Mastering Jenga Tower Physics: Calculating Forces and Layers

[maserati1969]

Homework Statement

With regards to a Jenga tower:
I am looking for the solution to Q4 in the image below. (also below the image are my 3a and 3b attempts and attempt at Q4

Relevant Equations

F = Mu.N where Mu is the coefficient of friction and N = normal force acting downwards (=mg)

My attempt at Q3 yielded:
3a) F = Mu.n = Mu x m x g
as a function of layer = Mu x m x g x n x i
where n = number of blocks per layer
and i = layer number.
3b) From - Fn = n x (LWH) x density x gravity x L/2 = F x d
where n = layer number
d = (n x L x H) / 2

My attempt at Q4 is there should be no difference as mu is constant (1) and whether you pull a block out parallel or perpendicular to length should make no difference to the force

 

[haruspex]

3a) F = Mu.n = Mu x m x g
as a function of layer = Mu x m x g x n x i
where n = number of blocks per layer
and i = layer number.

How are you numbering the layers?
How many normal forces are there on a given layer?
Does a block span a whole layer? Surely not. So what are the normal forces on one block?
As a block is slid out, is the normal force on it constant?

 

[maserati1969]

The layers are numbered as i = 1 from the bottom layer, 12 layers in total
There are 3 blocks per layer, laid side-by-side with each layer alternating in orientation (90 degrees).
Each block is 25 x 33 x 100cm and density is 0.4 kg/m^3.
So each block weighs 0.32373 N, and one layer therefore weighs 0.971 N.
How many forces on a given layer? Must be weight of layers above, weight on layer 1 = 0.971 x 11, on layer 2 must be 0.971 x 10 etc.
As a block is moved out the force on it must decrease proprtionally to the surface area remaining in the tower somehow.

 

[haruspex]

How many forces on a given layer? Must be weight of layers above

If that is the only force on it, why doesn't it accelerate downwards?

As a block is moved out the force on it must decrease proprtionally to the surface area remaining in the tower somehow.

Actually it is rather complicated. To answer accurately, you would have to think about how the other blocks deform slightly as they take more of the load. A simplistic view is that the block's share of the load is proportional to its contact area as a fraction of the remaining total area, so, yes, it does reduce. The point of noting that is it means you only have to consider the initial extraction force.

 

[maserati1969]

There is a reaction force on layer below equal to the load above.
Thanks Haruspex

 

[haruspex]

There is a reaction force on layer below equal to the load above.
Thanks Haruspex

So what is the friction on layer 2 of 3 total?

 

[maserati1969]

Well the coefficient of friction is 1.
Not sure what you mean by 'what is the friction' ?
Do you mean what is the force required to overcome friction?

 

[haruspex]

Well the coefficient of friction is 1.
Not sure what you mean by 'what is the friction' ?
Do you mean what is the force required to overcome friction?

How many normal forces are there on a middle layer? So how many friction forces?

 

[maserati1969]

3 normal forces due to 3 blocks, therefore 3 friction forces

 

[haruspex]

3 normal forces due to 3 blocks, therefore 3 friction forces

You’re still not getting it.
If I slide one block out of the middle layer of three layers, it is subject to normal forces above and below, so two frictional forces on the one block.
What do they add up to?

 

[maserati1969]

No I'm not, I turned 54 today and don't need quizzing like a child. Thanks.

 

[jbriggs444]

No I'm not, I turned 54 today and don't need quizzing like a child. Thanks.

It is the Physics Forums way. We do not provide answers. We provide guidance, trying to lead people to answers.

 

[maserati1969]

I have gone as far as I can with it.
Thanks for your help so far.

 

[maserati1969]

Would the magnitude of the normal force on the bottom of the layer as a function of layer number be something like:
Fn = m x g x (i +1) where i = 0 for the bottom layer?

 

[jbriggs444]

Would the magnitude of the normal force on the bottom of the layer as a function of layer number be something like:
Fn = m x g x (i +1) where i = 0 for the bottom layer?

Something like that, yes. But I am not sure how you are counting. Surely the bottom layer (lowest number for $i$) would have the highest normal force and the uppermost layer (highest number for $i$) would have the lowest.

Let $n$ be the number of layers. Then the normal force on the bottom of the bottom block would be $mgn$, right?

Can you then add $i$ to the formula?

 

[maserati1969]

I guess I could use m x g x (12 - i) ?

 

[maserati1969]

We can't use n & i as we would be multiplying by layer number twice?

 

[jbriggs444]

We can't use n & i as we would be multiplying by layer number twice?

As I was using the variables, $n$ is the total number of layers. That is 12, apparently. Meanwhile, $i$ is the layer number we are considering.

There is no problem using both $n$ and $i$ in the same formula.

 

[haruspex]

The first step with such a problem should be a free body diagram for the block being removed. I was remiss in not asking for one first up.
If it's not easy for you to post such, can you list all the forces on it?

 

[maserati1969]

I've figured it out. Thanks all.

 

[haruspex]

I've figured it out. Thanks all.

Are you able to verify your answer? If not, please post it, just in case.