A 350N box is pulled along a horizontal pathway by a rope which makes an angle of 42degrees with the pathway. If a force of 250N is applied on the rope, what is the coefficient of sliding friction?
M*g = 350 N.
Fn = Mg-Fap*sin42 = Normal force.
Fn = 350-250*sin42 =
Fap*Cos42-Fk = M*a.
250*Cos42-Fk = M*0 = 0, Fk = ?.
u = Fk/Fn.