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An 8.0 m, 240 N uniform ladder rests against a smooth wall. The coefficient of static friction between the ladder and the ground is 0.55, and the ladder makes a 50.0° angle with the ground. How far up the ladder can an 710 N person climb before the ladder begins to slip?

Respuesta :

Answer:

5.7 m

Explanation:

AD = length of the ladder = L = 8 m

AB = distance of the center of mass of the ladder = (0.5) L = (0.5) 8 = 4 m

AC = distance of person on the ladder from the bottom end = x

W = weight of the ladder = 240 N

[tex]F_{g}[/tex] = weight of the person = 710 N

F = force by the wall on the ladder

N = normal force by ground on the ladder = ?

Using equilibrium of force along the vertical direction

N = [tex]F_{g}[/tex] + W

N = 710 + 240

N = 950 N

μ = Coefficient of static friction = 0.55

f =static frictional force on the ladder

Static frictional force is given as

f = μ N

f = (0.55) (950)

f = 522.5 N

Force equation along the horizontal direction is given as

F = f

F = 522.5 N

using equilibrium of torque about point A

F Sin50 (AD) = W Cos50 (AB) + ([tex]F_{g}[/tex] Cos50 (AC))

(522.5) Sin50 (8) = (240) Cos50 (4) + (710) Cos50 (x)

x = 5.7 m

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Lanuel

The maximum distance a 710 N person climb before the ladder begins to slip is 5.66 meters.

Given the following data:

  • Length of ladder = 8.0 meter
  • Force = 240 Newton
  • Coefficient of static friction = 0.55
  • Angle of inclination = 50°
  • Weight of person = 710 Newton

To find how far (distance) up the ladder can a 710 N person climb before the ladder begins to slip:

First of all, we would determine the distance with respect to the center of mass of the ladder.

[tex]D_c = \frac{1}{2} \times length\\\\D_c = \frac{1}{2} \times 8\\\\D_c = 4 \;meters[/tex]

Next, we would solve for the normal force acting on the ladder:

[tex]N = W_p + W_L[/tex]

Where:

  • N is the normal force.
  • [tex]W_P[/tex] is the weight of the person.
  • [tex]W_L[/tex] is the weight of the ladder.

Substituting the given parameters into the formula, we have;

[tex]N = 710 + 240[/tex]

Normal force, N = 950 Newton

Also, we would solve for the force of static friction:

Mathematically, the force of static friction is given by the formula;

Fs = μFn

Where;

  • Fs represents the force of static friction.
  • μ represents the coefficient of friction.
  • Fn represents the normal force.

[tex]F_S = 0.55 \times 950[/tex]

Fs = 522.5 Newton.

Now, we can find how far (distance) up the ladder can a 710 N person climb before the ladder begins to slip:

[tex]F_ssin50(L) = W_LCos50(D_c) + W_Pcos50(D)\\\\522.5sin50(8) = 240Cos50(4) + 710cos50(D)\\\\4180 \times 0.7660 = 960\times 0.6428 + 456.39D\\\\3201.88 = 617.09 + 456.39D\\\\456.39D = 3201.88 - 617.09\\\\456.39D = 2584.79\\\\D = \frac{2584.79}{456.39}[/tex]

Max. distance, D = 5.66 meters.

Read more: https://brainly.com/question/13754413

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