Equilibrium and principle of moments

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SUMMARY

The discussion centers on the principle of moments in physics, specifically addressing a homework problem involving torque calculations. The equation Torque = F x D is applied, where the moments on either side of a pivot must balance. The confusion arises from the variable F, which represents a force that is relevant only in part (b) of the question, as it replaces a 5N force in that context. Participants clarify that F does not factor into the calculations for part (a), leading to the conclusion that the problem requires understanding the distinction between the two parts to solve correctly.

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  • Understanding of torque and its formula: Torque = F x D
  • Knowledge of equilibrium conditions in physics
  • Familiarity with the concept of moments about a pivot point
  • Ability to solve algebraic equations with multiple variables
NEXT STEPS
  • Study the concept of static equilibrium in physics
  • Learn how to apply the principle of moments in various scenarios
  • Explore problem-solving techniques for algebraic equations with multiple unknowns
  • Review examples of torque calculations in real-world applications
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This discussion is beneficial for students studying physics, particularly those tackling problems related to equilibrium and torque, as well as educators seeking to clarify concepts of moments and forces in their teaching.

Apple&Orange
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Homework Statement



upload_2016-5-14_20-34-7.png


Homework Equations



Torque = F x D
Manticlockwise = Mclockwise

The Attempt at a Solution



From what I understand, the moment on left hand side of the pivot point needs to be equal to the moment on right hand side, therefore...

(Fx.2)+(5x.14) = (7xd)

My initial thinking was to substitute 2N into F, so that both forces on either side of pivot point are equal to 7N. However, this was not correct because the resultant forces were not in equilibrium after checking.

The answer in the book was this:
[/B]
upload_2016-5-14_21-16-11.png


Could someone explain to me why F wasn't factored into the working? The way I'm interpreting this is that there is no force acting at point F. Would this be a correct assumption?
 
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Looks like a badly worded question to me.

If you read Question (b) it says F replaces the 5N force. So F doesn't exist for question (a) ?
 
Apple&Orange said:
(Fx.2)+(5x.14) = (7xd)

If F does exist for (a) then that's correct but you can't solve it. You have one equation and two unknowns so best you can do is write an equation for d in terms of F.
 
Apple&Orange said:
Could someone explain to me why F wasn't factored into the working?
Part (b) explains that F acting at a distance 200mm is used in (b) as the substitute for the 5N force at a distance 140mm.

So the diagram is doing double-duty, and F is relevant only to part (b).
 

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