A rubber chicken of mass 45.0 kg is pulled along a horizontal surface by a rope that makes an angle of 32.0 with the horizontal. If the coefficient of friction is 0.113 and the tension in the rope is 95.0 N, what is the acceleration of the object?

Answer 1

X: ma=Fcosα-F(fr)
Y: 0=-mg+N+Fsinα =>N=mg-Fsinα
F(fr) = μN=μ(mg-Fsinα)
ma= Fcosα- μ(mg-Fsinα)
a = { Fcosα- μ(mg-Fsinα)}/m