Power required to operate a ski tow

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SUMMARY

The power required to operate a ski tow on a slope with a 14.3-degree angle and a length of 350 meters is calculated to be approximately 30,700 Watts. The calculation involves determining the force exerted by the mass of the riders, which totals 49 riders at an average mass of 73.0 kg each, resulting in a force of 8,658.5 N due to gravity's component along the slope. The work done is computed using the formula W = F * s, leading to a total work of 3,127,356.9 J over a time of 101.74 seconds. The final power output is derived from the work done divided by the time taken.

PREREQUISITES
  • Understanding of basic physics concepts, particularly forces and work-energy principles.
  • Familiarity with trigonometric functions, specifically sine and cosine.
  • Knowledge of unit conversions, particularly between kilometers per hour and meters per second.
  • Ability to perform calculations involving mass, gravity, and force.
NEXT STEPS
  • Study the principles of work and energy in physics to deepen understanding of power calculations.
  • Learn about the effects of friction and other forces on mechanical systems, particularly in ski tow operations.
  • Explore advanced topics in dynamics, such as the impact of varying slope angles on power requirements.
  • Investigate the design considerations for ski tows, including safety factors and efficiency improvements.
USEFUL FOR

Physics students, mechanical engineers, ski resort operators, and anyone involved in the design or operation of ski tow systems will benefit from this discussion.

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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