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

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SUMMARY

The discussion centers on calculating the required force for a ski tow operating on a slope with a 15-degree incline. The correct force to overcome is determined to be 3430N * sin(15), which accounts for the gravitational force acting parallel to the slope. This contrasts with the initial assumption that the force should be calculated using the cosine function, which incorrectly considers the vertical component of the weight. The clarification emphasizes the importance of understanding the relationship between the components of forces in inclined planes.

PREREQUISITES
  • Understanding of Newton's laws of motion, particularly the first law.
  • Basic knowledge of trigonometric functions, specifically sine and cosine.
  • Familiarity with vector components in physics.
  • Concept of forces acting on inclined planes.
NEXT STEPS
  • Study the application of Newton's laws in real-world scenarios.
  • Learn about vector decomposition in physics, focusing on inclined planes.
  • Explore the role of trigonometric functions in force calculations.
  • Investigate the dynamics of ski tows and similar mechanical systems.
USEFUL FOR

This discussion is beneficial for physics students, engineers, and anyone involved in mechanical design or analysis of systems involving inclined planes and forces.

Helicobacter
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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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