Comparing Static Frictions

Question

In this figure, which of the static frictional forces will be more?

My aim isn't to solve this particular problem but to learn how is static friction distributed . Since each of the rough-surfaces are perfectly capable of providing the $-1N$ horizontal frictional force but why don't they ? This is kind of ambiguity that who will provide a bigger share in total static friction. And as the surface have different $\mu$, so we can't even invoke symmetry.

Answer 1

In the cases where you have static friction, the forces will always be defined by the looking at the system and applying the constraints(in other words $F_s\le \mu N$ will only give an upper bound). On the other hand when you are dealing with kinetic friction, it can be easily derived from the famous $F_k=\mu N$.

As an example, let's solve this problem(As they say, a good picture is worth a million words).

The fastest way to find the amount of the static friction forces will perhaps be this:

1. Look at the horizontal forces acting on $m_1$. There is only one such force, $F_{s_1}$. But we know that the object is stationary($\Rightarrow a=0$), ergo, $F_{s_1}=0$.

2. Now that $F_{s_1}$ is zero, let's look at the horizontal forces acting on $m_2$. We have:

$$F_{s_2}-F=0 \\ \Rightarrow F_{s_2}=F$$ Or in this case the friction in the bottom of the lower box is $1_N$ and the static friction between boxes is zero. $\square$

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One should perhaps check whether $F_s\le \mu N$ or not. In this case, we can easily verify that:

$$N_1=m_1 g \\ N_2=N_1+m_2 g=(m_1+m_2)g$$ $$\Rightarrow N_1\approx 98_N \\ N_2 \approx 196_N$$

$$0_N\le 0.75 \times 98_N \color{green}{\large{\checkmark}}\\ 1_N\le 1\times 196_N \color{green}{\large{\checkmark}} $$

Answer 2

All forces act in pairs, so let me start by matching them up:

Force on $M_1 = F = - M_1$ on Some force providing device

Surface on $M_1 = F_1 = -M_1$ on Surface

$M_2$ on $M_1 = F_2 = - M_1$ on $M_2$

The values for the forces horizontal components are found using... $F =$ Given (1 Newton) $$F_{sf} \le \mu_{sf} \cdot F_n $$ $$F_1 \le \mu_1(M_1+M_2)g$$ $$F_2 \le \mu_2 \cdot M_2 \cdot g$$ From this we can clearly see $F_1$ acts to reduce the velocity of, both blocks, $F_2$ is simply the transfer of $F$ and $F1$ into $M_2$. The amount of momentum per unit time transferred cannot exceed $F_2$, or $M_2$ will start to move relative to $M_1$. So $F_2$ could never be the force stopping the system from moving. Unless we change the setup.

Static friction, by definition only applies to a non moving system. So, we know that for M1 one of three cases must be true:

$F>F_1$ and $(F-F_1)<F_2$ - Thus : $M_1$ and $M_2$ act as a single mass that $F$ must accelerate at... $$a = {F-F_{kf1} \over M_1+M_2}$$

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$F>F_1 + F_2$ - Thus : $F$ must accelerate $M_1$ more than $M_2$ at a rate... $$a = {F-(F_{kf1}+F_{kf2}) \over M_1}$$

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$F<F_1$ - Thus : Nothing moves

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Added Part:

I am trying to say that $M_2$ cannot be a force stopping $M_1$ from moving unless $M_1$ was provided with more than enough force to move it anyway. This will not change regardless of the values for the masses or coefficients of static friction. $M_2$ May act as extra momentum that the force $F$ must pull, and its weight may be the contribution required to make the static friction with the surface greater than $F$. But, no matter, for this given diagram the static friction between the two blocks can never be the reason $F$ fails to pull them.

Extra Notes:

• A case for $M_2$ accelerating more than $M_1$ is not possible, unless something in the system is changed(like the force is put on $M_2$ and not $M_1$).

• If the static friction with the surface were to be zero the system would accelerate regardless of the static friction with the other block.

• If you really want to make this problem complicated, put the force on $M_2$ not $M_1$

• If $M_2$ is being held in place by some force, that would totally change the problem.

Answer 3

The contact forces with two blocks are $N_1 = m_1 g + m_2 g$ for the bottom block (to the floor) and $N_2 = m_2 g$ for the top block (to the 1st block).

The available traction is $F^\star_1 = \mu_1 (m_1+m_2)\,g$ and $F^\star_2 = \mu_2 m_2\, g$ or

$$ \begin{pmatrix}F_1^\star\\F_2^\star\end{pmatrix} = \begin{bmatrix}1&-1\\0&1\end{bmatrix} ^{-1} \begin{pmatrix}\mu_1 m_1\,g\\\mu_2 m_2\,g\end{pmatrix} = \begin{pmatrix}\mu_1 (m_1+m_2)\,g\\\mu_2 m_2\,g\end{pmatrix} $$

The balance of horizontal forces is

$$ \boxed{ \begin{pmatrix}P_1\\P_2\end{pmatrix} - \begin{bmatrix}1&-1\\0&1\end{bmatrix} \begin{pmatrix}F_1\\F_2\end{pmatrix} = \begin{pmatrix}m_1 \ddot{x}_1\\m_2 \ddot{x}_2\end{pmatrix}} $$

where $P_1$, $P_2$ are any applied forces on the blocks (in your case $P_1=1N,\; P_2=0N$) and $F_1$, $F_2$ are the friction forces. Here comes the fun part:

Assume blocks are sticking and solve for the required friction $F_1$, $F_2$ when $\ddot{x}_1=\ddot{x}_2=0$

$$ \begin{pmatrix}F_1\\F_2\end{pmatrix}_{stick} = \begin{bmatrix}1&-1\\0&1\end{bmatrix} ^{-1} \begin{pmatrix}P_1\\P_2\end{pmatrix} = \begin{pmatrix}P_1+P_2\\P_2\end{pmatrix} $$

Find the cases where required friction exceeds traction

$$ \begin{pmatrix}F_1\\F_2\end{pmatrix}_{stick} > \begin{pmatrix}F_1^\star\\F_2^\star\end{pmatrix} = \begin{pmatrix}\mu_1 (m_1+m_2)\,g\\\mu_2 m_2\,g\end{pmatrix} $$

For those cases set $F_i = F_i^\star$ otherwise set $\ddot{x}_i = \ddot{x}_{i-1}$ and solve the balance of horizontal forces.

Example 1, All slipping:

$$ \begin{pmatrix}P_1\\P_2\end{pmatrix} - \begin{bmatrix}1&-1\\0&1\end{bmatrix} \begin{pmatrix}\mu_1 (m_1+m_2)\,g\\\mu_2 m_2\,g\end{pmatrix} = \begin{pmatrix}m_1 \ddot{x}_1\\m_2 \ddot{x}_2\end{pmatrix} $$

to be solved for $\ddot{x}_1$ and $\ddot{x}_2$

Example 2, All sticking:

$$\begin{pmatrix}P_1\\P_2\end{pmatrix} - \begin{bmatrix}1&-1\\0&1\end{bmatrix} \begin{pmatrix}F_1\\F_2\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}$$

to be solved for $F_1$ and $F_2$

Example 3, Bottom slipping, top sticking:

$$ \begin{pmatrix}P_1\\P_2\end{pmatrix} - \begin{bmatrix}1&-1\\0&1\end{bmatrix} \begin{pmatrix}\mu_1 (m_1+m_2)\,g\\ F_2\end{pmatrix} = \begin{pmatrix}m_1 \ddot{x}_1\\m_2 \ddot{x}_1\end{pmatrix} $$

to be solved for $F_2$ and $\ddot{x}_1$.

The matrix $A=\begin{bmatrix}1&-1\\0&1\end{bmatrix}$ is the connectivity matrix, and it can be expanded if you have more blocks. See my full solution here of similar problem in more detail: https://physics.stackexchange.com/a/79182/392

Answer 4

The ground will provide all of the static friction. Imagine what would happen if the upper block contributed even a tiny amount to the static friction: It would have to move forward due to the reaction force. Having M2 inch along you pull M1 (which stays stationary) would be very strange indeed.

Static friction always acts to prevent relative motion. It acts from the ground because the ground is the first place where relative motion would kick in if you accelerated it (friction from the top would be able to prevent at least some of the relative motion)