Question

One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction between the blocks is the same as the coefficient of static friction between the lower block and the horizontal surface. A horizontal force is applied to the upper block, and its magnitude is slowly increased. When the force reaches 41.6 N, the upper block just begins to slide. The force is then removed from the upper block, and the blocks are returned to their original configuration. What is the magnitude of the horizontal force that should be applied to the lower block, so that it just begins to slide out from under the upper block?

Answer 1

The magnitud of the force is 124.8N.

Step-by-step 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`

`F=FrictionAB=Nab*`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`

`Nab=W=mg`

replacing this we have:

`F=`μs
`*Nab=`μs*
`mg=41.6N`

and μs
`=41.6N/(mg)`

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`

This is the new external F force that we are looking for:

`F=Friction1(ground)+Friction2(AB)`

we know Friction2(AB) because we found that in the previous block so:

`F=Friction1(ground)+mg`*μs

for the other friction we have to use the equation:

`Friction(ground)=N(ground)`*μs

from y axis we have:

`N(ground)-w-Normal(AB)=0`

`N(ground)=w+Normal(AB)`

we found the value of Normal(AB) with the previous block so:

`N(ground)=mg+mg=2mg`

and:

`Friction(ground)=2mg`*μs

`F=Friction(ground)+mg`*μs

`F=2mg`*μs+μs*
`mg=3mg`*μs

and since: μs*
`mg=41.6N`

the new F force would be:

`F=3mg`*μs
`=41.6*3=124.8N`