tofuedits1
contestada

PART ONE
A ladder rests against a vertical wall. There
is no friction between the wall and the ladder.
The coefficient of static friction between the
ladder and the ground is µ = 0.464 .
(USE THE PICTURE TO DETERMINE THE ANSWER)
Identify the set of equations which is correct.
ANSWER CHOICES:
1. A1, B2, C3
2. A2, B2, C1
3. A1, B1, C1
4. A1, B2, C2
5. A1, B1, C2
6. A2, B1, C3
7. A2, B1, C2
8. A1, B2, C1
9. A1, B1, C3
10. A2, B1, C1

PART TWO
Determine the smallest angle θ for which the
ladder remains stationary.
Answer in units of ◦

PART ONE A ladder rests against a vertical wall There is no friction between the wall and the ladder The coefficient of static friction between the ladder and t class=

Respuesta :

MathPhys

Answer:

1. A1, B2, C3

2. 47.1°

Explanation:

Sum of forces in the x direction:

∑Fₓ = ma

f − Fᵥᵥ = 0

f = Fᵥᵥ

Sum of forces in the y direction:

∑Fᵧ = ma

N − W = 0

N = W

Sum of moments about the base of the ladder:

∑τ = Iα

Fᵥᵥ h − W (b/2) = 0

Fᵥᵥ h = ½ W b

Fᵥᵥ (l sin θ) = ½ W (l cos θ)

l Fᵥᵥ sin θ = ½ l W cos θ

The correct set of equations is A1, B2, C3.

At the smallest angle θ, f = Nμ.  Substituting into the first equation, we get:

Nμ = Fᵥᵥ

Substituting the second equation into this equation, we get:

Wμ = Fᵥᵥ

Substituting this into the third equation, we get:

l (Wμ) sin θ = ½ l W cos θ

μ sin θ = ½ cos θ

tan θ = 1 / (2μ)

θ = atan(1 / (2μ))

θ = atan(1 / (2 × 0.464))

θ ≈ 47.1°

Equation of equilibrium is the static or dynamic equilibrium of all internal and external forces present in the system.

  • The set A1, B2, C3 is correct.
  • The smallest angle θ for which the ladder remains stationary is [tex]47.1^o[/tex]

Given information-

A ladder rests against a vertical wall.

The coefficient of static friction between the ladder and the ground is 0.464.

What is equation of equilibrium?

Equation of equilibrium is the static or dynamic equilibrium of all internal and external forces present in the system.

  • A) Set of equations which is correct.

From the equation of equilibrium of two bodies, the net force in x axis can be given as,

[tex]\sum F_x=0[/tex]

As the normal force and friction force acting on the x-axis. Thus,

[tex]-f_w+f=0[/tex]

[tex]f=f_w[/tex]

Thus option A1 is correct.

As the gravitational force(due to weight of the body) is acting in the vertical direction. Thus from the equation of equilibrium of two bodies, the net force in y axis can be given as,

[tex]\sum F_y=W[/tex]

As the normal force acting on the x-axis. Thus,

[tex]N=W[/tex]

Thus option B2 is correct.

Apply torque equation at the base of the ladder,

[tex]\sum \tau=Ia[/tex]

[tex]F_wh=\dfrac{1}{2} Wb[/tex]

Here the value of h and b can be changed in the form of I using the trigonometry formula. Thus,

[tex]F_w(I\sin \theta)=\dfrac{1}{2} W(\cos \theta[/tex]

Thus option C3 is correct.

Hence the set A1, B2, C3 is correct.

  • B) The smallest angle θ for which the ladder remains stationary-

The equation of option C3 is,

[tex]F_w(I\sin \theta)=\dfrac{1}{2} W(\cos \theta[/tex],

As normal force is can be given as,

[tex]F_w=W\mu[/tex]

Thus,

[tex]W\mu(I\sin \theta)=\dfrac{1}{2} W(\cos \theta[/tex]

[tex]\dfrac{\sin \theta}{\cos \theta} =\dfrac{1}{2\mu} \\\tan \theta =\dfrac{1}{2\times 0.464} \\\theta =\tan^-(1.0770)\\\theta=47.1^o[/tex]

Hence the smallest angle θ for which the ladder remains stationary is [tex]47.1^o[/tex]

Hence,

  • The set A1, B2, C3 is correct.
  • The smallest angle θ for which the ladder remains stationary is [tex]47.1^o[/tex]

Learn more about the equation of equilibrium here;

https://brainly.com/question/1967702

Otras preguntas