
Answer:
b) N = 560 N, c)  fr = 138.56 N, d)  μ = 0.247
Explanation:
a) In the attachment we can see the free body diagram of the system
b) Let's write Newton's second law on the y-axis
       N + T_y -W = 0
       N = W -T_y
let's use trigonometry for tension
       sin θ = T_y / T
       cos θ = Tₓ / T
       T_y = T sin θ
       Tₓ = T cos θ
we substitute
       N = W - T sin 30
we calculate
       N = 640 - 160 sin 30
       N = 560 N
c) as the system goes at constant speed the acceleration is zero
X axis
       Tₓ - fr = 0
        Tₓ = fr
we substitute and calculate
       fr = 160 cos 30
       fr = 138.56 N
d) the friction force has the formula
       fr = μ N
       μ = fr / N
we calculate
       μ = 138.56 / 560
       μ = 0.247