force is absent.Hence string will be slack.
.acceleration of block A=0
And for minimum contact force on A
i.e zero.This condition is achieved when force is applied String is taut ,hence tension force is present.
T-w(weight of A)=Mass of A multiplied by acceleration of A.As block A accelerates,it loses contact from ground so normal contact force becomes zero.So,not appears in equation.
Right?

 

[Chestermiller]

In case of maximum contact force
Normal contact force=weight
This case occurs when force 294 N is not applied,therefore tension force is absent.Hence string will be slack.
.acceleration of block A=0
And for minimum contact force on A
i.e zero.This condition is achieved when force is applied String is taut ,hence tension force is present.
T-w(weight of A)=Mass of A multiplied by acceleration of A.As block A accelerates,it loses contact from ground so normal contact force becomes zero.So,not appears in equation.
Right?

Well, this is not exactly what I had in mind, but your analysis for the case of zero contact force is exactly correct. And your conclusion regarding the contact force is correct. As soon as the tension force T = 147 N gets applied, the contact force changes discontinuously from 98 N immediately before application of the tension to 0 immediately after the application of the tension. This is because its acceleration is positive (5 m/s^2), its displacement above the table is ##\frac{1}{2}at^2##, and for all t > 0, this is positive.

Chet

 

[gracy]

Thanks,your answer are always beneficial for me.