# Determine the minimum horizontal force to disturb the equilibrium

Three blocks of 1N 2N and 3N are placed one over the other respectively. If the friction between the topmost 1N and the middle 2N block is twice the friction between the lowest 3N and floor then determine the minimum horizontal force to disturb the equilibrium. (I tried to solve, the photo contains my Solution.) • Few questions / clarifications. 1 Does friction in the question (unambiguously) mean coefficient of friction or friction force or maximum value of static friction ? 2 What is mentioned about the interface between middle block and lower block ? If that was a friction less interface, It would be easy to disturb the equilibrium. 3 Are we free to apply the force anywhere on the system or is it restricted to a particular block ? 4 Definition of horizontal equilibrium (movement of the whole thing or relative movement between the blocks)
– AJN
May 31 at 5:31
• IMO : acceleration doesn't need to appear in the calculations. Use maximumpossible friction force that the interfaces can develop. Maximum friction force (not verified): 1/2 interface $$2\cdot x\cdot 1N$$. 3/floor interface : $$x\cdot 6N$$
– AJN
May 31 at 5:35
• What is the rationale between equating the accelerations ? Why would the accelerations be the same ?
– AJN
May 31 at 5:37
• I have the same doubts too. As the question does not clear them. So, I tried to solve it by taking some assumption of my own. Like, the friction given here is the coefficient of friction, the 2nd box will apply internal force and will not affect the external force as the whole system would act as a body and the since the horizontal force is asked so I assumed that I force can be horizontally only. May 31 at 5:39
• Since the force will apply to the whole system then the acceleration should be the same. May 31 at 5:40

Your answer is correct. But I don't think acceleration needs to be involved for this case, rather it can be solved by statics as depicted in the figure below. From the figure, it is clear that in order to mobilize the middle block, the force must be equal to the larger friction force on the top and bottom faces, $$F \geq 6\mu N$$. 