There are six books in a stack, and each book weighs 5 N. The coefficient of static friction between the books is 0.2. With what horizontal force must one push to start sliding the top 5 books off the bottom one.

five books weigh 25 N

.2 *25 = 5N

Aren't there five books on top, so the normal force is 25N?

Friction=mu(normal force)

3 N

9utg34

To determine the horizontal force required to start sliding the top 5 books off the bottom one, we need to consider the static friction between the books.

The maximum static frictional force (F_max) that can act between the books is given by:

F_max = μ_s * N

Where μ_s is the coefficient of static friction and N is the normal force between the books. In this case, the normal force (N) is the weight of the top 5 books, which is:

N = 5 N/book * 5 books = 25 N

Given that the coefficient of static friction (μ_s) is 0.2, we can calculate the maximum static frictional force (F_max):

F_max = 0.2 * 25 N = 5 N

Therefore, the maximum force of 5 N is required to start sliding the top 5 books off the bottom one. It's important to note that this is the maximum force required, so a force of 3 N would indeed be insufficient to overcome the static friction between the books.