Question

26. A 40 kg block rests on an inclined plane with an angle of 30 degrees a) Draw the FBD representing the forces that act on the block Depict: Weight (Fw), Parallel force (Fp), Normal force (Fn), Frictional Force (F)

Answer 1

Because m = 40 kg since there is no motion perpendicular to the surface, we have F_n - mg Cos 30 degree = 0 F_n = mg Cos 30 degree = 40 times 9.8 times squareroot 3/2 = 339.47 Newton parallel force F_p = mg sin 30 degree - F_f Because F_f = mu F_n = mu mg Cos 30 degree = 0.3 times 339.47 = 101.84 N Therefore F_p = 40 times 9.8 times 1/2 - 101.84 = 196 - 101.84 = 94.16 Newton If a is the acceleration, there m a = 94.16 a = 94.16/40 = 2.35 m/sec^2