A 75 kg man starts climbing a ladder that leans against a wall. If the weight of the ladder is negligible, determine how far up the ladder the man can climb before the ladders starts to slip. The coefficient of friction on both surfaces is μS=0.25

x (max) = 0,25*L*tanα

Explanation:

Letá call  

A: point at the top of the ladder

B: the point at the bottom of the ladder

C: point where the man is up the ladder

L the length of the ladder

α angle between ladder and ground

"x" distance between C and B

Description

The following forces act on the ladder

Point A: horizontal (to the right)  reaction (T) of the wall against the     ladder

Point B : Vertical (upwards) reaction (R)  of ground against the ladder

               frictional horizontal ( to the left ) force (F)

Point C : Weight (vertical downwards)) of the man mg

mg = 75 Kg * 9,8 m/s²       mg = 735 [N]

Then in static conditions:

∑Fx = T - F = 0    ⇒   T = F

∑Fy = mg - R = 0       ⇒   735 - R = 0     ⇒  R = 735

∑Torques(b) = 0

Note: In point B the torque produced by forces R and F are equal to 0

Then:

∑Torques(b) = 0      

And the arm lever for each force is:

mg = 735

d₁ for mg     and d₂  for T

cos α = d₁/x     then    d₁ = x*cosα

sin α  = d₂ / L   then    d₂ = L*sinα

Then:

∑Torques(b) = 0     ⇒   735*x*cosα  - T*L*sinα = 0

735*x*cosα =  T*L*sinα

T = F then       735*x*cosα = F*L*sinα

F = (735)*x*cosα/L*sinα         cos α / sinα = cotgα = 1/tanα

F = (735)*x*cotanα/L     or   F = (735)*x/L*tanα

When F < μ* R  the ladder will stars slippering over the ground

μ(s) = 0,25           0,25*R = 735*x/L*tanα

x   = 0,25*R*tanα*L/735

But R = mg = 735 then

0,25*L*tanα = x

Then  x (max) = 0,25*L*tanα

because growing process had stoped after 4-5 year