Stack of Blocks?

Question

A stack of identical aluminum blocks of length l = 13.8 cm and thickness d = 2.0 cm are stacked the right ledge of a table. If we use 7 blocks, how far to the right of the edge of the table can the right edge of the top block maximally reach?

I tried using the equation x(n-1) = x(n) + L/(2n-2) where n is the number of blocks, and L is the length, and then I added 1/2 the length of a block since that equation gives the distance to the middle of the last block. However, that did not work.

What is the minimum height of a stack of blocks for which the left edge of the top block can be located to the right of the right edge of the table?

Since the density is given, couldn't I relate this answer somehow to the first one?

Thanks.

Answer 1

I had some fun conceptualizing this one.

I found a relationship between the overhang of the block under the following two assumptions:

The overhang of each block w/r/t the block below is a constant. I labeled this distance x. The bottom block overhangs the edge of the table by the same amount.

The second assumption I made is an approximation. I assumed that the thickness of the block being much less than the width of the block, the thickness may be ignored. If I get some time I may revisit and try to consider the width of the block to see what that solution looks like.

Under my assumptions, here's the solution:
The x is constant at
x=L/(N+1)

The distance of the right edge of the top block is
N*x

distance=
N*L/(N+1)

for 1 block
L/2

for 2 blocks
L*2/3

for 7 blocks
L*7/8

Here's the proof:
I summed the torques at the edge of the table to find the exact balance point. ( I left out the weight since it divides out)

(L-x)^2/(2*L)=
x^2/(2*L)+(2*x-L/2)+(3*x-L/2)
+(4*x-L/2)+(5*x-L/2)
+(6*x-L/2)+(7*x-L/2)


multiply by 2*L
(L-x)^2=x^2+4*L*x-L^2
+6*L*x-L^2+8*L*x-L^2
+10*L*x-L^2
+12*L*x-L^2
+14*L*x-L^2


(L-x)^2=x^2+54*L*X-6*L^2
L^2-2*x*L+x^2=
=x^2+54*L*x-6*L^2
7*L^2=56*L*x
7*L/56=x
L/8=x

note that this is
L/(N+1)=x

Here's an update.

I looked at the problem and included the thickness.

For block N, the torque is r*weight*cos(th)

Where th is the angle subtended by the hypotenues of a right triangle formed by a line vertical from the center of mass of the block which is N*x-L/2 horizontally from the edge of the table. The other side of the triangle is the horizontal line from the edge of the table to the vertical line. It intersects the vertical line at a point (2*N-1)/2 above the edge of the table. The hypotenuse is length r.

Using these dimensions, cos(th)=(N*x-L/2)/r
so the torque is
(N*x-L/2)*weight


Note that this is the same result as obtained by assuming zero thickness.

Looking at the torque of the bottom block, I get the result that
(L-x)^2/(2*L)*weight is the torque to the left
and x^2/(2*L)*weight is the torque to the right.

Again, this is consistent with the first solution.

Good question, thanks for asking


j