Aladder whose length l is 12 m and mass m is 45 kg rests against a wall. its upper end is a distance h of 9.3 m above the ground. the center of mass of the ladder is one-third of the way up the ladder. a fire fighter whose mass m is 72 kg climbs the ladder until his center of mass is halfway up. assume the wall but not the ground is frictionless. what forces are exerted on the ladder by the wall and by the ground? provide both the magnitude and direction of the ground force.

FN₁ = 1146.6N : Force exerted on the ladder by the floor , vertical and upward

FN₂ = 407.5 N : Force exerted on the ladder by the wall , horizontal and opposite to friction force between the floor and the ladder

Explanation:

The equilibrium equation are:

∑Fx=0

∑Fy=0

∑M = 0  

M = F*d  

Where:  

∑M : Algebraic sum of moments

M : moment  ( N*m)  

F : Force ( N)  

d :Perpendicular distance of the force to the point  ( meters )

Data

m =45 kg  : mass of the ladder

M =72 kg : mass of the fire fighter

g = 9.8 m/s²: acceleration due to gravity

L = 12 m : ladder length

h =  9.3 m: ladder height

L/3 = 12/3 = 4m Location of the center of mass of the ladder of the way up

L/2 = 12/2 = 6m Location of the center of mass of the  fire fighter

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

θ  : angle that makes  the  ladder  with the floor

sinθ = h/L = 9.3 m/12 m

θ =sin⁻¹( 9.3 / 12)

θ = 50.8°

Forces acting on the ladder

W₁ =m*g = 45 kg* 9.8 m/s² = 441 N: Weight of the ladder (vertical downward)

W₂ =M*g = 72 kg * 9.8 m/s² = 705.6 N : Weight of the fire fighter(vertical downward)  

FN₁ :Normal force that the floor exerts on the ladder (vertical upward)  (point A)

fs : friction force that the floor exerts on the ladder (horizontal and opposite the movement )(point A)  

FN₂ :  Normal Force that the wall exerts on the ladder ( horizontal and opposite to friction force between the floor and the ladder)

∑Fy=0

FN₁ -W₁ -W₂= 0

FN₁ = W₁ + W₂

FN₁ = 441N+ 705.6N

FN₁ = 1146.6N : Force exerted on the ladder by the wall (vertical and upward)

Calculation of the distances of the forces at the point A (contact point of the ladder on the floor)

d₁ = 4*cos 50.8° (m) = 2.53 m:  Distance from W₁ to the point A

d₂ =6*cos 50.8° (m)= 3.79 m  : Distance from W₂ to the point A

d₃ = 9.3 m : Distance from FN₂ to the point A

The equilibrium equation of the moments at the point A  (contact point of the ladder with the floor)  

∑MA = 0  

FN₂(d₃) - W₁( d₁) - W₂(d₂) = 0

FN₂(d₃) = W₁(d₁) + W₂(d₂)

FN₂(9.3) = (441 )(2.53) + (705.6)( 3.79 )

FN₂(9.3) = 1115.73 + 2674.2

FN₂ = (3790) / (9.3)  

FN₂ = 407.5 N : Force exerted on the ladder by the wall , horizontal and opposite to friction force between the floor and the ladder

spare some change and give it to the poor

explanation:

plz : ((