Why Is Calculating the Force on a Skier So Challenging?

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Calculating the force on a skier being pulled up a slope involves understanding the components of gravitational force and friction. The skier's weight has a component acting down the slope, while friction opposes the skier's motion. Since the skier is moving at a constant velocity, the net force must equal zero, meaning the tow bar's force must balance both the gravitational and frictional forces. The correct calculations require using the sine and cosine functions appropriately for the angle of the slope. The discussion highlights the importance of recognizing the forces at play in such scenarios.
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I've been working on it since yesterday. And i can't seen to get it right:


A skier is pulled up a slope at a constant velocity by a tow bar. The slope is inclined at 26.0° with respect to the horizontal. The force applied to the skier by the tow bar is parallel to the slope. The skier's mass is 60.0 kg, and the coefficient of kinetic friction between the skis and the snow is 0.130. Find the magnitude of the force that the tow bar exerts on the skier.



I calculated the following:

(.13)(60)(9.8)cos(26) but that was wrong

then i tried (.13)(60)(9.8)sin(26) but that was wrong too. Any advice?
 
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Weight of the skier has a component down the slope AND the friction acts in the opposite direction of motion i.e. downn the slope. And traveling at a constant velocity means what for the resultant force?
 
rock.freak667 said:
Weight of the skier has a component down the slope AND the friction acts in the opposite direction of motion i.e. downn the slope. And traveling at a constant velocity means what for the resultant force?

Ok, that helped alot. I figured out the correct answer thanks.

Now the other problem! (in the other thread)
 
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