# Static friction with ground when two blocks have different coefficients

Suppose I have two blocks of masses $$m_1=m_2=m$$ with coefficients of friction $$\mu_1=1$$ and $$\mu_2=2$$ respectively with the ground that are in contact. If I apply a force $$F=2mg$$ to $$m_1$$, what are the values of the static friction on each block respectively?

I know that the blocks won't slide as $$F\le \mu_1 m_1 g+\mu_2 m_2g=3mg$$, so $$f_1+f_2=F=2mg$$ $$f_1\le\mu_1m_1g=mg$$ $$f_2\le\mu_2 m_2 g=2mg$$ but there are multiple values of $$f_1$$ and $$f_2$$ that satisfy the system above, such as $$(0,2mg),(mg,mg),(0.5mg,1.5mg)$$ to name a few. Surely, the situation is deterministic and must have a single answer. Am I missing some extra constraint on the values of the frictions?

• From the FBD you obtain $~m\,a_1=F-f_1~,m\,a_2=f_1-f_2~$ thus for $~a_i=0~$ $~f_1=F~,f_2=f_1=F~$
– Eli
Commented May 25 at 13:43
• @Steeven Yes, I mistook this for a block on block problem. Commented May 25 at 16:25

## 4 Answers

Contrary to some comments, you can simply add the static forces together (For a rigidly connected system), as correctly claimed in the question. To prove this, consider a single block with a static fiction coefficient of $$\mu$$. The horizontal force required to get it moving is just over $$mg \mu$$. Cut the block in half orthogonal to the horizontal force. The result for the combined static friction of the two half blocks being pushed in tandem, remains the same as for the original uncut block (as it must). Each half block has static friction equal to $$(1/2)mg \mu$$ because each half block has mass $$m/2$$ and the total static friction is the sum of the static frictions of the half blocks.

I am going to assume (as you did not make it clear), that you intended F to be a horizontal force pushing M1 which in turn pushes M2 horizontally.

An example of a seemingly equally indeterministic situation, is two equal and opposite forces acting on a single static block. The opposing forces could be anything as along as they are equal in magnitude. The situation is resolved when we measure the compression on the block and then an exact answer is obtained.

If the two blocks are a rigid assembly, the force required to get the assembly moving is just over 30 Newtons (using a rounded figure of $$g = 10$$ for the gravitational acceleration) which is the sum of the static frictions as mentioned above. For any force less than 30N, the rigid system remains static and the sum of the two static friction forces is always equal to the input force and static friction of the fist block is always half that of the second block. For example, if the input force is 15N, the static friction exerted by the first block is 5N and the static friction exerted by the second block is 10N. This is completely deterministic.

If however, we place a spring between the two blocks, the situation gets more complicated, because now the first block is free to start moving before the second block. We now have to take dynamic friction into account. Lets say the dynamic friction is half the static friction. When a force of just over 10 newtons is applied the first block starts moving and now the friction switches from static to dynamic and the force accelerating the first block is $$F -m g \mu/2 = 10 - 1\times 10 \times 1/2 = 5$$ N. The first block continues to accelerate until the compression of the spring rises to 5N and then the system comes to a stop because the accelerating force is less than the static friction of the second block. If we now raise the Input force to just a small fraction over 25N the force exerted by the spring on the second block exceeds the static friction of the second block and the whole system starts accelerating at only 25N.

It turns out the minimum force required to get an elastic system of two blocks moving, is the static friction of the second block plus the dynamic friction of the first block. The system will now continue accelerating as long as the input force is greater than 15N. This is the sum of the dynamic friction of the two blocks (30/2). Any lower and the dynamic friction exceed the input force and the system will slow to a stop.

In Summary, for a rigidly connected system and using your notation:

For $$F\le \mu_1 mg + \mu_2 mg$$,

$$f_1+f_2= F, \ f_2 = F_1 \times 2$$

$$\rightarrow \ f_1 = F /3,\ f_2 = 2F/3$$

For $$F > \mu_1 mg + \mu_2 mg,\ f_1 = kF/3, \ f_2 = 2kF/3$$ (Moving),

where k is a fraction such that $$k\mu$$ is the dynamic friction.

For the none rigidly connected system:

For $$F \le \mu_1 m g,\ f_1 = F,\ f_2 = 0$$.

For $$(F > \mu_1 mg)\ \& \ (F \le \mu_2 mg),\ f_1 = k \mu_1 mg,\ f_2 = F- f1$$,

For $$F > \mu_2 mg + k \mu_1 mg, \ f1 = k \mu_1 mg, \ f_2 = k \mu_2 mg \$$ (Moving).

Am I missing some extra constraint on the values of the frictions?

Yes. Assuming ideal rigid body behavior, for $$mg the magnitudes of the applied force $$F$$ and the static friction force $$f_2$$, are related by

$$F-f_{2}=mg$$

See the FBDs below.

Static friction is self-adjusting and only exists in opposition to a net applied force, with that opposition limited to the maximum possible static friction force of $$\mu_{s}N$$.

Given that $$m_1$$ experiences no kinetic friction in the given range of $$F$$, the only possible friction is static, and that is capped at $$mg$$ in the negative $$x$$-direction. What’s more, we know that if $$m_2$$ were not present $$m_1$$ would slide. That tells us that $$m_2$$ must be exerting a force on $$m_1$$ preventing it from sliding.

Thus for $$mg, the net of the applied force $$F$$ in the positive $$x$$-direction and static friction force on $$m_2$$ in the negative $$x$$-direction will always be opposed by the limiting friction on $$m_1$$ in the negative $$x$$-direction.

In other words, as $$F$$ increases the static friction force on $$m_2$$ increases by an equal amount, such that the static friction on $$m_1$$ remains at the limiting friction force $$mg$$.

Thus, if $$F=2mg$$, $$f_{1}=f_{2}=mg$$ which is your second combination of values.

As for the first and third combinations, the third is impossible since $$m_2$$ can't experience any force until the limiting friction force on $$m_1$$ is exceeded, whereas the first combination implies zero friction acts on $$m_1$$.

Note that if $$F=3mg$$, $$f_{2}=2mg$$ and motion of both masses is impending.

Hope this helps.

If the first block can withstand the pushing force $$F$$ (if the friction $$f_1$$ can be large enough to achieve $$f_1=F$$), then there is no need for the second mass to exert any force, and so $$f_2$$ would be $$0$$.

This is not the case since $$f_1$$ cannot be large enough. But it does indicate for us that the force $$F$$ does not just distribute itself evenly onto each block. It first attacks the first block.

So, we might assume that the first block excerts the largest possible force, and that the second block only excerts what is needed extra. This is not a perfectly realistic model, but it might be the final constraint your task is expecting you to use.

• I see what you say but this doesn't seem like a convincing argument. Why should it be this way and not the other? Commented May 25 at 15:09
• @VulgarMechanick My argument would be that, if the first block's friction is not at its maximal value, then it would increase until it is, before it would not be able to hold the pushing force on its own. Commented May 25 at 16:12

The coefficient of friction represents a maximum: the maximumun frictional force before motion.

So the order of forces is important.

From left to right: $$F, \mu_1m_1g, \mu_2m_2g$$

where $$\mu_1 = 1$$, $$\mu_2 = 2$$ and $$m_1=m_2=m$$

A force is applied to $$m_2$$ only when $$F> mg$$.

All force greater than $$mg$$ is applied to $$m_2$$

If $$F = 2mg$$ then $$f_1 = f_2 = mg$$

If the force was applied to $$m_2$$ instead of $$m_1$$ then $$f_2 = 2mg$$ and $$f_1 = 0$$