On the bunny hill at a ski resort, a towrope pulls the skiers up the hill with c

Question

On the bunny hill at a ski resort, a towrope pulls the skiers up the hill with constant speed of 1.63 m/s. The slope of the hill is 10.5° with respect to the horizontal. A child is being pulled up the hill. The coefficients of static and kinetic friction between the child's skis and the snow are 0.161 and 0.101, respectively, and the child's mass is 56.9 kg, including the clothing and equipment.

What is the force with which the towrope has to pull on the child before movement, right as the child grabs the towrope?



What you need is the static friction force, right? So I'm finding the normal force (mgsin(theta)), then multiplying that by the coefficient of static friction. Maybe I'm making it too simple? I just don't know.

Explanation / Answer

Angle = 10.5 o The coefficients of static friction between the child's skis and the snow = 0.161 The child's mass including the clothing and equipment m = 56.9 kg The force with which the towrope has to pull on the child before movement = Static frictional force                         F = ( Normal force)                            = (mg cos )                            = 0.161 x 56.9 x 9.8 x sin 10.5                            = 16.36 N

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