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 τ = 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
