What is the minimum force T necessary to lift the wheel?

In summary, the conversation discusses a problem involving a retracting nose wheel assembly and the necessary torque T to lift the wheel. The force/moment is applied at point D and must be at least equal to the combined moment of the wheel and arm AO, which have a combined mass of 50kg with a center of mass at G. The required torque T can be determined through moments/torques through linkages involving CD and BC. The lifting moment must operate at D perpendicular to AD, with a moment arm of 0.8m.
  • #1
grayworks
2
0
Hi I'm very new so I'm not sure if this is the best place to put this in. I've got a problem that I'm having trouble working out and any help would be great.

'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 is 30degrees.

Cheers
 

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  • #2
The torque acts on a moment arm, and one must determine the necessary force at D (normal) the arm AO.

So the force/moment is applied at D, but that moment must be at least equal to the moment of wheel and arm AO. That information is given - "a combined mass of 50kg with a centre of mass at G."

This post belongs in the Engineering and Technology Homework Forum.
 
  • #3
k I've worked out the moment between A-G is 245Nm but I can't see how this relates to B-C
 
  • #4
grayworks said:
k I've worked out the moment between A-G is 245Nm but I can't see how this relates to B-C
One needs to show work. It would appear that one use 50 kg * 9.8 m/s2 * cos (60°), which is correct.

The applied moment must be in the opposite direction of that and of the same or greater magnitude for the wheel to move. The lifting moment operates at D perpendicular to AD, which has a moment arm 0.8 m.

Has one done moments/torques through linkages?

CD applies a force at D, and BC applies a force to CD.

The torque T, acts along BD, by virtue of BC and CD.
 

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