Induction (plank question)

[lovemake1]

1. The problem statement, all variables and given/known data

Suppose that n identical planks, each of length 1 metre, are stacked flat on top of each other, with each one shifted a little further to the right, creating an ever larger overhang. Prove the maximum span of this overhang is:
1/2(1 + 1/2 + 1/3 + ..... + 1/n-1) metre


2. Relevant equations



3. The attempt at a solution

Ok so I've got a mental image of what an overhang loooks like..

Top
space ----------------
spac ----------------
spa ----------------
sp ----------------
s ----------------
----------------
Bottom

So Fn = 1/2 (1 + 1/2 + 1/3 + ... + 1/n-1) metres

we have to define what Fn is..
and using the already given formula, im suppose to come up with one right ?
But thats where i am having problem.

Please let me know if im going in the right direction.
Helps apprecicated thanks

[tiny-tim]

hi lovemake1! :smile:

probably easiest to do it by induction …

start with n = 2, then assume it for general n and prove it for n+1 :wink:

[lovemake1]

hmm...
Yea I actually tried to solving by using induction method.. so here is where i am at
F[lower subscript(k + 1 )] = 1/2(1 + 1/2 + 1/3 + ... + 1/k)

the above equation is after i substitute k+1 in n.
But how do i represent this F[lower subscript (k +1)] in terms of Fk ? alone ?
Im not sure how else to proceed from here
Helps appreciated

[tiny-tim]

hi lovemake1! :smile:

(just got up :zzz: …)
hmm...
Yea I actually tried to solving by using induction method.. so here is where i am at
F[lower subscript(k + 1 )] = 1/2(1 + 1/2 + 1/3 + ... + 1/k)

the above equation is after i substitute k+1 in n.
But how do i represent this F[lower subscript (k +1)] in terms of Fk ? alone ?
Im not sure how else to proceed from here
Helps appreciated

(try using the X2 icon just above the Reply box :wink:)

pretend that the first k planks are nailed together :wink: …

then find how far the (k+1)th can go without toppling it! :smile:

[lovemake1]

I know that (k+1)st plank must have less than 0.5m of overhang in comparison to k th plank. Otherwise, the weight will shift and fall.

So...
Im not sure what where i would be using the formula of the maximum distance (k+1) can go without toppling.... in this induction problem..

Im througly confused to say the least. Can you give me some hint please?

[tiny-tim]

I know that (k+1)st plank must have less than 0.5m of overhang in comparison to k th plank. Otherwise, the weight will shift and fall.

That's when the top plank will fall over …

but before that happens, the others will fall over …

when is that? :smile:

[lovemake1]

hmm... as obvious it may seem... i dont see what im suppose to find...

sp ------- (k+2)
s ------- (k+1)
------- k

what am i exactly looking for ?
the point when k+1 topples over?

[tiny-tim]

no, pretend that the first k are nailed together …

finding when they topple should be easy :smile:

[lovemake1]

im not how to represent it mathematically.

spa----------- (k+1)
sp----------
s---------
--------

the first k element topple over when the sum of distance shifted to the right from k to (k+1) is > 0.50 right? [distance is s + sp + spa + etc...]

[tiny-tim]

i'm not sure what you mean :confused:

they'll topple when their centre of mass is over the end of the base

[lovemake1]

thats what i sort of meant. any distance more than 0.5m will shift centre of mass to the right and fall.
But how do we write it in mathematical equation?!
and when do i use this equation to solve the equation using induction?

Prove the maximum span of this overhang is:
1/2(1 + 1/2 + 1/3 + ..... + 1/n-1) metre

[tiny-tim]

you need to find where the centre of mass is :smile:

[lovemake1]

is it something like
k + (k + 1) / 2 ?

[tiny-tim]

is it something like
k + (k + 1) / 2 ?

"something like" ??

calculate it! :rolleyes:

[lovemake1]

ok heres my shot at it
centre of mass is

R = (m1*r1 + m2*r2)/(m1 + m2)
asumming all masses are equal.

[tiny-tim]

try an easy case first, k = 3 …

if the 2nd plank is 0.5 along, how far along does the 3rd plank have to be for the centre of mass to be just above the edge and for the planks to be about to topple?

[lovemake1]

if 2nd plank is placed...
space-------- <- any slight shift to the right will cause the 2nd plank to topple over.
0.5m--------
--------

btw when you said "2nd plank is 0.5long" you mean if it has an overhang of 0.5m right ?
cuz all panks are 1m :O

r = 1/2( 1 + 1/2 ) = 3m
if k plank has 0.5 overhang then the (k+1)th plank must have less than 2.5m ?

[tiny-tim]

(yes, but i said "along", not "long"! :biggrin:)

forget the kth plank, just try the 3rd one :wink:

[lovemake1]

it depends where 2nd plank is placed with relative to the base to determine how far 3rd plank has to be inorder to fall right ?

if its
spppaaa---------- something like this would have to occur
s-----------
-----------

centre of mass = (2nd plank distance + 3rd plank distance ) / 2
centre of mass must not go over the base which is distance of 1m ?

[lovemake1]

Im so lost, can anyone give me a big hint or first part of the equation

[tiny-tim]

for 3 planks, you know from the given answer that the top plank must be 1/2 metre along

so where must the middle plank be? :smile: