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?
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].