Six crates(a, b,c, d,e, f) at rest on level surfaces. the crates have differentmasses and the frictional coefficients [given as (\mu_{\rm s},\mu_{\rm k}) ] between the crates and the surfaces differ. thesame external force is applied to each crate, but none of thecrates move. rank the crates on the basis of thefrictional force acting on them. rank from largest to smallest. a. [500kg μ=(0.6,0.3)] b. [750kg μ=(0.4,0.3) ] c. [250kg μ=(0.2,0.1)] d.[600kg μ=(0.8,0.5)] e.[750kg μ=(0.6,0.5)] f. [1500kg μ=(0.3,0.1) ]

This question involves the concepts of frictional force and coefficient of static friction.

The ranking of the frictional force acting on the crates will be.

The frictional force can be calculated by the following simple formula:

where,

f = frictional force = ?

= coefficient of static friction (since boxes are not moving)

R = Normal Reaction = Weight of Box = mg

Now, we use this formula to calculate the frictional force acting on each box.

BOX A:

BOX B:

BOX C:

BOX D:

BOX E:

BOX F:

Therefore, the ranking will be:

Learn more about friction force here:

link

The attached picture shows the friction force.

all will have same frictional force on them

Explanation:

As we know that the crate is placed at rest on the rough floor

so here the forces must be balanced

So here the applied force must be equal to the friction force on the crate

So we will say that the friction force on the crate is static friction and it must be equal to the applied force on the crate

So all the crates have same external force which means that all the crates will have same frictional force on them

so we have

all will have same frictional force on them

hi i’m sorry to call us today so we will get back with y’all and we will have to do it haha omg was the day