Skier on a slope (Fnormal, Ffriction, etc)

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To determine the force exerted by the tow bar on the skier, the skier's mass is 61.4 kg, and the slope is inclined at 30.9 degrees. The coefficient of kinetic friction is 0.116, and the force of friction can be calculated using the equation Ff = uN, where N is the normal force. The normal force is derived from the skier's weight and the angle of the slope, leading to a calculation involving the gravitational force's horizontal component. The discussion suggests that the tow bar force equals the force of friction, resulting in a calculated force of approximately 59.89 N.
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Homework Statement



A skier is pulled up a slope at a constant velocity by a tow bar. The slope is inclined at 30.9 ° 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 61.4 kg, and the coefficient of kinetic friction between the skis and the snow is 0.116. Find the magnitude of the force that the tow bar exerts on the skier.

Homework Equations



Ff=uN

The Attempt at a Solution



so is it the tow bar force = force of friction? the answer is just u times horizontal component of gravitational force?
 
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or was it u times normal force? 61.4*9.8cos30.9*.116=59.8925 N?
 

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