Accelerating Incline - Conditions to prevent slipping

[Pi-Bond]

Homework Statement

(Kleppner & Kolenkow - Introduction to Mechanics - 2.17)
The first attached image shows a box on an incline which is accelerated at a rate a meters per second squared. $\mu$ is the coefficient of friction between the box and incline surface. The questions are in the image.

Homework Equations

Newton's Laws

The Attempt at a Solution

I can understand that there will be a range for the accelerations which will leave the box static on the incline. I'm not sure how the math is to be applied to get the maximum acceleration. I got the answer for minimum by resolving forces along the surface of the incline, and equating them. The answer I got was:

$a=g \frac{sin(\theta)-\mu cos(\theta)}{cos(\theta)+ \mu sin(\theta)}$

This gives the correct answer for the case given in the clues. I'm not sure how to proceed to find the maximum acceleration. Any hints?

 

[Doc Al]

Hint: Consider the direction of the friction force.

 

[Pi-Bond]

Of course! In the maximum case, the block is on the verge of drifting up the incline, so the friction force must be directing down the incline! So the maximum acceleration is given by-

$a=g \frac{sin(\theta)+\mu cos(\theta)}{cos(\theta)-\mu sin(\theta)}$

Thanks a lot!

 

[aLearner]

I got the same answer, but solved for μ. So

μ = $\frac{gsinθ - acosθ }{asinθ + gcosθ }$

but the correct answer is μ = tanθ
help?

 

[Nickie]

To find the maximum acceleration without slipping, you can use the concept of limiting friction. This means that the frictional force between the box and the incline will be equal to the maximum possible value of the frictional force, which is given by the equation F = \mu N, where \mu is the coefficient of friction and N is the normal force.

To determine the normal force, you can use the fact that the box is at rest, so the sum of forces in the y-direction must be equal to zero. This means that N = mg cos(\theta), where m is the mass of the box and g is the acceleration due to gravity.

Substituting this into the equation for the maximum frictional force, we get F_{max} = \mu mg cos(\theta).

Now, we can use Newton's second law to relate the maximum frictional force to the maximum acceleration: F_{max} = ma_{max}.

Substituting F_{max} into this equation, we get: \mu mg cos(\theta) = ma_{max}.

Solving for a_{max}, we get the final equation: a_{max} = \mu g cos(\theta).

This is the maximum acceleration that the incline can have without the box slipping. Any acceleration greater than this value will result in the box slipping down the incline.