Get that old nut of yours warmed up with this one.
Buffer  16/12/2021 11:31:07 
343 forum posts 155 photos  I found this in a book yesterday and to be honest I can't work it out. I know the answer as I have a good CAD program. But what about you, can you do it the old fashioned way? How high above the surface of the part does the ball protrude?

RobCox  16/12/2021 11:57:41 
59 forum posts 20 photos  99.4 thou if my maths is correct. If the slot width is 2W and the ball diameter = 2R and the angle of the slot sides to the vertical = A: The ball contacts the sides a height of R.sinA below the ball centre. The distance across the slot at the contact points is 2R.cosA, so the height below the top of the slot is (WR.cosA) / tanA, giving the height of the top of the ball above the top of the slot as H = R + R.sinA  (WR.cosA) / tanA For R = 0.5, W = 0.625 and A = 9075 = 15, that gives the answer above.
Shoot me down if I'm wrong.
Rob 
JasonB  16/12/2021 12:09:22 
Moderator 23042 forum posts 2769 photos 1 articles  0.414" My method 1. Work out how far below ctr of ball it makes contact 0.129 and also the horizontal distance 0.966 2. Work out how far down the taper the 0.966 width is 0.215 3 height is the 0.129 plus radius of 0.500 less the 0.215. Answer or at least my answer is 0.414 Edited By JasonB on 16/12/2021 12:14:16 
RobCox  16/12/2021 12:20:36 
59 forum posts 20 photos  My answer is clearly wrong, because half of 1.125 is not 0.625 Edited By RobCox on 16/12/2021 12:20:56 
JasonB  16/12/2021 12:25:22 
Moderator 23042 forum posts 2769 photos 1 articles  So is mine now I have done it in CAD, method was right but I got the distance down that the points of contact are wrong as I used TAN not SIN, won't post the right answer so others can have a go Edited By JasonB on 16/12/2021 12:28:52 
RobCox  16/12/2021 12:32:06 
59 forum posts 20 photos  Jason, I agree with your method, but I get the height of the contact below the top (x in your diagram) to be 0.0795/tan15 = 0.2968, which gives me a revised answer of 0.3326. Thank goodness I haven't got to worry about maths exams any more! 
JasonB  16/12/2021 12:55:36 
Moderator 23042 forum posts 2769 photos 1 articles  That's it.
Next 
not done it yet  16/12/2021 13:07:32 
6887 forum posts 20 photos  Nothing other than a cursory glance, but I think it needs a couple of simultaneous equations to sort out the contact point. Might look at it later… 
Martin Connelly  16/12/2021 13:08:11 
2180 forum posts 227 photos  This is my take on the solution. It requires solving the equation to find angle alpha using trig identities which I did not include on the page as it is messy and seems a bit unnecessary to show here. The point is it gets the same result a different way from those above. Martin C 
Jon Lawes  16/12/2021 15:31:09 
984 forum posts  I can't help but feel inferior to the brains knocking around on this forum. I'd almost certainly have had to google the technique; I am for some reason unable to retain mathematics techniques and such.
Well done to the examiner and to those who attempted an answer! 
Buffer  16/12/2021 17:01:28 
343 forum posts 155 photos  Well as Jason asked for it, here it is. I take no credit for this and I can't do it without CAD. If the ladder just touches the cube how far up the wall does the ladder touch the wall? Good luck you'll need it.

Brian Wood  16/12/2021 18:06:54 
2579 forum posts 39 photos  I cheated and solved it by construction! The answer there is 9.625 feet Brian 
JA  16/12/2021 18:25:50 
1401 forum posts 81 photos  Agreed. I have met Buffer's problem before and it was deemed unsolvable. I have just written the equations down: three simulanteous quadratics. My evening meal calls. I am sure someone will solve it by the time I next look at the website. JA 
Bob Mc  16/12/2021 18:26:57 
231 forum posts 48 photos  Ladder box wall problem... It can be done, I spent many a happy hour trying to solve this one some years ago, and although I didn't find a solution myself the problem can be solved, here is one solution. 
Pete Rimmer  16/12/2021 18:47:24 
1256 forum posts 69 photos  I coudn't do the boxandladder one but for sure it has two possible solutions. Only the visual representation (and the fact that noone would use such a lowangle ladder) suggests that the base of the triangle is less than the height. Mathematically there are two values of X where the line meets the cube. 
JA  16/12/2021 20:07:50 
1401 forum posts 81 photos  I have just looked at the link. It is the same two solutions turned through ninety degrees. It would be interesting see if complex numbers would make the solving easier. Too old for that now. JA 
DMB  16/12/2021 21:09:08 
1350 forum posts 1 photos  With the first diagram, I didn't actually try maths but thought "ballpark"  the height of a vertical radius above the flat top is approximately 2/3 of the total radius. Total radius is 1/2", (16/32) so 2/3 is about 10, or 5/16, so it seems my guess was quite close! 
pgk pgk  16/12/2021 21:40:47 
2603 forum posts 293 photos  Surely the first problem has a simple approach. pgk 
duncan webster  16/12/2021 22:17:37 
4119 forum posts 66 photos  My answer to problem 2 is 9.677 ft, but I had to use a Bisection routine to get to it, otherwise I think you have to solve a 4th power equation. I've checked my answer on CAD 
RobCox  16/12/2021 22:31:58 
59 forum posts 20 photos  I could try and pretend I'm really whizzy with maths by giving a solution, but I'm not, so I'll instead refer you to a youtube channel I've watched from time to time. The channel creator's name is Presh Talwalker and he solves this exact problem in his video " the ladder and box problem  a classic problem". Me, I just got bogged down in quadratic equations that looked like they might turn into quartics... 
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