I push a stack of six books (each weighting 4 kg) horizontally across a table to the right with a kinetic coefficient of friction uk = 0.5. I apply a force of 120 N.

a) What is the acceleration of the books?
b) what is the force of the rightmost book on its neighbor?

Next, I stack the books vertically and push them horizontally across a table to the right by applying a force only to the bottom book.. The kinetic coefficient of friction of the books with the table is uk=0.5 and the static coefficient of friction between the books is us=0.3.

c) What is the maximum force I can apply to the bottom book to move the books as a single unit, before books starts to fall off the stack?

I solved for a) and b) already. For a) I got 5 m/s^2 and for b) I got 40N.

But I am stuck on part c.

I don't really understand how to start because I know that the normal force would be constantly changing, but I don't know how does the kinetic friction and static friction plays in for finding the maximum force that I can apply to the books.

Can someone please explain to me how to finish the problem? Thank you.

How fast can the stack accelerate without the books above the first one slipping?

weight down on bottom book = 5 * 4 kg * g
maximum static friction force between top 5 and bottom book = 5 * 4 * g * 0.3

so max force = mass of top 5 * a

5*4*g*.3 = 5*4 * a
so
a = .3 g
that gives us the maximum acceleration before we get a slip
now what force gave that acceleration?
F = m a
F = 6 * 4 * a
F = 24 * .3 g
if g = 9.81 m/s^2
F = 70.6 Newtons

shouldn't you take into consideration the kinetic friction, the less force you need according to this method you would need more force if you have 100 books.

To solve part c) of the problem, we need to consider the maximum force that can be applied to the bottom book before the books start to fall off the stack.

First, let's analyze the forces acting on the bottom book. When you push the bottom book to the right with a force, there are a few forces we need to consider:

1. The friction force between the bottom book and the table: This force opposes the motion and is given by the kinetic coefficient of friction (uk) multiplied by the normal force acting on the bottom book.

2. The force of static friction between the bottom book and the book above it: If this force is not exceeded, the books will remain stationary with respect to each other.

3. The force of gravity: This force acts vertically downwards on the bottom book. It depends on the weight of the book, which is the mass multiplied by the acceleration due to gravity (9.8 m/s^2).

Now, let's break down the steps to find the maximum force you can apply to the bottom book:

Step 1: Find the normal force acting on the bottom book.
The normal force is equal to the weight of the bottom book (mg), where m is the mass of the book and g is the acceleration due to gravity. In this case, since each book weighs 4 kg, the normal force is (4 kg) × (9.8 m/s^2) = 39.2 N.

Step 2: Calculate the force of friction between the bottom book and the table.
The friction force is given by the kinetic coefficient of friction (uk) multiplied by the normal force. In this case, uk = 0.5, so the force of friction is (0.5) × (39.2 N) = 19.6 N.

Step 3: Determine the maximum force of static friction between the bottom book and the book above it.
The maximum force of static friction can be found by multiplying the static coefficient of friction (us) by the normal force. Here, us = 0.3, so the maximum force of static friction is (0.3) × (39.2 N) = 11.76 N.

Step 4: Find the net force required to overcome the friction forces.
To move the books as a single unit, the net force applied to the bottom book must overcome both the force of friction between the bottom book and the table and the force of static friction between the bottom book and the book above it.

The net force required is the sum of the forces opposing the motion, which is the force of friction plus the force of static friction:
Net force = force of friction + force of static friction
= 19.6 N + 11.76 N
= 31.36 N

Therefore, the maximum force you can apply to the bottom book before the books start to fall off the stack is 31.36 N.