Kirk K. and Arnold S. push on opposite sides of a 21.2 kg, 1.384 m wide uniform

Question

Kirk K. and Arnold S. push on opposite sides of a 21.2 kg, 1.384 m wide uniform door hinged in the standard way as shown in top-down view. (The dimension into the page is irrelevant, and the door can be thought of as a uniform rod.) Kirk applies a force of 272 N at the far edge of the door from the hinge at a 53.13 degree angle as shown, whereas Arnold applies a 639 N force 0.300 m away from the hinge

If clockwise is positive, what is the initial angular acceleration of the door in rad/s2 ? (i.e. Who wins?)

Explanation / Answer

Torque applied by Arnold = F*r = 639*.300 = 191.7 N-m

Torque applied by Kirk = 272*1.384*sin(53.13) = 301.15 N-m (We take only the vertical component of the force since the horizontal component is along the door, thus providing no torque at all).

Since the directions are opposite, resultant torque = 109.45 N-m

So, angular acceleration = 109.45/(Moment of Inertia) = 109.45/(21.2*1.384*13.84/3) = 8.085

{Moment of inertia for a rod is M*L*L/3 = 21.2*1.384*1.384/3}

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