A heavy sled is being pulled by two people as shown in the figure. The coefficie

Question

A heavy sled is being pulled by two people as shown in the figure. The coefficient of static friction between the sled and the ground is s = 0.603, and the kinetic friction coefficient is k = 0.459. The combined mass of the sled and its load is m = 261 kg. The ropes are separated by an angle = 23°, and they make an angle = 31.1° with the horizontal. Assuming both ropes pull equally hard, what is the minimum rope tension required to get the sled moving?

              1.? N

               2. ? M/S^2

Explanation / Answer

The normal force
Fn = mg - 2Fsin = 261kg * 9.8m/s² - 2 * F * sin31.1º
Fn = 2557.8N - 1.033*F

static friction Ff = µ*Fn = 0.603 * (2557.8N - 1.033F) = 1542.35N - 0.3678F

kinetic friction Ff = µ*Fn = 0.459 * (2557.8N - 1.033F) = 1174.03N - 0.47F

The horizontal component in the direction of motion of the two ropes must overcome the friction:
2 * F * cos * cos(/2) = 1174.03N - 0.47F
1.033F = 1174.03N - 0.47F
F = 1174.03N / 1.503 = 781.12 N

The friction WAS Ff = 1542.35N - 0.3678*781.12 = 1255.05 N,
so that IS the applied horizontal force.
The friction NOW IS Ff = (1174.03N - 0.47*781.12) = 806.90 N
so the net force
Fnet = 1255.05N - 806.90N = 448.15 N

acceleration a = Fnet / mass = 448.15N / 261kg = 1.71 m/s²

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