the force required to lift the cubes into the air

Question

On the horizontal table, there are ten identical little cubes in a row.

The cubes are not bonded together in any way. Squeeze the entire chain with the two outermost cubes in your palms. Which force F should you use to squeeze the cubes off the table and split them apart? Each cube has a mass of m, and the coefficient of friction between cubes is.

Answer 1

I do not really understand the question and given that nobody has answer it yet,

I am probably not alone.

So let me just ignore the question and tell you simply what happens

when you try to lift an array of blocks between your hands.

Let

n be the number of blocks,

m be the mass of each block,

w be the width of each block,

h be the height of each block,

f be the coefficient friction between the blocks,

g be gravitational acceleration,

F be the force applied by your hands.

There are two ways in which holding the blocks together may fail:

1) They are too slippery and fall down.

2) They open up, typically in the middle and fall down.

Both ways imply a constraint on the force F.

1) F must prevent slipping.

Each block experiences horizontal force F causing a force of friction Ff.

This force of friction must overcome the force of gravity mg.

So

Ff > mg

F > mg/f

2) F must prevent splitting.

For simplicity assume that n is even, so that the array of blocks have a tendency to

split right in the middle.

Let k = n/2 - 1

The slit normally looks like this:

The two blocks directly in your hands stay there, but in-between the two arrays of length k

break up in the middle.

Consider the left array of length k blocks.

Let B be the bottom left corner of the array; that point is shared with the block in your left hand.

Let T be the top left corner of the array; that point is also shared with the block in your left hand.

The failure involves the top point T rotating around the bottom point B.

The torque of the weight of the array causes the rotation and

the torque of the force F works against the rotation.

Therefore to prevent this type of failure, the torque of F must be larger than the torque of the weight.

The force F acts in the middle of the height h; therefore its torque about B is

(h/2)×F

The weight of the array is kmg and acts in the middle of the array; therefore its torque is

(kw/2)×kmg

The condition of preventing this type of failure is

(h/2)×F > (kw/2)×kmg

F > k²mgw/h

The conclusion:

In the usual situation where the friction is sufficiently large, so that

F > mg/f

the force F

- must increase linearly with the weight of each block,

- must increase linearly with the width of each block,

- can decrease linearly with the height of each block,

- must increase quadratically with the number of blocks.