Torque Calculation. Short- Am I correct?

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SUMMARY

The torque calculation for a steering wheel with a diameter of 300mm and forces of 20N and 10N acting at the 3 o'clock and 9 o'clock positions respectively results in a torque of 4.5 kgm when calculated about the center of the wheel. The correct approach involves using the radius (0.15m) and adding the torques from both forces, as they act in the same rotational direction. It is crucial to specify the point of reference for torque calculations, as torque is a vector quantity. Understanding the relationship between force application points and the axis of rotation is essential for accurate torque assessments.

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  • Knowledge of basic physics principles related to force and distance
  • Familiarity with the concept of radius in circular objects
  • Ability to perform unit conversions (e.g., mm to m)
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mazz1801
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Homework Statement



What is the torque on the steering wheel?

Homework Equations



I don't know how to draw on here.

Steering wheel is circular.
It has a Diameter of 300mm
A force of 20N acts downwards toward the 3 oclock position
A force of 10N acts Upwards toward the 9 oclock position


The Attempt at a Solution


Torque=Force X Distance

300mm=0.3m

Force=20-10=10
Torque=10 X (0.3)
Torque=3Nm


This is my first torque calculation and I am working unadvised. I would like to know if I am doing it correctly by using the diameter instead of the radius. Also If I have made any mistakes could somebody please explain what I have done wrong! Thank you in Advance!
 
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Hello there !When you are talking about Torque you should ALWAYS mention your relevant point ( ie Torque about a point A on the rim of the steering wheel or at its centre etc ).Don't forget that Torque is a vector !

Torque about the centre of the wheel :

T= 20 x 0.15 + 10*0.15 = 30*0.15 = 4.5 KGm !

Both forces are trying to rotate the wheel to the same direction so you should NOT subtract them !
 
Last edited:
Ok so I was completely off the mark on that one!

You have used a radius perpendicular to the application point of the torque.
And you have resolved the torque for both forces and added them...
 
Exactly . if you are confused about the Forces ( ie when they look that they are in the opposite directions ) you should try "rotating" one of the forces until you reach the other one so that you can decide whether you should subtract or add them.
 
Thank you for the tip! You are very helpful!
What about if the point of application of the force is away from where you want to calculate the force? For example a spanner moving a nut... Would it have no effect or would I use the entire length of the spanner up to the centre of the nut?
 
If you grabbed the spanner from its end then you would use its entire length , if you held it from its middle then you would use L/2 etc . The distance that you will need is the distance from the axis of rotation ( so in the wheel it was the radius , and in this case of the spanner it is its whole length because the axis of rotation is passing perpendicular to the nut and passing through it ).
 
That's why steering wheels in public buses are huge because the wheels of the bus are also huge . The bigger the distance from the axis of rotation the easier it is to rotate anything. Try , opening a door by pushing it from it's fixed end...LOL . Or try steering a lorry with a 20 cm steering wheel . You got the point i hope :P
 
Thanks for your help!
I am pretty sure I understand it now!
 

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