a 20 kg ladder leaning against a frictionless wall and resting on a frictionless horizontal surface...?

Question

To keep the ladder from slipping, the bottom of the ladder is tied to the wall with a thin wire. The tension in the wire is 29.4 N. The wire will break if the tension exceeds 200 N.

How far up the ladder can a 86 kg person climb? (Measure along the length of the ladder.)

http://www.webassign.net/tipler/12_63.gif

Answer 1

The ladder
20*g*L*1.5/(2*L)
15*g
The wire
T*L*5/L
T*5
The reactive force of the floor (vertical equal to weight)
-20*g*L*1.5/L
-30*g
Sum these together and they equal zero
g*(15-30)+T*5=0
T=3*g
using 9.8 for g
T=29.4

Now, using these equations, introduce the 86kg person

this adds a torque of
86*L*1.5/(x*L)
where x is the proportion of L that the man is from the top
at the bottom x=1
Halfway up, x=2
3/4 of the way up x=4
and so on.

the torque is

86*g*1.5/x
129*g/x
The torque at the bottom is now
-106*g*L*1.5/L
-159*g

the wire is still T*5

solve for x when T reaches 200

T*5+g*(15-159)+g*129/x=0

200*5+9.8*(15-159)+9.8*129/x=0
411.2=1264.2/x

x=1264.2/411.2
x=1264.2/411.2
x=3.07

That means that the person can climb to
L/3.07 from the top
or L*(1-1/3.07) from the bottom

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