Answer:
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
