Respuesta :
Answer:
A = 10.83 in² ,  w = 3.29 in
Explanation:
For this exercise we must use the rotational equilibrium condition
    Σ τ = 0
in this case they give us the external torque Ï„ = 3800 lb in.
    τ - τ’= 0
    τ = τ'
where τ‘ is the torque exerted by the brake shoe that is given by the friction force
   τ’ = fr r sin θ
indicates that θ= 120º and the radius is half the diameter of the drum
   r = 6 in
the friction force is given by the expression
    fr = μ N
substitute
    τ = μ N r sin θ     (1)
to find the normal let's use Newton's second law on the perpendicular pressure axis
    Σ F = 0
    N - f = 0
    N = f
The applied force can be found using the definition of pressure
     P = f / A
where A is the area of ​​the footing
     f = P A
let's substitute
     N = P A
let's substitute in 1
     τ = μ P A r sin θ
     [tex]A= \frac{\tau }{\mu \ P \ r \ sin\ \theta }[/tex]
let's calculate
     A = 3800 / (0.45 150 6 sin 120)
     A = 10.83 in²
to find the size of the shoe we must assume a specific shape, suppose the shoe is square
     A =w²
     w = √A
     w = 3.29 in
