Tension and acceleration of a sled?

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Homework Help Overview

The problem involves determining the minimum rope tension required to move a heavy sled being pulled by two people, considering the coefficients of static and kinetic friction, the mass of the sled, and the angles at which the ropes are positioned.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the forces acting on the sled and the need to find the force required to initiate movement. There is uncertainty about the diagram's accuracy and the correct angles to consider for the forces involved.

Discussion Status

Some participants suggest starting with a single force applied at one angle before addressing the division of that force between the two ropes. Others emphasize the need to clarify the angles involved and the approach to solving the problem in stages.

Contextual Notes

Participants note the potential confusion regarding the angles φ and θ, and there is an acknowledgment that the problem may need to be approached in parts to avoid complications.

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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