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

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To calculate the tension in the ski lift tow rope for a skier of mass 65.0 kg being pulled up a slope at a constant speed on a 26.0-degree incline, the key principle is that the forces are balanced due to the lack of acceleration. The tension in the rope equals the component of the gravitational force acting parallel to the slope, which can be expressed as T = mg sin(θ). The correct calculation involves using the gravitational force and the sine of the slope angle to find the tension. The initial calculation of 916.24 N was incorrect, as the proper formula and understanding of forces were clarified in the discussion. This explanation helped solidify the understanding of the tension calculation in this scenario.
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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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