Tension in Rope Problem: Calculating Tension in a Ski Lift Tow Rope

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SUMMARY

The discussion focuses on calculating the tension in a ski lift tow rope for a skier with a mass of 65.0 kg being pulled up a slope at a constant speed. The slope has an angle of 26.0 degrees above the horizontal, and friction is ignored. The correct formula for tension is derived as |T_{rope}| = m|g|sin(θ), where θ is the slope angle. The initial calculation of 916.24 N was incorrect due to misunderstanding the balance of forces involved.

PREREQUISITES
  • Understanding of Newton's Second Law (F=ma)
  • Knowledge of free body diagrams (FBD)
  • Basic trigonometry, specifically sine functions
  • Concept of gravitational force components
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the mechanics of tension in ropes and forces acting on objects on inclined planes.

physicsgirlie26
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tension in rope problem. please help!

A skier of mass 65.0 is pulled up a snow-covered slope at constant speed by a tow rope that is parallel to the ground. The ground slopes upward at a constant angle of 26.0 above the horizontal and you can ignore friction.

Calculate the tension in the tow rope.



I know you use the F=ma and you also need to draw a fbd which i did. I calculated the answer to be 916.24 N but that isn't right. I used T= m2a2x+m2gsin(theta).

Thank you!
 
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Assuming that there is no friction, since the skier is not accelerating, there is no unbalanced force. The normal force is balanced by the component of the gravitational fforce perpendicular to the incline, the pulling force of the rope is balanced by the component of the gravitational force parallel to the incline. The pulling force imparted on the skier by the rope is the tension in the rope. Hence, |T_{rope}|= |mg_{parallel}|, which is, |T_{rope}|=m|g|sin(\psi), where psi is twenty six degrees in this situation, of course.
 
THANK YOU SOOOO MUCH FOR YOUR HELP. Your explanation helped me understand it better than what I knew.
:)
 

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