Let's say a mass of 5 kg is kept on a 10 kg block which is kept on the ground.The friction coefficient between the blocks is extremely large say 10 while the friction coefficient is 0 between the 10 kg block and the ground.A force of 100 N is applied on the 5kg block.What will happen?

Now of course the limiting friction between the blocks (500 N) is greater than the force applied (100 N).Thus the 5kg block must not move.The magnitude of the friction between the blocks comes out to be 100 N.This will supposedly move the 10 kg block.However it is not possible that the 10 kg block moves while the 5 kg block doesn't.So my question is will this force cause any motion?

• I don't think more than a minute playing with a couple of actual objects would be needed to reveal the error in the premise of the question, which is that friction prevents relative motion, not necessarily absolute motion. Jul 31, 2023 at 1:45
• The actual friction force is only 66.66 N, not 100 H. Jul 31, 2023 at 11:43

It is not a paradox if you consider the two blocks moving together. Friction may prevent them from moving against each other, but there is literally nothing in the definition of the problem that would prevent the system of blocks from sliding on the table.

Moreover, both blocks are going to accelerate in the direction of the force together with the value of $$a = \frac{F}{10+5}$$

Now due to Newton's 3rd law the friction force $$P$$ acting on the 5kg block in the reverse direction from $$F$$, is also acting on the 10kg block in equal and opposite fashion as shown below.

Applying Netwon's 2nd law to each of the blocks you have

\begin{aligned} F - P & = 5 a\\ P & = 10 a \\ \end{aligned}

Which produces the solution $$P = \frac{10}{10+5} F$$ and $$a = \frac{1}{10+5} F$$

The coefficient of friction $$\mu$$ between the blocks only plays a role if the force required to move the blocks together is more than the available traction $$P > \mu 5$$.

If that is the case, then the two blocks will have different acceleration, but the value of $$P$$ will be known

\begin{aligned} F - P & = 5\, a_{\rm top}\\ P & = 10\,a_{\rm botttom} \\ \end{aligned}

where $$P = \mu 5$$.

Thus the 5kg block must not move.

This the 5kg block must not move with respect to the surface on which it is resting. That surface is the top of the 10kg block, so the 15kg system composed of the two blocks will move together, with the lower block seeing $$66 \frac 2 3 \mathrm N$$ of force.

In this example one must remember that static friction can have values up to a maximum of $$500\,\rm N$$ depending on the situation.

System: two blocks

As long as the external force $$F\le500\,\rm N$$ the two blocks act as one, ie just as if it was one block, and the external frictional force between the lower block and the ground is zero.

When $$F>500\,\rm N$$ the internal frictional forces are not sufficient to prevent relative motion between the two blocks and the internal frictional forces are then taken to have a magnitude of $$500\,\rm N$$.

You can also change your frame to frame of 5 kg and apply newton's 2nd law on 10kg to compare limiting friction and pseudo force. If limiting friction is greater than pseudo force, then there will be no relative motion and vice versa. If $$F_{lim}>F_{pseu}$$, then force on 10kg is is equal to pseudo force because limiting friction is variable.

No , the best approach for solving such problems is

1. First assume no slipping & check whether the friction actually satisfies the required friction amount & if it satisfies then the block does not slip & moves with a common acceleration now u can assume both to be a same mass and check with the friction & pulling force on the ground whether the system moves or does not actually move.
2. Now if the friction force is lesser than the required then the block slips. Now assume the lower block to be the observation frame and find the lower block's acceleration and the first block is to be observed. Apply the pseudo force in the direction opposite to the acceleration of the lower block & apply the limiting friction & using F=MA find the relative acceleration & now solve it in the ground frame for all details.

In this problem there is no relative acceleration between the blocks

Magnitude of friction (ie.100N) is to be considered as an internal force here. The logic behind doing what you did is to find out if the force exceeds the limiting friction value but since it is 500N, we have to consider the two blocks as a singular system. Had the force applied been greater than 500N, the situation would have been different where we don't consider them as a single system.

form the FBD

$$m\,a_m=F-F\mu\\ M\,a_M=+F\mu$$

thus the center of mass acceleration

$$a_{CM}=\frac{m\,a_m+M\,a_M}{m+M}~\overset{\rm EOM's}{=}\frac {F}{m+M}$$