How do I calculate the torque required to lift an aeroplane's nose wheel?

  • Thread starter Aviation phys
  • Start date
  • Tags
    Lift Torque
In summary, the question asks for the torque required to lift the nose wheel of an airplane using a retracting nose wheel assembly. The wheel and arm have a combined mass of 50kg and are at a 30 degree angle. Using the equation T = r x F x Sinθ, the force at D must be determined in order to lift point G through moment arm AG. This force must balance the opposite force at C on moment arm BC. When D is directly under B, BC and CD form an isoceles triangle with base BD. The force at C would be the weight of the wheel plus gravity acting to pull CD down, which is equivalent to 490 N. The torque must be able to balance this opposite force in
  • #1
Aviation phys
3
0

Homework Statement


Question refers to the torque required to lift the nose wheel of an aeroplane

A retracting nose wheel assembly is raised by the application of Torque T applied to link BC through a shaft 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 θ is 30 degrees

Homework Equations



I have attached the figure that came with this question

T= r x F x Sinθ where T = Torque, r = Length of the Arm, F = Magnitude of the force and θ = the angle between the two arms, but i am unsure if this is what i use to find Torque required to lift

The Attempt at a Solution



I have set up my own free body diagram representing this question however the part i am struggling with is setting up the equations required and getting started.
parts i am having problems with are for the Moment Arm (D) part of the equation, is this length BC and how do i determine the Force part of the equation, I know the mass is 50kg and acceleration is Gravity = 9.8 ? any help that could point me in the right direction would be greatly appreciated
 

Attachments

  • Fig 1.1.jpg
    Fig 1.1.jpg
    11.2 KB · Views: 803
Last edited:
Physics news on Phys.org
  • #2
Here is my attempt so far

F = 50 x 9.8
= 490 N

T = 500mm x 490 x sin(30)
T = 122.500

I have no idea to tell if this is on ther right track or not
 
  • #3
One has to determine the force applied by arm CD acting at D, which lifts G through moment arm AG.

The force at D applies an opposite force at C on moment arm BC.

The torque at B must balance FC acting at C of moment arm BC. When D is directly under B, BC and CD are the two legs of an isoceles triangle with base BD.
 
  • #4
Ok so would the Force acting at C Be the Weight of the wheel plus gravity acting to pull CD down? in this case 490 N? therefore the torque must be able to balance 490 N of opposite Force? I am still unsure how to determine the force required by Arm CD acting at D to lift G, and once i determine this Force how to translate it to Torque.
 

What is torque?

Torque is a measure of the force that causes an object to rotate around an axis. It is calculated by multiplying the force applied by the distance from the axis of rotation.

How is torque related to lifting?

In order to lift an object, a certain amount of torque is required to overcome the force of gravity acting on the object. The amount of torque needed depends on the weight of the object and the distance from the axis of rotation.

What factors affect the torque required to lift an object?

The torque required to lift an object is affected by the weight of the object, the distance from the axis of rotation, and the angle at which the force is applied. Additionally, the duration of the force and the friction between the object and the surface it is being lifted from can also affect the torque required.

How can torque required to lift an object be calculated?

The torque required to lift an object can be calculated using the formula T = F x d, where T is torque, F is the force applied, and d is the distance from the axis of rotation. It is important to ensure that the units of force and distance are consistent, such as using Newtons for force and meters for distance.

Can torque be reduced when lifting an object?

Yes, torque can be reduced when lifting an object by either decreasing the weight of the object or increasing the distance from the axis of rotation. This is why using longer levers or pulleys can make lifting heavy objects easier, as it decreases the amount of torque required.

Similar threads

  • Mechanical Engineering
Replies
2
Views
2K
  • Mechanical Engineering
Replies
3
Views
762
  • Mechanical Engineering
Replies
13
Views
2K
Replies
15
Views
2K
  • Mechanical Engineering
Replies
2
Views
2K
  • Mechanical Engineering
Replies
21
Views
584
  • Introductory Physics Homework Help
Replies
7
Views
211
  • Mechanical Engineering
Replies
8
Views
4K
  • Mechanical Engineering
Replies
19
Views
12K
Back
Top