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 μs = 0.619, and the kinetic friction coefficient is μk = 0.459. The combined mass of the sled and its load is m = 366 kg. The ropes are separated by an angle φ = 22°, and they make an angle θ = 31.8° 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?

Answer 1

along the vertical direction , force equation is given as

F_n + 2T Cos(φ/2) Sinθ = mg

F_n = mg - 2T Cos(φ/2) Sinθ                                                  eq-1

static frictional force is given as

f_s = u_s F_n = u_s (mg - 2T Cos(φ/2) Sinθ )

along the horizontal direction , force equation is given as

2T Cos(φ/2) Cosθ = f_s

using eq-1

2T Cos(φ/2) Cosθ = u_s (mg - 2T Cos(φ/2) Sinθ )

2T Cos(22/2) Cos31.8 = (0.619) ((366 x 9.8) - 2T Cos(22/2) Sin31.8 )

T = 961.6 N