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 = 0.587, and the kinetic friction coefficient is k = 0.419. The combined mass of the sled and its load is m = 366 kg. The ropes are separated by an angle = 24°, and they make an angle = 30.4° 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

Static Friction Between the sled and Ground us = 0.587
Mass of Sled = 366 Kg
Let the Tension in each string be T

Minimum force needed to move sled = us * (m*g - 2 * T* cos(/2) * sin())
Minimum force needed to move sled = 0.587 *( 366 * 9.8 - 2*T*cos(24/2) * sin(30.4))
Minimum force needed to move sled = 2105 - 0.587 T N

Total Tension which would move the sled, 2 * T* cos(/2) * cos()
2 * T* cos(24/2) * cos(30.4) = 2105 - 0.587 T
1.68 T =  2105 - 0.587 T
T = 928.54 N

Minimum Rope Tension required = 2*T = 2 * 928.54
Minimum Rope Tension required = 1857 N

(b)
Kinetic Friction uk = 0.419
Frictional Force = uk * (m*g - 2 * T* cos(/2) * sin())= 0.419 *( 366 * 9.8 - 1857) N
Frictional Force = 725 N

Acceleration = (Force - Frictional Force)/ mass
Acceleration = (1857 - 725)/ 366 m/s^2
Sled's Acceleration = 3.1 m/s^2

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