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 mu_s = 0.595, and the kinetic friction coefficient is mu_k = 0.427. The combined mass of the sled and its load is m = 276 kg. The ropes are separated by an angle = 26degree, and they make an angle theta = 30.8degree with the horizontal. Assuming both ropes pull equally hard, what is the minimum rope tension required to get the sled moving? If this rope tension is maintained after the sled starts moving, what is the sled's acceleration?

Explanation / Answer

a) The normal force = Fn = mg - 2Fsin = 276 kg * 9.8m/s² - 2 * F * sin30.8º
Fn = 2704.8 N - 1.024*F
static friction Ff = µ*Fn = 0.595 * ( 2704.8 N - 1.024*F) = 1609.4 N - 0.61F

The horizontal component in the direction of motion of the two ropes must overcome the friction:
2 * F * cos * cos(/2) = 1609.4 N - 0.61F
1.67F = 1609.4 N - 0.61F
2.28F = 1609.4 N
F = 705.88 N to get the sled moving

b) The friction was Ff = 1609.4 N - 0.61*705.88 = 1178.8 N,
so that is the applied horizontal force.

The friction NOW is Ff = 0.427 * (2704.8 N - 1.024*705.88 N) = 846.3 N
so the net force = Fnet = 1178.8 N - 846.3 N = 332.5 N

acceleration a = Fnet / mass = 332.5N/ 276kg = 1.205 m/s²

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