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

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The discussion centers on calculating the required force for a ski tow on a slope, specifically addressing the confusion between using cosine and sine in the force equations. One participant argues that the force should be calculated using -3430N * cos(15) due to the direction of the tow's motion and the balance of forces with no acceleration. However, the counterargument emphasizes that the ski tow must overcome the gravitational force acting down the slope, which is represented by the sine function, leading to the correct calculation of 3430N * sin(15). The key takeaway is that the force opposing the tow's motion is related to the sine of the angle, not the cosine, due to the geometry of the slope. Understanding the distinction between these components is crucial for accurate force calculations in this context.
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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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Helicobacter said:
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!
 
HallsofIvy said:
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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