Question

A ladder 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.

Answer 1

Answer:

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

Answer 2

Force excreted by floor on ladder is 1146.6 N (vertically upward) and force excreted by wall on ladder is 407.5 N (horizontally towards the normal force).

What is force?

Force is the effect of pull or push due to which the object having a mass changes its velocity.

The force is of two types-

• Push-When the force applied in the direction of motion of the object, then the force is called the push force.

• Pull- When the force applied in the opposite direction of motion of the object, then the force is called the pull force.

The length of the ladder is 12 meter and the mass of the ladder is 45 kg.Its upper end is a distance h of 9.3 and the mass of fire fighter is 72 kg.

The sine angle for the ladder can be given as,

$\sin\theta=\dfrac{9.3}{12}\\\theta=50.8^o$

The forces acting on the system is normal force and force due to the weight of the two bodies.

The summation of the vertical forces is equal to the zero to keep it at rest. Therefore,

$F_n-45\times9.8-72\times9.8\\F_n=1146.6 \rm N$

By the equilibrium of the momentum for the system,

$F_{N2}\times(9.3)+45(9.8)\times(4\cos50.8)+72(9.8)\times(6\cos50.8)\\F_{N2}=407.5\rm N$

Thus, the force excreted by the floor on ladder is 1146.6 N (vertically upward) and the force excreted by the wall on ladder is 407.5 N (horizontally towards the normal force).

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