physics question

Question

A boy pushes a 25.6 kg lawnmower making it go forward at a constant velocity. The handle of the lawn mower makes an angle of Θ = 22.0° below the horizontal. The boy pushes on the handle with a force of 135 N. Ans: Draw, Fax = 125 N, Fay = 50.6 N, Fg = 251 N, Ff = 125 N, FN = 301 N

a. Draw an FBD of the lawnmower.

b. What are the horizontal and vertical components of his applied force?

c. Determine the numerical values of all the forces on your FBD (Fg, FN, Fa, Ff) and label them.

Answer 1

b) Fax = 135cos(22) = 125 N

Fay = 135sin(22) = 50.6 N

c) Fg = mg = 25.6(9.81) = 251 N

FN = Fg + Fay = 301 N

Fa = 135 N

Ff = Fax = 125 N