Recent content by maserati1969

I've figured it out. Thanks all.

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

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

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?

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

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

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

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?

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

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...

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...

Apologies, I'll post this as new

Hi again BvU, thanks again for this information. Can you help with Question 4 below:

Yes, I see, they are more or less the same question. Thanks for the note though.

Great, that makes perfect sense. Thanks BvU