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

http://img283.imageshack.us/img283/7131/ladderquestion2tz8.th.png [Broken]

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<1

w = 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)^0.5

Next 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 = 0.8R

Cancelling 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): :surprised

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 [Broken]

Which i cancelled 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 [Broken]

(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?

http://img283.imageshack.us/img283/7131/ladderquestion2tz8.th.png [Broken]

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<1

w = 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)^0.5

Next 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 = 0.8R

Cancelling 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): :surprised

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 [Broken]

Which i cancelled 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 [Broken]

(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?

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