Friction forces and sliding slabs

I have 2 questions, one generalizing the other.

Question 1: Suppose we have 2 slabs resting horizontally on a table. Assume there is friction between the 2 slabs as well as between the bottom slab and the table and that all friction coefficients are different. Now we apply a horizontal force to the top slab. How do we figure out the direction of movement of the bottom slab?

Question 2: We now have a stack of n slabs resting horizontally on the table. All surfaces in contact have friction, and all friction coefficients are different. If we apply a horizontal force to the top slab, how can we predict the direction of movement of each slab in the stack?

For the first question, I am guessing that depending on the amount of force and the values of the friction coefficients, there can be multiple scenarios: the slabs won't move till the force overcomes the first static friction coefficient, then they might move together, then one might move in one direction, and the other... well, how does the bottom slab move exactly? That's where I am confused. It seems I have to know the direction of movement to set the sign of the friction force between the bottom slab and the table correctly, but I am not sure how to establish the force transmitted by the top slab to the bottom one. Is it just the friction force between top and bottom slab?

For the second question, I'd gladly apply the same method as for the first question repetitively, but lo and behold, I haven't solved the first question yet...

This is not for a "homework" and I am not a student trying to get his/her homework answered :-) Thanks!

• First remove friction. Which way things are going to move? Apply friction to oppose this motion. That is how you figure out the direction of friction. – ja72 Oct 1 '13 at 1:23

Lets say we have $k$ blocks, numbered $i=1 \ldots k$ with 1 on the bottom and $k$ on the top. The top block has an applied force $\mathcal{P}$ and each block has mass $m_i$ and friction coefficient with the previous block (or the ground) $\mu_i$. Also the movement of each block is characterized by the acceleration $\ddot{x}_i$. In matrix form the above define

$$P=\begin{pmatrix}0\\ 0\\ \vdots\\ 0\\ \mathcal{P} \end{pmatrix}$$ $$m=\begin{bmatrix}m_{1}\\ & m_{2}\\ & & \ddots\\ & & & m_{k-1}\\ & & & & m_{k} \end{bmatrix}$$ $$\mu=\begin{bmatrix}\mu_{1}\\ & \mu_{2}\\ & & \ddots\\ & & & \mu_{k-1}\\ & & & & \mu_{k} \end{bmatrix}$$ $$\ddot{x}=\begin{pmatrix}\ddot{x}_{1}\\ \ddot{x}_{2}\\ \vdots\\ \ddot{x}_{k-1}\\ \ddot{x}_{k} \end{pmatrix}$$

The weight on each block is $m_i g$ and the contact force with the previous block (or the ground) is $N_i$. Also the friction limit is $F_i \leq \mu_i N_i$. In matrix form the above is

$$N=\begin{pmatrix}N_{1}\\ N_{2}\\ \vdots\\ N_{k-1}\\ N_{k} \end{pmatrix}$$

$$F \leq \begin{bmatrix}\mu_{1}\\ & \mu_{2}\\ & & \ddots\\ & & & \mu_{k-1}\\ & & & & \mu_{k} \end{bmatrix}\begin{pmatrix}N_{1}\\ N_{2}\\ \vdots\\ N_{k-1}\\ N_{k} \end{pmatrix}=\begin{pmatrix}\mu_{1}N_{1}\\ \mu_{2}N_{2}\\ \vdots\\ \mu_{k-1}N_{k-1}\\ \mu_{k}N_{k} \end{pmatrix}$$

Why do we need all this? To to make the equation of motion for the $i$-th block, which is $P_i - F_i + F_{i+1} = m_i \ddot{x}_i$

Look at the free body diagram above. By convention the i-th friction opposes the motion which is to the right. The friction from the above block is reacted upon this block and applied to the left. That is why the sum of the fores is $P_i + F_{i+1} - F_i$.

The balance in matrix form, using an adjacency matrix is

$$A=\begin{bmatrix}1 & -1\\ & 1 & -1\\ & & \ddots & \ddots\\ & & & 1 & -1\\ & & & & 1 \end{bmatrix}$$ $$P-A\, F=m\ddot{x}$$

which expands out to

$$\begin{pmatrix}0\\ 0\\ \vdots\\ 0\\ \mathcal{P} \end{pmatrix}+\begin{pmatrix}F_{2}-F_{1}\\ F_{3}-F_{2}\\ \vdots\\ F_{k}-F_{k-1}\\ -F_{k} \end{pmatrix}=\begin{pmatrix}m_{1}\ddot{x}_{1}\\ m_{2}\ddot{x}_{2}\\ \vdots\\ m_{k-1}\ddot{x}_{k-1}\\ m_{k}\ddot{x}_{k} \end{pmatrix}$$

Now the contact normal force is derived from the blocks above it with

$$A\,N = m\,g$$ $$N = A^{-1} m\,g$$ $$\begin{pmatrix}N_{1}\\ N_{2}\\ \vdots\\ N_{k-1}\\ N_{k} \end{pmatrix}=\begin{bmatrix}1 & 1 & 1 & 1 & 1\\ & 1 & 1 & 1 & 1\\ & & \ddots & \vdots & \vdots\\ & & & 1 & 1\\ & & & & 1 \end{bmatrix}\begin{pmatrix}m_{1}g\\ m_{2}g\\ \vdots\\ m_{k-1}g\\ m_{k}g \end{pmatrix}$$

So all together $$P - \left( A\,\mu A^{-1}\right) m\, g=m\ddot{x}$$

or with $\mu_{SYS}=A\,\mu A^{-1}$

$$\mu_{SYS}=\begin{bmatrix}1 & -1\\ & 1 & -1\\ & & \ddots & \ddots\\ & & & 1 & -1\\ & & & & 1 \end{bmatrix}\begin{bmatrix}\mu_{1}\\ & \mu_{2}\\ & & \ddots\\ & & & \mu_{k-1}\\ & & & & \mu_{k} \end{bmatrix}\begin{bmatrix}1 & 1 & 1 & 1 & 1\\ & 1 & 1 & 1 & 1\\ & & \ddots & \vdots & \vdots\\ & & & 1 & 1\\ & & & & 1 \end{bmatrix} \\ \mu_{SYS}=\begin{bmatrix}\mu_{1} & \mu_{1}-\mu_{2} & \cdots & \mu_{1}-\mu_{2} & \mu_{1}-\mu_{2}\\ & \mu_{2} & \cdots & \mu_{2}-\mu_{3} & \mu_{2}-\mu_{3}\\ & & \ddots & \vdots & \vdots\\ & & & \mu_{k-1} & \mu_{k-1}-\mu_{k}\\ & & & & \mu_{k} \end{bmatrix}$$

$$P -\mu_{SYS} m\, g=m\ddot{x}$$ $$\ddot{x} = m^{-1} \left(P-\mu_{SYS} m\, g \right)$$

So this is the motion once with have slipping. We need to reverse the equations and find the traction required when $\ddot{x}=0$ which ends up being

$$\mu_i \geq \frac{\mathcal{P}}{g (\sum_{j=i}^k m_j)}$$

When the above is not satisfied the contact is slipping. Otherwise system will have $\ddot{x}_i=0$ for when the contact sticks.

Block Matrix Solution

Here are the steps needed to solve the above system

1. Stick all contacts with $\ddot{x}=0$ and find the friction needed $F^{\star}=A^{-1}P$. For example $$F^{\star}=\begin{bmatrix}1 & 1 & \cdots & 1 & 1\\ & 1 & \cdots & 1 & 1\\ & & \ddots & \vdots & \vdots\\ & & & 1 & 1\\ & & & & 1 \end{bmatrix}\begin{pmatrix}0\\ 0\\ \vdots\\ 0\\ \mathcal{P} \end{pmatrix}=\begin{pmatrix}\mathcal{P}\\ \mathcal{P}\\ \vdots\\ \mathcal{P}\\ \mathcal{P} \end{pmatrix}$$
2. Compose the system mass matrix $M=A^{-1}m$ such that the horizontal equations of motion are $\boxed{F^{\star}=M\ddot{x}+F}$
3. Compare friction needed to available traction with $F^{\star}<\mu N$. Construct two projection matrices $T$ and $U$ with $k$ rows and values as follows: For each block $i$ that is sliding add a column to $U$ with the i-th row element equal to 1 and all others 0. For each block $i$ that is sticking add a column to $T$ with the i-th row element equal to 1 and all others 0. For example if only the last element (top) slides then \begin{aligned} T&=\begin{bmatrix}1\\ & 1\\ & & \ddots\\ & & & 1\\ & & & 0 \end{bmatrix}&U&=\begin{bmatrix}0\\ 0\\ \vdots\\ 0\\ 1 \end{bmatrix} \end{aligned}
4. Define the known motions (sticking blocks) with $T^{\top}\ddot{x}=0$ and the known friction (sliding blocks) with $f=U^{\top}F=U^{\top}\mu N$. With the example above then \begin{aligned} \begin{pmatrix}0\\ 0\\ \vdots\\ 0 \end{pmatrix}&=\begin{bmatrix}1\\ & 1\\ & & \ddots\\ & & & 1\\ & & & 0 \end{bmatrix}^{\top}\begin{pmatrix}\ddot{x}_{1}\\ \ddot{x}_{2}\\ \vdots\\ \ddot{x}_{k-1}\\ \ddot{x}_{k} \end{pmatrix}=\begin{pmatrix}\ddot{x}_{1}\\ \ddot{x}_{2}\\ \vdots\\ \ddot{x}_{k-1} \end{pmatrix}\\f&=\begin{bmatrix}0\\ 0\\ \vdots\\ 0\\ 1 \end{bmatrix}^{\top}\begin{pmatrix}\mu_{1}N_{1}\\ \mu_{2}N_{2}\\ \vdots\\ \mu_{k-1}N_{k-1}\\ \mu_{k}N_{k} \end{pmatrix}=\begin{pmatrix}\mu_{k}N_{k}\end{pmatrix} \end{aligned}
5. Define the unknown motions vector $a$ and unknown forces vector $R$ such that the block motion is $\ddot{x}=U\, a$ and the block friction $F=T\, R+M\, U\left(U^{\top}M\, U\right)^{-1}f$. Note that $U^{\top}F=f$ and $T^{\top}M^{-1}F=\left(T^{\top}M^{-1}T\right)\, R$.
6. The horizontal equations of motion are $\boxed{ F^{\star}=T\, R+M\, U\left(a+\left(U^{\top}M\, U\right)^{-1}f\right)}$ with $R$ and $a$ as unknowns.
7. Project to the sliding blocks with $U^{\top}F^{\star}=U^{\top}M\, U\left(a+\left(U^{\top}M\, U\right)^{-1}f\right)$ } $\boxed{a=\left(U^{\top}M\, U\right)^{-1}\left(U^{\top}F^{\star}-f\right)}$
8. Project to the sticking blocks with $T^{\top}M^{-1}F^{\star}=\left(T^{\top}M^{-1}T\right)\, R$ } $\boxed{R=\left(T^{\top}M^{-1}T\right)^{-1}T^{\top}M^{-1}F^{\star}}$
9. Back substitute the projections to get $\ddot{x}=U\, a$ and $F=F^\star - M \ddot{x}$.
• I think W could be dropped: you are not using later in the derivation. Nice derivation :-) – Frank Oct 1 '13 at 3:40
• Can you explain a bit more how that works in terms of directions of the friction forces? I see that you are using Pi - Fi + Fi+1 = mi d2(xi)/dt2, which seems to assume that Fi+1 has the same direction as Pi. Is that always going to work? – Frank Oct 1 '13 at 3:49
• The equation with $\ddot{x}$ is never applicable right? – Brian Moths Oct 1 '13 at 12:54
• @Frank W is used to derive N. The contact normal force is the sum of the weights of the blocks above it. – ja72 Oct 1 '13 at 14:51
• I'll put it in Mathematica and try it for a couple of cases where I think I know the solution. Thanks a lot! – Frank Oct 1 '13 at 17:34

Just going by the definition of friction , it tends to oppose relative motion between two objects or in other words it opposes the TENDENCY of motion.

Ans 1)So if you apply horizontal force to the right to the top slab, the top slab will definitely have friction lets say $fr_1$ in the backward direction(left). Now to the lower slab this $fr_1$ acts in the opposite direction with the same magnitude.(Remember Newton's 3rd Law).To the lower surface of the lower block $V_{relative}$ with respect to ground frame is to the right. Again friction should oppose relative motion .Hence $fr_2$ acts in the backwards(left direction). Hence $fr_1$ and $fr_2$ act in the right and left directions respectively for lower slab and $fr_1$ acts in left direction for upper slab.

Applying the same concept we get quite an interesting answer for the n slabs .(Think !)

• I think this is only half correct - IMHO you need to create scenarios depending on the values of the friction coefficients. For example, if the friction with the table is very small when the friction between the 2 slabs is high, depending on the force applied, the two slabs could move together towards the right. Also, agreed that there is fr1 and fr2, but depending on their magnitude, the net acceleration of the bottom slab could be either right or left, IMHO. – Frank Oct 1 '13 at 3:33

Let's consider the case where there are two blocks first. In this case, there are two interfaces, the first one is below the top block, and it is described by a coefficient of static friction $\mu_{ts}$. The second one is below the bottom block and is described by a coefficient of static friction $\mu_{bs}$.

If you apply a force $F$ to the top block, and if no motion is to happen, then the force from the bottom block on the top block must cancel $F$. The source of this force must be friction. The strength of friction can be as large as $m_t g \mu_{ts}$ where $m_t$ is the mass of the top block. Also, if no motion is to happen, the ground must provide a frictional force $F$ on the bottom block. This force can be as large as $(m_t + m_b) g \mu_{bs}$, where $m_b$ is the mass of the bottom block.

If you increase $F$ from zero, eventually something will move. The weakest interface will break, where "weakest" means it has the lowest maximum $F$ it can sustain. After the slip occurs, the force need to be supplied by the other interfaces decrease. Say if the top interface is weaker then after the top interface breaks, the forces needed to be supplied by the other interfaces will be $m_T g \mu_{Tk}$, where now $\mu_{Tk}$ is the coefficient of kinetic friction for the interface. Thus no additional slipping will occur (as long as we ignore inertia.)

Now lets talk about the case of many blocks. Let $m_n$ be the mass of the $n$th block from the top. Let me $\mu_{ns}$ be the coefficient of static friction for the interface below the $n$th block. Then the maximum force that can be supported by the $n$th interface is $F_n = \sum_{i=1}^n m_i g \mu_{ns}$.

If you increase $F$ from zero, eventually something will move. The movement will occur at the interface that has the lowest maximum force it can support, that is, the interface with the lowest $F_n$. After that, the force at the other interfaces will drop to $\frac{\mu_{nk}}{\mu_{ns}}F_n$, and no further interfaces will break. This means all of the blocks above the broken interface will move at the same speed and all the blocks below the broken interface will remain stationary.

• I can't find any fault in what you are saying, but it doesn't IMHO answer the question: what direction does the bottom slab move, in the case of 2 slabs? – Frank Oct 1 '13 at 3:35
• I am not sure that your reasoning for the n slabs case is valid: if you continue increasing the force after the weakest interface "broke", why wouldn't you reach a point where the second weakest interface can break? – Frank Oct 1 '13 at 3:36
• @Frank, for your first comment about the case of two slabs, just reread the bottom paragraph with $n=2$, so the bottom slab doesn't move unless the bottom interface is the weaker one and the force is sufficiently high to break it. For your second comment, the forces at the interface decrease to $\frac{\mu_{nk}}{\mu_{ns}}F_n$ which is too weak to get any of the other interfaces to break, so only a maximum of one interface will break. – Brian Moths Oct 1 '13 at 12:57
• Staying stationary feels incorrect to me: first, in the case of 2 slabs, if the bottom slabs was moving in unison with the top block when F was not strong enough, why would the bottom slab suddenly stop? Second, even if an interface "breaks", it doesn't mean the friction coefficient there drops to zero. It changes to the kinetic friction coefficient, and there is still a force transmitted to the bottom slab via that friction. So, there are forces on the bottom slab, and hence possibly an acceleration. IMHO the bottom slab becomes stationary only if the forces on it precisely cancel. – Frank Oct 1 '13 at 14:18