Need help finding maximum force

Question

 A 3 kg box is sitting atop a 4 kg box on level ground. You begin pushing horizontally on the bottom 4 kg box to start the stack moving. The coefficients of friction between also surfaces are μs= 0.25 and μk= 0.12.

What is the maximum force you can exert without the top box starting to slide off?

I found that the minimum force to exert for the stack to move is 17.2N

Answer 1

Angel, here's the physics of the problem

1. Before the blocks start moving, the force applied to the lower block, would be acting against the static frictional force (equal to the total weight of the two blocks times the static coefficient of friction). This will determine the minimal needed for blocks to start moving

2. As the blocks just start moving (so far w/o the 2-nd block sliding) , the force applied to the lower block would now be working against the kinetic friction due to the total weight of the two blocks.

3. The acceleration of the blocks now will be determined through the second Newtonian Law as the net force ( the applied force to the lower block minus the force of kinetic friction) divided by the total mass of two blocks.

4. Now let's look at the upper block. Before it would start sliding, It is pushed forward by the static frictional force created by the weight of the upper block only. That would determine the maximum acceleration that force could sustain before the upper block would start sliding.

5. Finally, match the acceleration found in P3 with the acceleration from P.4. That should solve for the max. force that can be applied to the lower block for the upper one to get on the verge of sliding.

Hope that would help you with the strategy of solving the problem, the rest is to carefully use the data given.

Good luck,

Best,

Dr.Ariel B.

Answer 2

Let F_p = the force pushing the bottom box,

m₁ = 4 kg = the mass of the bottom box, and

m₂ = 3 kg = the mass of the top box.

Net force on the stack

F_net = F_p – μ_k(m₁ + m₂)(g) = F_p – 0.12(7)(9.8) = F_p – 8.23 N

Acceleration

a = F_net /(m₁ + m₂) = (1/7)F_net

The top box will start to slide when m₂a = (3/7)F_net exceeds the static frictional force given by

μ_s(m₂)(g) = 0.25(3)(9.8) = 7.35 N

This means

(F_net)_max = (7/3)7.35 = 17.15 =  (F_p)_max – 8.232 which means that

(F_p)_max = 17.15 + 8.23 = 25.38 N

Answer

The maximum force you can exert without the top box starting to slide off is

25.38 N