Static Mechanics: Aircraft Nosewheel Linkage Torque

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The discussion focuses on analyzing the forces and torques in a linkage system involving an aircraft's nosewheel. Participants emphasize the need for free body diagrams (FBDs) for each moving part to accurately represent the forces acting on them. It is assumed that links BC and CD are massless, and the torque at point M must balance the forces to prevent rotation of the assembly. A vertical force at point G, representing the weight of the combined mass of 50 kg, is constant, while the moment M varies as the wheel is raised. The problem requires calculating the moment M at a specific position during the mechanism's stroke.
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Homework Statement
Hi I'm stuck at how to draw the free body diagram for thise nose wheel question
Relevant Equations
Moment equal 0
see attached

mmexport1760866157266.webp


IMG_20251019_172253.webp
 
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A free body diagram shows the forces on a single rigid body. Here you have a linkage, so you need an FBD for each moving part. This will involve defining unknowns for forces and torques between connected bodies, each such force being represented in the FBDs for both bodies.
What are the moving rigid bodies in this problem? Pick one and list the forces on it. Then post your attempt at its FBD.
 
I think it is safe to assume that links BC and CD are massless because no masses are given for them. Assume that the torque at M gives rise to a force ##F_D## at D that is just right to prevent the combined (arm + wheel) assembly AO from rotating either clockwise or counterclockwise.
 
Consider the external force and moment acting on the whole mechanism first.
One is the represented M, which is necessary to raise the combined mass of 50 kg, which can be represented as a vertical force applied at point G.

Note that the value of that vertical force will be constant, while the value of the necessary moment M will change as the wheel is being raised, reason for which the problem asks for the value of moment M at that specific position of the mechanism's stroke.

Please, see this example:
https://www.school-mechademic.com/blog/static-force-analysis-on-a-4-bar-link-mechanism

:cool:
 
Last edited:
ok I think I got it thanks everyone
 

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Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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