Tension and acceleration of a sled?

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TheNeezoMan
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Homework Statement


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 = 0.579, and the kinetic friction coefficient is μk = 0.435. The combined mass of the sled and its load is m = 336 kg. The ropes are separated by an angle φ = 25°, and they make an angle θ = 30.1° with the horizontal. Assuming both ropes pull equally hard, what is the minimum rope tension required to get the sled moving?

Sled Diagram.jpeg

Homework Equations


T-mg=ma (I think...)

The Attempt at a Solution



So, say person 1 exerts tension T1 and person 2 T2, so T1=T2=T ...So they both exert a force of 2T.

Untitled.png

I am not even sure if this diagram is right?
I also think wemust must find Ff,s = μs * Fg

I don't know where to go from here or I if this is where I should even be

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I think you'll want to start by seeing what single force applied at angle θ = 30.1° will get the sled moving. After that you can deal with how to divide that force between two ropes with the given angular spread.
 
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If I am going to divide an angle wouldn't it be φ = 25°?
 
TheNeezoMan said:
If I am going to divide an angle wouldn't it be φ = 25°?
Dealing with φ will come later. Treat this as two separate problems: 1) Find the force required to move the sled if the force is applied to the sled at angle θ; 2) Take the force from (1) and split it between two ropes separated by angle φ.
 
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