A young parent is dragging a 65 kg (640 N) sled (this includes the mass of two kids) across some snow on flat ground, by means of a rope attached to the sled. The rope is at an angle of 30 degrees with respect to the ground and the tension in the rope is 160 N. The sled is moving at a constant velocity of 1.5 m/s. (a) Draw and label all forces acting on the kids + sled system. Indicate the relative size of each force by scaling the length of each force arrow appropriately
(b) Calculate the normal force acting on the system
(c) Calculate the force of friction acting on the system.
(d) Calculate the coefficient of friction between the sled and the snow.

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

graph question answered .. you think better when you are relaxed and happy