Effort-Saving Angle for Pulling Sled on Slope: Challenging Physics Homework

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The discussion revolves around determining the optimal angle, φ, at which a boy should pull his sled up a slope to minimize effort, given a coefficient of friction of 0.10. The user presents a mathematical approach involving force equations and relationships between normal force, gravitational force, and friction. They derive a formula for φ based on the slope angle θ and the coefficient of friction. The user also questions whether further clarification of their work would be helpful for understanding. The thread emphasizes the importance of correctly applying physics principles to solve the problem efficiently.
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Homework Statement


A boy pulls his sled up a snowy slope of angle θ from the horizontal. If the coefficient of friction between the slope and sled is 0.10, at what angle, φ, from the slope should the boy pull the sled so that he exerts the least effort?


Homework Equations



N/A

The Attempt at a Solution



Well, I don’t think this is the correct approach to the problem but here it goes:

∑Fy = Fn + Fpsin(φ) - mgcos(θ) = 0 and ∑Fx = Fpcos(φ) - mgsin(θ) - μFn = 0 which, through some algebra, Fp = (mgcos(θ) - Fn)/sin(φ) = (mgsin(θ) + μFn)/cos(φ) and dividing the two equations results with 1 = tan(φ)(mgsin(θ) + μFn)/(mgcos(θ) - Fn) ⇒ cot(φ) = (mg sin(θ) + μFn)/(mgcos(θ) - Fn). Also, since the net force acting one the object is zero, the net work is zero as well; therefore, mgh = μFn*s or similarly mgsin(θ) = μFn. Applying this relationship I find cot(φ) = (2μFn)/(μFncot(θ) - Fn) = 2μ/(μcot(θ) - 1) which ultimately yields φ = cot^(-1)[2μ/(μcot(θ) - 1)] = tan^(-1)[(μcot(θ) - 1)/2μ].

Thanks.
 
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