# Maximum static friction between two bodies being forced in opposite directions

When a body is lying on the floor, the following is true.

The maximum static friction is Frmax=μ*Fn, where Fn is the normal force and μ the static friction coefficient.

If a force F1 is applied to the body in a first direction, the body will move if F1 is bigger than Frmax.

However, what happens if the body is not lying on the floor but on another body that is being forced in the opposite direction?

In which case does a movement between the bodies occur?

F1 + F2 > Frmax

F1 or F2 > Frmax

|F1 - F2| > Frmax?

As long as static friction is present F1 and F2 will be transmitted to the respective other body and cancel each other. But is this transmission relevant for overcoming the static friction?

Fr = F1-F2 (if Fr is defined parallel to F1) So it should be the third formula in my opinion. But that feels wrong. Why would it be harder to push something off a table if someone is trying to move the table at the same time? Additionally, when using superposition, there would then be a partial friciton greater than Frmax.

Bonus question: Why is every website, book, whatever that I find always looking only at one force? Is this question that trivial or that irrelevant?

• What is $F_2$? is it the friction force by the ground on block b or an external force ? Oct 29, 2020 at 12:39
• @Bhavay an external force. Situation 2 is located in a vacuum with no gravity and no ground. (But there is an external Force Fn making Frmax > 0) Oct 29, 2020 at 13:02
• In absence of gravity , there is no friction btw the blocks, as there will be no normal forces btw the blocks. Oct 29, 2020 at 13:11
• @Bhavay there is now: imgur.com/a/DMm3VYO Oct 29, 2020 at 13:21

Let the upper block have mass $$m$$ and the lower block have mass $$M$$.

Suppose $$F_1>F_2$$. If friction is sufficient to hold the blocks together then they will move to the right with acceleration

$$a = \dfrac{F_1-F_2}{M+m}$$

The net force on the upper block must therefore be $$ma$$, so if the frictional force is $$Fr$$ acting to the left on the upper block then we have

$$F_1 - Fr = ma = (F_1-F_2) \dfrac m {M+m} \\ \Rightarrow Fr = F_1 - (F_1-F_2) \dfrac m {M+m} = \dfrac {F_1M + F_2m}{M+m}$$

We can check this by calculating the net force on the lower block. By Newton's third law the frictional force on the lower block is $$Fr$$ acting to the right, so the net force on the lower block is $$Fr-F_2$$:

$$Fr - F_2 = \dfrac {F_1M + F_2m}{M+m} - F_2 = \dfrac {F_1M - F_2M}{M+m} = Ma$$

as expected.

So the blocks will slide over one another if

$$\dfrac {F_1M + F_2m}{M+m} > Fr_{max} = \mu mg$$

If $$F_2 > F_1$$ then a similar calculation gives the same condition.

• So if m = M then m(F1+F2)/2m > Frmax <=> (F1 + F2)/2 > Frmax? Then if F2 = 0 it's 1/2 * F1 > Frmax, even though it should be F1 > Frmax? That can't be right. Am I missing something? Oct 29, 2020 at 15:34
• @Modulus If $m=M$ and $F_2=0$ and the blocks do not slide then the net force on each block must be $\frac {F_1}{2}$ so $Fr=\frac{F_1}{2}$. The blocks will slide over each other if $\frac{F_1}{2}>Fr_{max}$. Remember the lower block is not fixed to the ground. The condition only becomes $F_1>Fr_{max}$ in the limit when $M>>m$. Oct 29, 2020 at 16:00
• Can you take a look at physics.stackexchange.com/q/589933/278169 ? I was trying to solve the problem there by figuring out if the criteria for overcoming static friction can be solved by superposition and just ignoring one of the forces, but apparently thats not the case. And now I've still not gotten anywhere. But thanks anyways. Oct 29, 2020 at 16:29