Solving Two Boxes & Friction Motion Problem

[JoeTrumpet]

I understand how to get the question right, but my friend posed an alternate method that is wrong and I was wondering why his method doesn't work.

Homework Statement

The coefficient of static friction is 0.75 between two blocks stacked on one another. The coefficient of kinetic friction between the lower block and the floor is 0.20. Force F vector acting on the upper block to the right causes both blocks to cross a distance of 4.0 m, starting from rest. What is the least amount of time in which this motion can be completed without the top block sliding on the lower block?

The upper box is 4kg and the lower box is 3kg.

Homework Equations

Upper box: Fxtotal = F - fs = m1*a (my guess is I made a mistake here since it wasn't pertinent to my solution)
Lower box: Fxtotal = fs - fk = m2*a

where a1 = a2 = a because of motion together.

Upper box: Fytotal = N = m1*g
Lower box: Fytotal = N = m1*g + m2*g

The Attempt at a Solution

I simply solved the lower equation for a and plugged it into the proper kinematic equation. However, my friend wrote from intuition F - fk = (m1 + m2)*a, which happens to be a combination of my two Fxtotal equations, and said that to maximize the distance F would have to equal fs, thus getting fs - fk = (m1 + m2)*a, which clearly would yield a response different from mine. I was wondering if anyone could clear up for us why it's wrong. I figure it has to either do with his assumption that F = fs or with my first equation and his overall equation (or both).

Thanks!

 

[LowlyPion]

I understand how to get the question right, but my friend posed an alternate method that is wrong and I was wondering why his method doesn't work.

Homework Statement

The coefficient of static friction is 0.75 between two blocks stacked on one another. The coefficient of kinetic friction between the lower block and the floor is 0.20. Force F vector acting on the upper block to the right causes both blocks to cross a distance of 4.0 m, starting from rest. What is the least amount of time in which this motion can be completed without the top block sliding on the lower block?

The upper box is 4kg and the lower box is 3kg.

Homework Equations

Upper box: Fxtotal = F - fs = m1*a (my guess is I made a mistake here since it wasn't pertinent to my solution)
Lower box: Fxtotal = fs - fk = m2*a

where a1 = a2 = a because of motion together.

Upper box: Fytotal = N = m1*g
Lower box: Fytotal = N = m1*g + m2*g

The Attempt at a Solution

I simply solved the lower equation for a and plugged it into the proper kinematic equation. However, my friend wrote from intuition F - fk = (m1 + m2)*a, which happens to be a combination of my two Fxtotal equations, and said that to maximize the distance F would have to equal fs, thus getting fs - fk = (m1 + m2)*a, which clearly would yield a response different from mine. I was wondering if anyone could clear up for us why it's wrong. I figure it has to either do with his assumption that F = fs or with my first equation and his overall equation (or both).

Thanks!

Your equation on the lower box overlooks that whatever acceleration there is must include the mass of both boxes.

The clearer way to view it for me is to look at what the maximum force may be applied. And that is simply .75*m1*g

From that you subtract the frictional force which is the combined weight times .2.

So the net force available then is .75*m1*g - .2*g*(m1 + m2)
That in turn equals the force available to accelerate the two boxes as a system, namely

.75*m1*g - .2*g*(m1 + m2) = a * (m1 + m2)

Then figure your time the usual way with kinematics.

 

[baba89]

Hello, it's great to see that you and your friend are both thinking about different ways to solve this problem. Let's take a closer look at your friend's method and see why it may not work.

First, let's review the equations that we are using for this problem. The equations for the upper and lower boxes are correct, but I would like to point out that the equations for the forces in the x-direction should be Fxtotal = F - fs = m1*a1 and Fxtotal = fs - fk = m2*a2, where a1 and a2 represent the respective accelerations of the upper and lower boxes. This is because each box experiences a different acceleration due to the force F, since they have different masses.

Now, let's look at your friend's method. They have assumed that the force F is equal to the maximum static friction force, fs. This is not necessarily true. In this problem, we are looking for the least amount of time in which the motion can be completed without the top block sliding on the lower block. This means that the force F must be less than or equal to the maximum static friction force, but it does not have to be equal to it. In fact, the force F could be much smaller than fs and the motion could still be completed without sliding.

Additionally, your friend's overall equation, F - fk = (m1 + m2)*a, is incorrect. This equation assumes that the force F is the only force acting on both boxes, which is not the case. The weight of the upper box, m1*g, is also acting on the lower box, so the equation should be F - fk - m1*g = (m1 + m2)*a. This equation is essentially the same as your solution, just written in a different form.

In conclusion, your friend's method is not entirely correct because they have made assumptions about the force F and their overall equation is missing a term. It's great to explore different methods of solving a problem, but it's important to carefully consider the assumptions and equations used in each method. Keep up the good work!