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A person is pushing a lawnmower of mass m D 38 kg and with h D 0:75 m, d D 0:25 m, `A D 0:28 m, and `B D 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.

Answer :

Answer:

The acceleration of the mower will be "4.7 m/s²".

Explanation:

Balance of vertical force will be:

⇒  [tex]Ra + Rb = mg[/tex]

For wheel to take off at A,

⇒  [tex]Ra = 0[/tex]

Hence,

[tex]Rb=mg[/tex]

Balancing moments about G will be:

⇒  [tex]F\times h = Rb\times LB[/tex]

As we know,

Force, F = [tex]\frac{Rb\times LB }{h}[/tex]

On putting the values, we get

⇒           = [tex]\frac{38\times 9.81\times 0.36}{0.75}[/tex]

⇒           = [tex]178.9 \ N[/tex]

Now,

Acceleration, a = [tex]\frac{F}{m}[/tex]

⇒                       = [tex]\frac{178.9}{38}[/tex]

⇒                       = [tex]4.7 \ m/s^2[/tex]

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