Askidder dragging of 520 kg walk-through forest at a constant speed of 3.5 m/s. if the skidder is applying a force of 1850 n what is the coefficient of friction between the log and the ground

The coefficient of friction between the log and the ground is, μₓ = 0.3630

Explanation:

Given data,

The skidder dragging the mass, m = 520 kg

The acceleration of the mass, a = 3.5 m/s

The force applied by the skidder, Fₓ = 1850 N

The kinetic frictional force is given by,

                       Fₓ = μₓ η

                      μₓ = Fₓ / η

Where  μₓ - the coefficient of the kinetic friction  

            η is the normal force acting on the mass due to gravitational force.

                          η = mg

                              = 520 x 9.8

                               = 5096 N

Substituting the values in the above equation,

                           μₓ = 1850 N / 5096 N

                                = 0.3630

Hence, the coefficient of friction between the log and the ground is, μₓ = 0.3630

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0 = 0.40 / 0 = 0

= 0.20 = 9.80 ⁄2 =?

=0 , = = 0.20 =0.50 0 0.40 ⁄

=0−122 =0−12(9.80 ⁄2)(0.50)2 =1.2