A stack of books (conflicting answers)

[ashum]

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

For the five books:
m * a_max = u_s * m * g
a_max = u_s * g
a_max = 0.2 * 10 m/s^2 = 2 m/s^2

For the system:
F_max = M * a_max
F_max = 3 kg * 2 m/s^2 = 6N

Was there a mistake in the approach I took? Apparently, the correct answer is 5N, since you only need to find F_s between the five books and the sixth book which is u_s * m * g. However, this would mean each book gets an applied force of 5/6 N. With a mass of 0.5 kg each, each book would only accelerate at 1.67 m/s^2, which is not equal to a_max (2 m/s^2). Any thoughts?

 

[Kazza_765]

Ahh, the equation F=ma should be written as F_net=ma. So in this case, the force you are applying is not F_net, F_net also needs to take into account the friction that is resisting the motion.

You should assume that there will be no acceleration, and you want to find the minimum force that will just barely start the books to move. For this to occur, the force applied has to be equal to the resisting force.

F = u_s * m * g

Note here that since we are letting the applied force equal the resisting force, F_net = 0, hence there is no acceleration.