What Torque is Needed to Lift a Retracting Nose Wheel at 30 Degrees?

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SUMMARY

The discussion focuses on calculating the torque (T) required to lift a retracting nose wheel assembly at a 30-degree angle. The assembly has a combined mass of 50 kg, with the center of mass located at point G. The torque applied to arm BC generates a force at point C, which must be resolved into components acting along arms AG and CD to determine the necessary lifting force. The solution involves analyzing the forces and torques around point A to ensure the applied torque overcomes the gravitational force acting on the mass.

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Homework Statement


A retracting nose wheel assembly is raised by the application of a Torque T applied to link BC through a shaft at B. The wheel and arm AO have a combined mass of 50kg with a centre of mass at G. Find the value of T necessary to lift the wheel when D is directly under B at which position angle theta is 30 degrees.


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The Attempt at a Solution



Having difficulties getting started. Would like some assistance to get started. Feeling very lost at the moment.
 

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Well, the torque applied to arm BC is going to cause a force tangent to the arm at C, with a magnitude of torque*length_arm. A component of that force will travel along arm CD. You now have a force at point D, with a vector pointing towards C.

From here, you break 'that' force into components, one is acting radially AG, one is acting tangentially. That force*length_AD must be greater than the torque that the masses create around point A.

If it's me, I first see what force at D is needed to overcome the mass. Then I see what force along arm CD is needed to get that particular component. Then backwards again through BC.
 

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