Answer:
The magnitud of the force is 124.8N.
Explanation:
First we have to find the value of the static friction coefficient, when the external force F is applied to upper block (i will call it A Block) we have a free body diagram as the one shown in the figure i attached, so since this block has no aceleration in any direction the force F should be equal to the friction force between A and B block, one we noticed this we can use the equation for the Friction force to find the coefficient:
![0=F-FrictionAB](https://tex.z-dn.net/?f=0%3DF-FrictionAB)
μs
and again, since the block has no acceleration the normal between A and B block should be equal to the weigth of the first block, so we have:
![0=Nab-W](https://tex.z-dn.net/?f=0%3DNab-W)
![Nab=W=mg](https://tex.z-dn.net/?f=Nab%3DW%3Dmg)
replacing this we have:
μs
μs*![mg=41.6N](https://tex.z-dn.net/?f=mg%3D41.6N)
and μs![=41.6N/(mg)](https://tex.z-dn.net/?f=%3D41.6N%2F%28mg%29)
now it's time to see the free body diagram for the b block, if we now apply the F force to the B block the diagram should look like in the figure.
the color of the arrow gives you an idea of where the force comes from, the blue ones comes from the B block, the red ones from the A block and the brown ones from the ground.
now for the B block you can see two friction forces, one for the ground and one for the A block, both of these directed bacwards, and two normal forces, again one for the ground and one for the A block but the normal force for the A block is aiming downwards.
again we use the fact that the block is not accelerating in any direction so the sum of the forces in x and y direction have to be 0, so:
![F-Friction1(ground)-Friction2(AB)=0](https://tex.z-dn.net/?f=F-Friction1%28ground%29-Friction2%28AB%29%3D0)
This is the new external F force that we are looking for:
![F=Friction1(ground)+Friction2(AB)](https://tex.z-dn.net/?f=F%3DFriction1%28ground%29%2BFriction2%28AB%29)
we know Friction2(AB) because we found that in the previous block so:
*μs
for the other friction we have to use the equation:
*μs
from y axis we have:
![N(ground)-w-Normal(AB)=0](https://tex.z-dn.net/?f=N%28ground%29-w-Normal%28AB%29%3D0)
![N(ground)=w+Normal(AB)](https://tex.z-dn.net/?f=N%28ground%29%3Dw%2BNormal%28AB%29)
we found the value of Normal(AB) with the previous block so:
![N(ground)=mg+mg=2mg](https://tex.z-dn.net/?f=N%28ground%29%3Dmg%2Bmg%3D2mg)
and:
*μs
*μs
*μs+μs*
*μs
and since: μs*![mg=41.6N](https://tex.z-dn.net/?f=mg%3D41.6N)
the new F force would be:
*μs![=41.6*3=124.8N](https://tex.z-dn.net/?f=%3D41.6%2A3%3D124.8N)