Question

A 35 kg lawnmower is pushed using a handle that makes an angle of 32 degrees with the horizontal. The lawnmower is undergoing acceleration. The coefficient of sliding friction between the lawnmower and the ground is 0.225. The force used to push the lawnmower 200N. A) What is the force normal to the plane? B) What is the force of friction? C) What is the net force? D) What is the acceleration given to the lawnmower?

Answer 1

m = 35 kg

angle with the horizontal, theta = 32 deg

F = 200 N

(a) Resolve the force into horizontal and vertical components.

Horizontal force, Fx = F*cos 32 = 200*cos32 = 169.6 N

Vertical force, Fy = F*sin32 = 200*sin32 = 106 N

So, normal force to the plane, R = Fy + mg = 106 + 35*9.8 = 449 N

(b) Force of friction, Ff = mu*R = 0.225*449 = 101 N

(c) Net force, F = Fx – Ff = 169.6 – 101 = 68.6 N

(d) Acceleration of the lawnmower = F / m = 68.6 / 35 = 1.96 m/s^2