Upgrade Your Space with a High-Quality Leaning Ladder for Easy Concurrent Use

[IamVector]

Homework Statement

A ladder leans against a friction less wall. If the
coefficient of friction with the ground is μ, what is the smallest angle the ladder can
make with the ground and not slip?

Relevant Equations

answer came out quite well by balancing torque about N1 and and equating N2 to friction force
N2 sin θ = mg(l/2) cos θ =⇒ N2 = mg/(2 tan θ) - (1)

This is also the value of the friction force F. The condition F ≤ μN1 = μmg
therefore becomes

mg/(2 tan θ) ≤ μmg =⇒ tan θ ≥1/2μ - (2)

ans : tan θ ≥1/2μ , but after that there is a second approach which say : Note that the total force exerted on the ladder by the floor points up at an angle

given by tan β = N1/F = (mg)/(mg/2 tan θ ) = 2 tan θ.

this force does not point along the ladder
This is the direction that causes the lines of the three forces on the ladder to be concurrent.

This theorem provides a quick way to solve the ladder problem in the more general
case where the center of mass is a fraction f of the way up. In this case, the concurrency
theorem tells us that the slope of the total force from the floor is (1/f )tan θ

how we this (1/f) tanθ and can anyone explain me the second approach ??

for concurrent

 

[kuruman]

For a system in equilibrium, if the sum of torques is zero about one point, it is zero about any other point. Note that the sum of torques about your concurrent point is zero. Actually, all three torques are zero about that point.