# Maximum Acceleration of sliding crates

## Homework Statement

Three identical 4 kg crates are stacked one on top of the other. You want to accelerate them as much as possible by pushing on the lower crate, but the forces of static friction between the surfaces of the crates max out at 25 N. What is the maximum acceleration you can achieve before the upper crates start to slide?

f_max/m

## The Attempt at a Solution

So I attemped this solution by taking the max force of 25 N and diving it by the mass of the creates (4kg) and got 6.25 m/s^2 but that is not being accepted by my online homework. I also did 4*3 for 12 and did 25/12 and it still didn't work and I'm not sure what I'm doing wrong.

tnich
Homework Helper

## Homework Statement

Three identical 4 kg crates are stacked one on top of the other. You want to accelerate them as much as possible by pushing on the lower crate, but the forces of static friction between the surfaces of the crates max out at 25 N. What is the maximum acceleration you can achieve before the upper crates start to slide?

f_max/m

## The Attempt at a Solution

So I attemped this solution by taking the max force of 25 N and diving it by the mass of the creates (4kg) and got 6.25 m/s^2 but that is not being accepted by my online homework. I also did 4*3 for 12 and did 25/12 and it still didn't work and I'm not sure what I'm doing wrong.
Between which two crates will the force of static friction be the largest? Between which will it be the least?

tnich
Homework Helper
Between which two crates will the force of static friction be the largest? Between which will it be the least?
Something bothers me about this problem. If the maximum force of static friction is between the bottom two boxes, then the coefficient of static friction is ##c_f=25N/[(12kg)(g)]##. So the maximum acceleration that can be applied to the bottom box before the next box starts slipping is ##a=F/m=(12kg)(g)c_f/12kg=g c_f##. Assuming that ##c_f## is the same between the top two boxes, the maximum acceleration that can be applied to the second box from the top before the top box starts slipping is ##a=(4kg)(g)c_f/4kg=g c_f##, the same acceleration as for the bottom box. So why is the question asking when the top boxes start to slide?
Maybe we are misinterpreting the (somewhat vague) statement "the forces of static friction between the surfaces of the crates max out at 25 N".

haruspex
Homework Helper
Gold Member
2020 Award
If the maximum force of static friction is between the bottom two boxes, then the coefficient of static friction is ##c_f=\frac{25N}{(12kg)(g)}
No, there are only two boxes on top of the one being pushed.
Assuming that ##c_f## is the same between the top two boxes
I would not assume that. Rather, accept the statement that each frictional force is up to 25N. But it doesn't matter to the answer.

haruspex
Homework Helper
Gold Member
2020 Award
So I attemped this solution by taking the max force of 25 N and diving it by the mass of the creates (4kg)
What are all the forces acting on the middle crate?

What are all the forces acting on the middle crate?
It's not given thats all the question provides.

haruspex