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

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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 = 0.571, and the kinetic friction coefficient is mu k = 0.459. The combined mass of the sled and its load is m = 396 kg. The ropes are separated by an angle = 21 degree, and they make an angle theta = 29.7 degree 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? Incorrect.

Explanation / Answer

Let angle phi = A = 21

& angle theta = B = 29.7

Let the tension be T

Let the point of contact of sled and ropes be O

And midpoint of the rod is P

Then taking components of tenions along OP

Net force = 2TCos(A/2)

The other component perpendicular to OP will cancel each other

Now taking vertical of Net force:

2TCos(A/2)Sin(B) + N = mg

Horizontal compponent:

2TCos(A/2)Cos(B) = u(static)*N

From these two equations:

2TCos(A/2) = u*mg/(CosB + u*SinB)

=>T = (0.571*396*9.8)/[2*(Cos29.7 + 0.571*Sin29.7)*(Cos(21/2))]

=> T = 978.55N

Now when the block starts moving for this Tension, friction acting will be kinetic friction

Therefore,

2TCos(A/2)CosB - u(kinetic)*N = ma

N = mg - 2TCos(A/2)Sin(B) = 2927.37 N

=> 1671.5 - 0.459*2927.37 = 396*a

=> a = 0.83 m/s^2

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