# Friction Block-on Block

Been studying block-on block problems. Attaching a picture of a sample problem. Pls do not answer the answer based on the picture, it is just a reference to the conception question to be asked.

In the first step to solving these questions, we always find the F max such that the blocks can move together. And for any force lesser than that, the blocks are said to move together with no relative motion. And obviously, if the limiting friction of the ground on the first block is greater than the force applied(assuming it is applied on the first block),the system doesnt' move. However,for the sake of the question let me assume that ground friction is zero. I do not understand why there is no condition that the F applied on the first block should be greater than the limiting friction between the surfaces of the blocks.How does the system move for a force which is lesser than the limiting friction between the surfaces of the block?

• I don't understand the second to last sentence of your post. Are you saying there is no condition where F can cause the top block to slide on the bottom block? Jan 5, 2021 at 19:09

When you apply a force less than maximum value of static friction between blocks will accelerate the lower block and the system will move without slipping (between surfaces of block).

When you draw free body diagram of lower block, you will see that it only has 1 horizontal force which is due to static friction between the blocks. Hence there is no way that lower block doesn't move when you apply a force on the lower block.

• Thanks for the reply!But could u elaborate a bit more, didnt quite get what u mean. Jan 5, 2021 at 19:00
• @VarunSudhir I updated my answer. Does it make sense now ? Jan 5, 2021 at 19:06
• Yup, thanks a ton! Jan 6, 2021 at 15:59

If the force on the first block does not exceed the maximum friction between the blocks and if the ground were frictionless, both blocks would move at the same speed. However if the force exceeded the maximum friction between the blocks the lower block will slide along the ground, and the upper block will also slide along the lower block. This is because the lower block will experience a force equal to the maximum friction between the blocks, and the upper block will experience a force equal to the total applied force, which exceeds the maximum friction between the blocks

I do not understand why there is no condition that the F applied on the first block should be greater than the limiting friction between the surfaces of the blocks.

Nothing says the applied force $$F$$ can't be greater than the maximum possible static friction force of $$\mu_{s}mg$$, or in this case greater than 29.4 N. But when it is greater the friction force between the blocks changes from static friction to kinetic friction and the top block slides on the bottom block. You then have a new situation where the top and bottom blocks experience different accelerations.

Up until the maximum possible static friction force is reached, the actual static friction force always matches the applied force and the two blocks stick together and move as one with the same acceleration. The transition from static to kinetic friction is illustrated on the friction plot shown here:

http://hyperphysics.phy-astr.gsu.edu/hbase/frict2.html#kin

How does the system move for a force which is lesser than the limiting friction between the surfaces of the block?

My undergraduate mechanics professor always said to the class DRAW A FREE BODY DIAGRAM!! And yes, he did yell it.

To analyze the motion of the two block system when $$F$$ is less than the maximum possible static friction force, you draw a free body diagram (FBD) showing all the external forces acting upon the two block system. See FIG 1 below.

The only external forces acting on the two block system are (1) the downward force of gravity on the two blocks, (2) the upward normal reaction force of the frictionless ground surface on block 1 and (3) the horizontal applied force $$F$$ on block 2. Note that the the equal and opposite static friction forces (equal to 10 N in this example) at the interface between the blocks are internal forces and are therefore not part of the FBD.

The vertical forces (1) and (2) are equal for a net vertical force of zero. The only net external force acting on the two block system is therefore the applied horizontal force $$F$$.

Then, from Newton's second law, the acceleration of the two block combination is

$$a_{x}=\frac{F}{(m_{1}+m_2)}$$

Hope this helps.