| Posted on Saturday, 08 November, 2003 - 02:37 pm: |
Just had a question from school about modelling how far a ladder on a table, resting on a wall, can move away from the wall without the table toppling, i know that i had to assume that the wall was smooth, but i cant work out why i had to make that assumption! please help!
| Posted on Saturday, 08 November, 2003 - 02:44 pm: |
If the wall was not smooth then it would contribute a frictional force to the ladder, as well as a normal reaction. This would make the equations more difficult to solve.
| Posted on Saturday, 08 November, 2003 - 03:08 pm: |
Does the answer actually depend on the coefficient of friction? If so, then you can't do the question because you haven't been told the coefficient of friction. If not, then you might as well assume there is no friction (coefficient of friction = 0). Of course, you have to work out which is the case. I can't tell you because I don't quite understand what the question is.
| Posted on Sunday, 09 November, 2003 - 02:10 am: |
I am actually wondering what exactly the question is.
however if i get you correctly then it seems that you are given some situation and then you need to make some assumptions regarding it.
One thing to note here is assumptions may be made or may not be made depending which rules you are going to apply.
We will simply assume for now that you are going to apply Newtonian Mechanics here.
In which case,you need to make following assumptions,
1> COG of the ladder is known (and if not known , then assume that mass is uniformly distributed and COG is in the center)
2> the ladder must be sufficiently long so that its COG lies outside the boundaries of the table
3> When the ladder falls, it exerts enough force to topple the table (This assumption basically is not necessary because the ladder may or may not topple the table and whether it is toppled or not can be seen through calculations.However in your question it says that the table is toppled and hence we will assume so)
4> The table should not slip.(Again this is not necessary but the question specifically mentions topples hence it is required)
I am not sure whether the wall being smooth is necessary for this one.
love arun
| Posted on Sunday, 09 November, 2003 - 02:01 pm: |
I'm pretty sure the answer does depend on the coefficient of friction with the wall, since this introduces an extra force, as Philip said. Physically, we'd expect a ladder to be less likely to topple a table when it's resting against a rough wall, because the friction provides some resistance to the ladder's tendancy to rotate that therefore doesn't need to be provided by the (rough) surface of the table. Hence you need either to include it in your calculations (which would make them considerably more complicated) or assume it to be zero, which is probably more sensible in a simple model.
I'm not sure what Arun's condition (2) is about, as the table can still topple even if the centre of mass lies above it. It will topple in the opposite sense of rotation from that of the ladder, because the ladder is pushing on it away from the wall. Condition (3) is essentially the same as saying that the ladder will not slip against the table, because (besides the table's toppling) there's no other way it can fall.
Matthew.
| Posted on Monday, 10 November, 2003 - 02:52 pm: |
Hmm i had taken it for granted that the ladder is placed at the center of the table in which case the COG must be outside the boundary of the table.
However,if it is placed on the edge of the table then COG is not important as you suggest.
love arun
| Posted on Monday, 10 November, 2003 - 05:45 pm: |
I kinda had the same question in physics. If you ignore the table, and include the wall having a frictional force you must use moments on the ladder in order for the forces on the ladder to be equal, which complicates things.
In the image i attached, look at how the second diagram would result in the ladder moving from the wall due to the reaction of the wall if you didnt use moments. The diagrams illustrate why you were told to assume the wall is smooth like in the first diagram, so you dont end up with a more complicated equation for the rest of the question.
| Posted on Monday, 10 November, 2003 - 05:55 pm: |
argh, i made a mess...but fixed it, here it is
| Posted on Monday, 10 November, 2003 - 06:02 pm: |
Well the answer is pretty simple.
If the wall had friction, then the ladder can be pulled away longer than in the case where the wall is frictionless.(The frictionless wall provides a limiting condition and especially a lower limit on the length the ladder may be pulled away).Such limiting conditions are to be assumed only if the friction on the wall is not known. If it is known, much better estimate of the length, the ladder can be pulled away can be made.
love arun
| Posted on Tuesday, 11 November, 2003 - 10:35 pm: |
James, when you write 'If you ... include the wall having a frictional force you must use moments' you don't mean to imply that the problem with a smooth wall can be done without applying the principle of moments, do you?
Matthew.
| Posted on Tuesday, 11 November, 2003 - 10:51 pm: |
I meant just the forces acting on the ladder, not the table, that of course has to be solved using moments.
I dont know if it makes any difference ignoring moments on the ladder in this way.
| Posted on Wednesday, 12 November, 2003 - 10:32 pm: |
Well, if the moments acting on the ladder aren't balanced, it will turn: they must be balanced for it to be in equilibrium. How did you work out the friction on diagram 1 without using moments?
| Posted on Wednesday, 12 November, 2003 - 11:43 pm: |
The friction is the same as the reaction of the wall on the ladder, in the opposite direction (since there is equilibrium vertically and horizontally).
I can (i think) shown mathematically that this is actually the same as if you take moments, i have just worked it out.
| Posted on Friday, 14 November, 2003 - 11:28 am: |
The friction is the same as the reaction on the wall, but how can you work out that reaction force without using moments? You can work out the reaction on the ground as being equal to the weight, but by applying the fact that the reaction on the floor is equal to the friction you have only one equation to find two unknowns.
| Posted on Friday, 14 November, 2003 - 05:33 pm: |
Here is how i showed that it is the same: Let the length of the ladder =L and the angle that the ladder makes with the horizontal =q R1= reaction from wall R2= reaction from ground Taking moments about the point where the ladder meets the ground. L×R1 sinq = L/2 W cosq 2R1 sinq = W cosq W=2R1 tanq Taking moments about the centre of the ladder: L/2 R1 sinq+L/2 F sinq- L/2 R2 cosq = 0 sinq(R1+F)-R2 cosq = 0 R2=tanq(R1+F) Taking moments at top of ladder L/2 W cosq+L F sinq- L R2 sinq = 0 W cosq+2F sinq-2 R2 sinq = 0 2F=(2R2 sinq-W cosq)/sinq 2F=2R2-W cotq Substituting in we get, 2F=2tanq(R1+F)-2R1 tanq×cotq F=tanq(R1+F)-R1 F=R1 tanq+F tanq-R1 F(1-tanq)=R1(tanq-1) F=R1(tanq-1)/(1-tanq) F=-R1 :) Therefore it is the same as when you just consider theequilibrium vertically and horizontally separately. Moments are only applicable here when the horizontal and vertical components aren't in equilibrium, like in the second diagram.
| Posted on Friday, 14 November, 2003 - 10:27 pm: |
That's an interesting proof, but it only shows that we can derive the condition F=-R-(1) from the principle of moments, not that we could derive the principle of moments from F=-R1 . If you can start from diagram 1, (knowing only L, W and theta ) and work out F and R2 without using moments (which is what you would need to do to continue with the toppling problem), could you explain how?
I know I'm being awkward about this, but I really can't see how you could solve this problem without moments, and I'd like to find out how you have.
Matthew.
| Posted on Friday, 14 November, 2003 - 11:40 pm: |
Well assuming the wall is smooth, as in the question, the reaction from the floor, is equal the the weight. If it wasnt, then there would be an acceleration vertically.
To find the friction, yes you do need to use moments (given only L, W and theta ) because the reaction of the wall on the ladder will be dependant on the angle of the ladder, however you can say that the friction force = the reaction from the wall(again, otherwise there would be an acceleration horizontally).
For some reason, up until just now, I had overlooked the fact that we need to find the frictional force to solve the problem and assumed therefore that the only force you need to work out is the reaction from the ground (which doesnt require moments).
This is why I misunderstood what you were asking before, sorry.
| Posted on Saturday, 15 November, 2003 - 06:15 pm: |
That's OK - I just wanted to clear it up.