A mass of 26 kg lies on a horizontal surface with a coefficient of static fricti

Question

A mass of 26 kg lies on a horizontal surface with a coefficient of static friction of 0.67 and a coefficient of kinetic friction of 0.38. If a force of 39.1 Newtons is applied parallel to the surface (+x direction if you prefer), what is the magnitude of the static frictional force on the object in Newtons?

Explanation / Answer

By F(net) = F(applied) - F(gravity) - F(friction) =>as v = constant=>a = 0=>F(net) = 0 => F(applied) = F(gravity) + F(friction) =>F(a) = mgsin? + µmg =>F(a) = mg[sin? + µ] =>F(a) = 26 x 10 [10/v{(24)^2 + (10)^2} + 0.2] (as given tan? = 10/24) =>F(a) = 152 N

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