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Determining the force for culculating a ski tow's perf. on a slope

  • #1
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Logically the req. force should be -3430N * cos(15) because it is the direction of the tows motion and it is the force req. to balance (Fy-net = 0, since a=0) - but it is 3430N * sin(15). Why?
 

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  • #2
HallsofIvy
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Logically the req. force should be -3430N * cos(15) because it is the direction of the tows motion and it is the force req. to balance (Fy-net = 0, since a=0) - but it is 3430N * sin(15). Why?
Why would you think that "Logically the req. force should be -3430N * cos(15)"? The ski tow has to overcome the force back down the slope of the hill and that is parallel to the "opposite side" to the 15 degree angle in the right triangle you have. "opposite side/hypotenus" is sine, not cosine!
 
  • #3
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Why would you think that "Logically the req. force should be -3430N * cos(15)"?
Thanks for your feedback HallsofIvy!
My thought is that the net force in the direction of inclination is zero because the tow has constant velocity and no acceleration (Newton's 1st law). The component of passenger-weight in this direction is 3430N*cos(15). In order to balance this out to 0N you need to have an opposite force of it, which is -3430N*cos(15). Why am I wrong? Why is the req. force "the force back down the slope of the hill" and not the other (my suggested) component vector of the weight force?
 
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