Why Is Calculating the Force on a Skier So Challenging?

  • Thread starter spidey12
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In summary, the skier's mass, the angle of the slope, and the coefficient of kinetic friction are given. The task is to find the magnitude of the force exerted on the skier by the tow bar. After several attempts, the correct calculation was determined with the help of considering the components of weight and friction and the constant velocity of the skier.
  • #1
spidey12
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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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  • #2
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?
 
  • #3
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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