Question

A 22.5 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The bottom of the ladder rests on a floor with a rough surface where the coefficient of static friction is 0.11. The angle between the horizontal and the ladder is θ . The person wants to climb up the ladder a distance of 0.65 m along the ladder from the ladder’s foot.

What is the minimum angle θmin (between the horizontal and the ladder) so that the person can reach a distance of 0.65 m without having the ladder slip?
The acceleration of gravity is 9.8 m/s².
Answer in units of °.

Answer 1

62°

Step-by-step explanation:

Variables:

x=0.65m (distance up ladder)

L=2.5m (ladder length)

m = 22.5kg (mass)

g = 9.8 (gravity)

μ = 0.11 (static friction)

N = Normal Force (Weight in this context)

Formulas:

Weight(W)=mg

Friction Force(F)=μN

Torque(T)=
`Frsin`θ

Substitute Force and length in each torque we're calculating

`T_(counterclockwise) =` (μN)*L*sinθ

`T_(clockwise) =` mg*x*cosθ

They need to be equal:

`T_(counterclockwise) = T_(clockwise)`

(μN)*L*sinθ = mg*x*cosθ

`(sin)/(cos)=tan`

tanθ= (mg*x) / (μ*N*L)

Since W=mg=N, they cancel out.

tanθ= (x) / (μ*L)

Inverse tangent

θ=
`tan^(-1)`(x / (μ*L))

Plug in variable:

θ=
`tan^(-1)`(0.65 / (0.11*2.5))

θ=
`tan^(-1)`(2.3636)

θ= 62.0676° ≈ 62°

Hope this helps ;)