# Question Video: Determing the Relation between the Coefficients of Static Friction in a Two-Blocks System Physics

The figure shows two blocks, block A and block B. The blocks have equal masses. Block A is on a rough surface, and block B is on block A. A force πΉ acts on block B, parallel to the surface. Initially, block B does not move, and πΉ is steadily increased. When πΉ has increased enough to make block B start to move, both blocks move in the direction of πΉ at the same speed. Which of the following is a correct statement about the relation of πβ, the coefficient of static friction between block A and the surface, to πβ, the coefficient of static friction between the blocks? [A] πβ = πβ [B] πβ < πβ [C] πβ > πβ

04:06

### Video Transcript

The figure shows two blocks, block A and block B. The blocks have equal masses. Block A is on a rough surface, and block B is on block A. A force πΉ acts on block B, parallel to the surface. Initially, block B does not move, and πΉ is steadily increased. When πΉ has increased enough to make block B start to move, both blocks move in the direction of πΉ at the same speed. Which of the following is a correct statement about the relation of π sub one, the coefficient of static friction between block A and the surface, to π sub two, the coefficient of static friction between the blocks? (A) π sub one is equal to π sub two. (B) π sub one is less than π sub two. (C) π sub one is greater than π sub two.

Alright, in this example, we have these two blocks: block B rests on block A and then block A is resting on this rough surface. While the two blocks are stationary, an applied force πΉ is exerted on block B, parallel to the interface between the blocks. At first, nothing happens. But then, as πΉ is increased, eventually block B does start to move. At that exact instant though, block A is also set in motion. And both blocks move to the left at the same speed weβre told. So, this is interesting. Even though the force πΉ is being applied to block B, what ends up happening is both blocks move as a unit to the left.

Knowing this, we want to make a comparison between the coefficient of static friction that exists between the two blocks, thatβs called π sub two, and also the coefficient of static friction that exists between block A and the rough surface; that is called π sub one. Basically, we want to make a comparison between these two coefficients of static friction. As we think about whatβs happening physically in this situation, we see that even though weβre applying a force just to block B, apparently, the frictional force between block B and block A is strong enough that the whole entire mass of the two blocks together begins to move before block B starts to slide across the top of block A.

We could say then that this static frictional force here that acts between block A and the surface is weaker than the static frictional force that acts between the two blocks. We can write out equations for these two forces of static friction. We can recall that the magnitude of the static frictional force acting on some object thatβs at rest on a flat surface is equal to that objectβs mass times the acceleration due to gravity times the coefficient of static friction between the object and the surface it rests on. If we call the frictional force in our scenario that exists between block A and the surface πΉ sub one, then we know thatβs equal to the mass of block A plus the mass of block B, because block B rests on top of block A, multiplied by π times π sub one, the coefficient of static friction between block A and the surface. Compared to this, if we call the static frictional force between the two blocks πΉ sub two, then thatβs equal simply to the mass of block B times π times π sub two.

In our problem statement, we were told that as these blocks move, they move as a unit. We saw that this implies that the frictional force between the blocks must be greater than that between block A and the surface. In other words, πΉ sub two must be greater than πΉ sub one. We can then replace πΉ sub two in this inequality with this expression and πΉ sub one with this one. And once weβve done that, letβs recall an important bit of information given to us earlier on. We werenβt told what the masses of blocks B and A are, but we were told that theyβre equal. Off to the side then, we can write that π sub B is equal to π sub A, and we can call them both just π. If we make that substitution in our inequality, then we have this expression where π and π are common to both sides.

If we then divide both sides by π times π, then this cancels out those terms on either side of the equality while keeping its direction the same. And we find that π sub two is greater than two times π sub one. This means that even if we double π sub one, itβs still less than π sub two. Looking at our three answer options, we see that this corresponds to option (B) π sub one is less than π sub two.