Static Friction and Frictional Force Ranking Task of Crate

[andyman21]

Below are six crates at rest on level surfaces. The crates have different masses and the frictional coefficients [the first value is the static friction and the second is the kinetic friction] between the crates and the surfaces differ. The same external force is applied to each crate, but none of the crates move.
Box 1= 600kg (Static friction-0.8)(Kinectic friction-0.5)
Box 2=750kg (Static friction-0.6)(Kinetic Friction-0.5
Box 3=1500kg(Static friction-0.3)(Kinetic friction-0.1)
Box 4=500kg(Static Friction-0,6)(Kinectic friction-0,3)
Box 5=750kg(Static Friction-0.4)(Kinectic Friction-0.3)
Box 6=250kg(Static Friction-0.2)(Kinetic Friction-0.1)

I need to rank the crates on the basis of the frictional force acting on them

The equation I would use here is the static friction equation (Static Friction is less than or equal to the miu times the normal force)

Since the boxes arent moving I would just multiply the mass of the box times the coefficient of the static friction (For example the first box would be 600x0.8 and then rank those numbers i get from greatest to least correct? However I tried this and it doesn't like my answer. Am i doing something incorrect?

 

[PhanthomJay]

The same external force is applied to each crate, but none of the crates move. I need to rank the crates on the basis of the frictional force acting on them.The equation I would use here is the static friction equation (Static Friction is less than or equal to the miu times the normal force)

If the crates don't move, and the external force applied to each is the same, and the static friction is less than or equal to uN, use Newton 1 to calculate the friction force acting on each one. Whch one, if any, has the greater friction force acting on it?

 

[andyman21]

I calculated for example the first box to be 600kg x the static coeffiecient which is 0.8. I have done this for all the boxes and my order from least to greatest was the 600 kg box being the greatest frictional force acting on it, then stacking box 2 and 3 because they had the same frictional force, and then stacking boxes 4 and 5 becuase they also had the same number and finally the least box was box 6 . To get the frictional force on these boxes that do not move i should use the static coefficient number given rather than the kinetic friction. So i did this for all these boxes and it did not like my order of answers I guess i am just still confused

 

[andyman21]

that order is from greatest to least, my apologies

 

[PhanthomJay]

Although you clearly stated in your first post that

Static Friction is less than or equal to the miu times the normal force

, you since have stated that

To get the frictional force on these boxes that do not move i should use the static coefficient number given...

Do you see the difference between these 2 statements? Apply Newton's first Law to each box. I don't understand how you can 'order' the choices, unless I've misunderstood the problem statement or you have written it incorrectly.

 

[SammyS]

The force of static friction is given by the following relation:

$F_\text{static friction }\le\mu_sN\,.$​

Why is there an inequality sign, ≤, rather than an equal sign, = ?

Can the force of static friction be greater than the external force being applied.