A \(20 \text{ kg}\) wheel consisting of a tire of radius \(R\) and rim of radius \(r\) makes contact with two rough surfaces. The coefficient of kinetic friction between the rim and the surface it contacts is \(\mu_{k_1} = 0.2\) and the coefficient of static friction between these two surfaces is \(\mu_{s_1}= 0.4 \). The coefficient of kinetic friction between the tire and the surface it contacts is \(μ_{k_2} = 0.3\) and the coefficient of static friction between these two surfaces is \(\mu_{s_2} = 0.6\). The tire makes exactly one revolution as it moves at a constant speed between the positions shown in the figure. The normal forces, \(N_{1}\) and \(N_{2}\), are equal and \(R=2r\).

What is the magnitude of the force \(F\) (in Newtons) that must be applied at the center of the wheel to keep it rolling at constant speed?

Give your answer to 1 decimal place.

Use \(g = 9.8 \text{ m/s}^2\) for the acceleration of gravity.

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