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 .-0.611, and the kinetic friction coefficient is 0.395. The combined mass of the sled and its load is m = 381 kg. The ropes are separated by an angle -28, and they make an angle = 31.4 with the horizontal. Assuming both ropes pull equally hard, what is the minimum rope tension required to get the sled moving? Number 1641.84 N If this rope tension is maintained after the sled starts moving, what is the sled's acceleration? Number 4.99 m/s Incorrect. rect.

Explanation / Answer

Let the Tension in the rope be T.
Net Tension due to Pull at an angle by the two men, Tnet = 2*T*cos(/2)

Force with which it is being pulled, F = Tnet * cos()

Normal Force, = m*g - Tnet * sin()

For Minimum Rope tension to get the sled moving,
Tnet * cos() = us * (m*g - Tnet * sin())
2*T*cos(/2) * cos() = us * (m*g - 2*T*cos(/2) * sin())

Substituting values in the above eq,
2*T*cos(14) * cos(31.4) = 0.611 * (381*9.8 - 2*T*cos(14) * sin(31.4))
T = 1003.2 N

Acceleration,
2*T*cos(/2) * cos() - uk * (m*g - 2*T*cos(/2) * sin()) = m*a
2*1003.2*cos(14) * cos(31.4) - 0.395 * (381*9.8 - 2*1003.2*cos(14) * sin(31.4)) = 381*a
a = 1.54 m/s^2

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