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.
my answer is 3N but i don't think that's right.
does anyone know what the answer is ..?
five books weigh 25 N
.2 *25 = 5N
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.
Aren't there five books on top, so the normal force is 25N?
Friction=mu(normal force)