Why Is Calculating the Force on a Skier So Challenging?

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SUMMARY

The challenge of calculating the force exerted on a skier being pulled up a slope at a constant velocity involves understanding the forces acting on the skier. The skier has a mass of 60.0 kg, and the slope is inclined at 26.0°. The coefficient of kinetic friction is 0.130. The correct approach requires considering both the gravitational force component acting down the slope and the frictional force opposing the motion. The resultant force must equal zero due to the constant velocity condition, leading to the conclusion that the force exerted by the tow bar must counteract both the gravitational and frictional forces.

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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 a lot. I figured out the correct answer thanks.

Now the other problem! (in the other thread)
 

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