Apainter climbs a ladder leaning against a smooth frictionless wall, which makes an angle to the floor of 53.0 ◦ . at some height the ladder will be on the verge of sliding against the floor. the coefficient of friction between the ladder and the floor is µ = 0.450. the ladder is 4.00 m long and has a mass of 30.0 kg.

a) find the maximum distance the 80kg painter can climb the ladder just before it slips.

d = 2.5 m

Explanation:

Data

m₁= 30 kg : mass of the lader

m₂ = 80kg  : mass of the painter

g= 9.8 m/s²: acceleration due to gravity

L = 4.00 m ladder length

α = 53° : angle that makes  the  ladder  with the floor

µ = 0 : coefficient of friction between the ladder and the wall

µ = 0.450 :coefficient of friction between the ladder and the floor

Forces acting on the ladder

W₁ =30 kg*9.8 m/s²= 294 N : Weight of the ladder (vertical downward)  

W₂ = 80kg*9.8 m/s²= 784 N : Weight of the painter (vertical downward)

FN :Normal force that the floor exerts on the ladder (vertical upward)  (point A)

f : friction force that the floor exerts on the ladder (horizontal to the left) (point A)

N : Forces that the wall exerts on the ladder (horizontal to the right)

Equilibrium  of the forces in Y

∑Fy=0

FN-W₁-W₂ = 0

FN= W₁+W₂

FN=294 N+784 N

FN = 1078 N

Calculation of the friction force between the ladder and the floor

f = µ*FN = 0.450 *1078 N =485.1 N

Equilibrium  of the forces in X

∑Fx=0

N -f = 0

N = f = 485.1 N

The equilibrium equation of the moments at the point contact point of the ladder with the floor:

∑MA = 0  

MA = F*d  

Where:  

∑MA : Algebraic sum of moments in the the point (A) (contact point of the ladder with the wall)  

MA : moment in the point A ( N*m)  

F : Force ( N)  

d :Perpendicular distance of the force to the point A ( m )

Calculation of the distances of the forces at the point A

d₁ = (L/2 )*cosα  =  (4/2 )*cos53° = 1.2 m: Distance from W₁ to the point A

d₂ = d*cosα= d*cos53° :  Distance from W₂ to the point A

d₃ = L *sinα =  (4)*sin53° =3.19 m : Distance from N to the point A

Equilibrium of the moments at the point A

∑MA = 0  

N(d₃)-W₁( d₁)- W₂(d₂) = 0

W₂(d₂) = N( d₃)-W₁( d₁)

784(d*cos53°) = 485.1( 3.19) -294( 1.2)

(471.8 ) d = 1547.469 - 352.8

d = (1194.7) / (471.8 )

d = 2.5 m

i think the answer is a and b

i wish you luck