• Sink41
In summary, the shortest possible length of the ladder that reaches a height of 2 meters on a quarter circle with radius 1 and an extra length of 1 on top, with a weight of 40 and a coefficient of friction of 0.8 between the ground and the ladder is 1.81 meters. This is calculated by considering torque and net force equations, and using the maximum force of friction as 40 times the coefficient of friction.
Sink41
Basically wondering what shortest possible length of this ladder is, one that reaches 2 meters high. The ground its on is quarter of a circle with radius 1 with an extra length of 1 on top. Weight is 40 (so 40*gravity for force) and µ (coefficient of friction) between ground and ladder is 0.8

I sort of worked out an answer, but its two complex for me to work out so i thought i'd post here

i started off by modeling the curve bit as y = -(1 - (x^2))^0.5 ; 0<x<1w = 1 - x

h = 1 - y (y is negative)The length can therefore be written as

L = (h^2 + w^2)^0.5

L = ((1-y)^2 + (1-x)^2)^0.5Next i worked out angles:

tan(theta1) = dy/dx

dy/dx = x/(1 - (x^2))^0.5

tan(theta1) = x/(1 - (x^2))^0.5
tan(theta2) = (1 - y)/(1 - x)

Forces must be the same in all directions:

S + Rsin(theta1) = Fcos(theta1) (x axis)

40g = Rcos(theta1) + Fsin(theta1) (y axis)

g = 9.8 (gravity)Magnitude of rotation around base of ladder must be zero:

(1/2)(40g)(L)cos(theta2) = (L)(S)sin(theta2)Friction = µR

F = 0.8RCancelling down i got two equations with L and x... unfortunatly one is so complicated that i have no idea how to put them together. I'm also worried that i made a mistake somewhere, since it got so complicated, is there possibly an easier/less complicated way of working it out than i did?

First equation, pretty easy to see where i got this (replaced y with x):

L = ((1 + (1 - (x^2))^0.5)^2 + (1 - x)^2)^0.5

Second equation (possibly incorrect):

0 = ((196(1-x))/(1+(1-(x^2))^0.5))/(0.8-(x/(1-(x^2))^0.5)) - ((392)/(1+((0.8x)/((1-x)^2))))

or written in paint:

http://img20.imageshack.us/img20/8880/equationcy1.th.png

Which i canceled down to (possibly even more incorrect)

0.8x^2 + 2.8x -1.6 + x(1 - x^2)^0.5 - 0.6(1 - x^2)^0.5 = 0

So i have no idea how to put those together, or work out when x is zero. Putting them into a graphing program i got:

http://img474.imageshack.us/img474/117/crazygraphhg7.th.png

(blue is first red is second)

this means there would be two possible places at which the ladder would be at limiting equilibrium... unsure whether this should happen.So can anyone help?

Last edited by a moderator:
If you calculate torque about the base of the ladder, and the net force in the horizontal axis, you should get two equations, which, with the information you already have, can be used to answer this question.

NateTG said:
If you calculate torque about the base of the ladder, and the net force in the horizontal axis, you should get two equations, which, with the information you already have, can be used to answer this question.

I already have both of those?

torque:

(1/2)(40g)(L)cos(theta2) = (L)(S)sin(theta2)horizontal:

S + Rsin(theta1) = Fcos(theta1) (x axis)I just tried cancelling everything down again and got the two equations again:

http://img20.imageshack.us/img20/8880/equationcy1.th.png

andL = ((1 + (1 - (x^2))^0.5)^2 + (1 - x)^2)^0.5I don't know how to put them together... or how to solve the first one.

Last edited by a moderator:
The maximum force of friction on the bottom of the ladder is:
$$40 \mu g$$

The only other horizontal fore acting on the ladder is the normal force on the wall. So, if I read your notation you should have:
$$F_h=40 \mu g+S=0$$
so
$$S=40 \mu g$$

NateTG said:
The maximum force of friction on the bottom of the ladder is:
$$40 \mu g$$

The only other horizontal fore acting on the ladder is the normal force on the wall. So, if I read your notation you should have:
$$F_h=40 \mu g+S=0$$
so
$$S=40 \mu g$$

Doesnt the normal force come out at a 90degre angle to whatever an object is resting on? Then R would change as you moved the ladder up the slope. I am confused whether you ment maximum force of friction as in when the ladder is at the bottom of the slope or whether you are saying that R is always 40g.

...
The normal force exerted by the floor on the ladder, and gravity are the only two forces acting on the lader in the vertical direction. Since this is a statics problem, they must cancel - so they're of equal magnitude in opposite directions.

NateTG said:
...
The normal force exerted by the floor on the ladder, and gravity are the only two forces acting on the lader in the vertical direction. Since this is a statics problem, they must cancel - so they're of equal magnitude in opposite directions.

Ok, got an answer of 1.81 thanks for the help

## 1. What is the concept of "Ladder and circle question"?

The "Ladder and circle question" is a metaphorical concept used in problem-solving and decision-making. It involves visualizing a ladder as a linear progression towards a specific goal or solution, and a circle as a cyclical process of reevaluating and adjusting that goal or solution.

## 2. How can the "Ladder and circle question" be applied in scientific research?

In scientific research, the "Ladder and circle question" can be used to approach a problem or hypothesis in a systematic way. Researchers can start by defining their goal or question (ladder) and then continually evaluate and adjust their methods and findings (circle) until they reach a satisfactory solution or conclusion.

## 3. What are the benefits of using the "Ladder and circle question" in problem-solving?

Using the "Ladder and circle question" can help individuals break down complex problems into manageable steps and continuously reassess their approach. It can also prevent individuals from getting stuck on one solution and encourage them to consider alternative perspectives and solutions.

## 4. Can the "Ladder and circle question" be used in other areas besides scientific research?

Yes, the "Ladder and circle question" can be applied in various fields, including business, education, and personal development. It is a useful tool for any situation that requires critical thinking and problem-solving.

## 5. Are there any limitations to using the "Ladder and circle question"?

While the "Ladder and circle question" can be a helpful approach to problem-solving, it may not be suitable for every situation. Some problems may not have a clear goal or may require a more creative approach. Additionally, it is essential to balance the linear thinking of the ladder with the cyclical thinking of the circle to avoid getting stuck in a loop.

• Calculus and Beyond Homework Help
Replies
4
Views
378
• Calculus and Beyond Homework Help
Replies
2
Views
653
• Calculus and Beyond Homework Help
Replies
8
Views
904
• Calculus and Beyond Homework Help
Replies
5
Views
818
• Calculus and Beyond Homework Help
Replies
5
Views
914
• Calculus and Beyond Homework Help
Replies
23
Views
1K
• Calculus and Beyond Homework Help
Replies
12
Views
1K
• Mechanics
Replies
2
Views
674
• Calculus and Beyond Homework Help
Replies
8
Views
835
• Calculus and Beyond Homework Help
Replies
9
Views
2K