Power required to operate a ski tow

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Homework Help Overview

The problem involves calculating the power required to operate a ski tow on a slope, given specific parameters such as the slope angle, length, speed of the rope, and the number of riders with their average mass. The original poster attempts to derive the necessary power using force equations and work-energy principles.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the forces acting on the ski tow, including the components of force in both the x and y directions. The original poster questions the necessity of including the perpendicular force in their calculations. Others suggest focusing on the component of force that does work along the slope.

Discussion Status

Participants are actively engaging with the problem, providing feedback on the calculations and clarifying the role of different forces. There is a recognition of the need to refine the approach to focus on the relevant forces affecting the power calculation.

Contextual Notes

There is a mention of the gravitational constant (g) being overlooked in one of the calculations, indicating a potential area for further exploration. The discussion reflects a collaborative effort to address assumptions and clarify the setup of the problem.

ph123
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"A ski tow operates on a slope of angle 14.3 of length 350m. The rope moves at a speed of 12.4km/hr and provides power for 49 riders at one time, with an average mass per ride of 73.0kg. Estimate the power required to operate the tow."

The eqn I used to sum the forces in the x-direction is:

EFx= Fcos(theta) - mgsin(theta) = 0
F = (mgsin(theta))/cos(theta) = 8935.3N

12.4km/h = 3.44m/s

d/v=t
350m/3.44m/s = t = 101.74s

W = F*s = (8935.3N)(350m)
W = 3127356.9 J

Power = W/dt = 3127356.9J/101.74s
P=3.07*10^4 W

However, this is not right. I feel like I need to include the force in the y-direction in my work equation, but then again perpendicular forces do now work. Does anyone have any idea what I may be overlooking here?
 
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As I think you suspect, the component of the force perpendicular to the slope does no work. Get rid of it.
 
Leaving, then,

F=mgsin(theta)
F=(49)(73.0kg)sin(14.3) = 8658.5N
 
Yes (though you forgot g in the second line).
 

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