Prisms

no one ?

Do you need help with this or are you posting this as a riddle?

i'm posting this as a riddle. The reason why i posted it in this section and not in the Riddle section is because i thought that this was more appropriate here.

So long as you're material isn't fragile (crushable or compressible) I'd expect a formulea like for μ > 0 and a < sin μ, f = mg / μ cos a

So given both you' re surface is't frictionless and the angle is less than the coefficient of static friction, your force will be weight, increased by your coefficient of static friction and increased as your angle a moves away from 0 to 90 degrees.

good job

zerone, stop posting tbs challs
and
g__day, your 'solution' is wrong

greenbite - feel free to add detail. As I see it

if a = 0 (i.e. prism is a block or sheet of paper) cos a = 1 so force must be mg (tension is same on both sides of teh block) divided by coefficient of friction μ.

So if m = 1kg and μ was .5 f = 1 * 9.81 / .5 = about 20 Netwons to hold a 1 kg mass - seems correct.

If a = 90 -> grip is impossible (cos a = 0, divide by zero so answer consistent).

If a = 0 and surface was almost frictionless e.g. μ = 0.01 in the above example you'd need 100 times the weight to hold the block, so workings seem correct so far.

If the angle went past the sin of the coeffiecnet of friction no force could hold the block.

I could easily have made a simple mistake, but the core thinking seems reasonable - please add soem more detail!

I think gD has it basically right. If F>mg/(2 μ cos a) the block stays up. The two comes from F being applied to each side. On a side note, what is a "tbs chall"?

tbs(=the black sheep). It's a cool site with lots of challenges (javascript, php, programming, logic, crypto, SCIENCE)
And this is one of the sience challenges.

The black sheep

Well i also send in the solution g__day posted and the auther of that challenge (and admin fo that site) said it's wrong. I thought it was right to, but hey if the auther says it's not...

It could easily be wrong on the triggers . limits where the formulea applies e.g. I maybe should have used a sin vs a cosine - must check this!

Okay I stuffed up in two or three ways that complicate the solution.

Point 1

Its hard to tell from the diagram but I assumed a = a', meaning f and f' are equal in magnitude, if not my answer is slightly more complex.

Point 2

If by geometry the two forces are of equal magnitude but opposite directions - this halves the force required, which I stated, but then forgot to do - d'oh!

Point 3

This point is far more complex, rather than just saying f = f' = mg/2μ cos a (note the divide by 2 now ) if you split f and f' into its component fx and fy and f'x and f'y sub-vectors then you realise that fy and f'y are pushing the Prism downwards - with a force equivalent to f sin a' and f' sin a respectively. So they add to the weight you have to lift!

Oops I didn't account for this additional weight. So the lift is actually transfered by an internal counter-force within the prism based upon the horizontal forces applied * (lessened by) a friction co-efficient you have to overcome.

So the lift comes by and equal and opposite force inside the prism whose x components resist crush pressure and -fx and -f'x cancel each other but -fy and -f'y create lift (or more correctly traction) if this is to work!

This is a bit like the science Romans used to build their aquaducts and river bridges from stones by the way.

* * * *

So this becomes solve to stay aloft (horizontal traction must exceed or equal vertical downward force. Lets describe all the forces now, decomposed into vertical and horizontal vectors.

Horizontal grip must equal or exceed vertical weight, alah:

μ (F cos a' + F' cos a) >= mg + F sin a' + F' sin a

if a = a' then F has the same magnitude as F' so the formulea becomes

2μ F cos a >= mg + 2 F sin a

F (2μ cos a - 2 sin a) >= mg

F >= mg/2(μ cos a - sin a)

This appears much more reasonable and valid. When a is small the cos term dominates but as a increases there reaches a point where sin a = μ cos a (so if μ = 1 then a = 45 degrees) and force must be infinite to hold the object. Also if μ is small alot more force is needed to hold any prism and the angle of slipage decreases from 45 dramatically.

Now if a <> a' then F <> F' in magnitude one can only say

μ (F cos a' + F' cos a) >= mg + F sin a' + F' sin a or

F(μ cos a' - sin a') + F'(μ cos a - sin a) >= mg

but to be in equilibrium the x-plane forces must exactly cancel each other (else the prism dangles at an absurb angle and things get more complex still - but this is not stated in the picture ) so we can rule it out and state a new ruling

F cos a' = F' cos a, otherwise the prism would slide left or right!

so substitute this new bit of knowledge in:

2μ (F cos a') >= mg + F sin a' + F' sin a is the best general form

Now if the base is an isoceles triangle it kinda infers a = a', so I'd say this reduces to F sin a' = F' sin a, meaning

2μ (F cos a') >= mg + 2F sin a'

but as a = a'

2μ F cos a >= mg + 2F sin a

2F (μ cos a - sin a) >= mg

once again F >= mg / 2(μ cos a - sin a)

Nice catch on the downward force vector of the puzzle. Well done.

Thank you,

Amazing what simple observations you can fail to make when looking at a flat 2-d puzzle. But visualising this in 3d and then using maths I was taught over 26 years ago a simple force diagram revealed the whoopsie and what a good puzzle this was. When I suddenly realised hang on - both forces are pushiing this down - so why does it stay up - I had even more appreciation of the puzzle.

So of course than answer to the first part is simply

μ cos a > sin a or rearranged the prism can be held given μ > tan a or a < arctan μ

Nice puzzle!