# Maximum Acceleration of sliding crates

• Darkapostle8
In summary, the maximum acceleration that can be achieved before the upper crates start to slide is 6.25 m/s^2.
Darkapostle8

## 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.

Darkapostle8 said:

## 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 said:
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".

tnich said:
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.
tnich said:
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.

Darkapostle8 said:
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?

haruspex said:
What are all the forces acting on the middle crate?
It's not given that's all the question provides.

Darkapostle8 said:
It's not given that's all the question provides.
I'm not asking for the magnitudes, just descriptions of them.

## 1. What is maximum acceleration of sliding crates?

The maximum acceleration of sliding crates refers to the maximum rate at which the velocity of a crate can change as it slides across a surface. It is typically measured in meters per second squared (m/s²).

## 2. How is maximum acceleration of sliding crates calculated?

The maximum acceleration of sliding crates can be calculated using the equation a = F/m, where a is acceleration, F is the force applied to the crate, and m is the mass of the crate. The force applied can be determined by factors such as the friction between the crate and the surface, and the angle of the surface.

## 3. What factors affect the maximum acceleration of sliding crates?

The maximum acceleration of sliding crates can be affected by several factors, including the mass of the crate, the force applied to it, the angle and surface of the slide, and the presence of any external forces such as air resistance.

## 4. How does the maximum acceleration of sliding crates relate to its speed?

The maximum acceleration of sliding crates is directly related to its speed, as it determines how quickly the crate can change its velocity. A higher maximum acceleration will result in a faster change in velocity, and therefore a higher speed.

## 5. Can the maximum acceleration of sliding crates be greater than the acceleration due to gravity?

Yes, the maximum acceleration of sliding crates can be greater than the acceleration due to gravity. This is because the maximum acceleration takes into account the force applied to the crate, while the acceleration due to gravity is a constant force acting on the crate regardless of its mass or the surface it is sliding on.

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