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Tension and acceleration of a sled?

  1. Sep 27, 2015 #1
    1. The problem statement, all variables and given/known data
    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
    2. Relevant equations
    T-mg=ma (I think......)

    3. 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.

    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

  2. jcsd
  3. Sep 27, 2015 #2


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    Staff: Mentor

    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.
  4. Sep 27, 2015 #3
    If I am going to divide an angle wouldn't it be φ = 25°?
  5. Sep 27, 2015 #4


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    Staff: Mentor

    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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