A person is pushing a lawnmower of mass m = 38 kg and with h=0.75 m, d = 0.25 m,

Question

A person is pushing a lawnmower of mass m = 38 kg and with h=0.75 m, d = 0.25 m, LA = 0.28 m, and LB = 0.36 m. Assuming that the force exerted on the lawnmower by the person is completely horizontal and that the mass center of the lawnmower is at G,
and neglecting the rotational inertia of the wheels, determine the minimum value of this force that causes the rear wheels (labeled A) to lift off the ground. In addition, determine the corresponding acceleration of the mower.

h is the distance from the center of the mass of motor (labeled G) to where the person is holding the mower. It is the vertical height.

d is the vertical distance down from the height of the center of the mass to the ground.

LA is the horizontal distance from the mass center G to the center of the back wheel, labeled A.

LB is the horizontal distance from the mass center G to the center of the front wheel, labeled B.

Explanation / Answer

Vertical forces balance: Ra + Rb = mg

For wheel at A to lift off, Ra = 0.

Hence, Rb = mg

Balancing moments about G: F*h = Rb*LB

Force F = Rb*LB / h = mg*LB/h = 38*9.81 *0.36 / 0.75 = 178.9 N

Acceleration = F / m = 178.9 / 38 = 4.7 m/s^2

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