Solving Gravitational Torque Homework Equations

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SUMMARY

This discussion focuses on solving gravitational torque equations related to a beam in equilibrium. The key equation presented is torque = r * Fsin(x), where T1 and T2 are the tension forces acting on the beam. The participants clarify that the net torque and net forces in both x and y directions must sum to zero, emphasizing the importance of accurately identifying all forces acting on the beam, including gravity and tension. The relationship between the forces T2 and Mg is established as T2 = Mg, confirming the equilibrium condition of the system.

PREREQUISITES
  • Understanding of torque and its calculation using the formula torque = r * Fsin(x)
  • Knowledge of forces acting on a beam in static equilibrium
  • Familiarity with the concepts of tension in cables and their impact on structures
  • Basic principles of physics, particularly related to gravity and equilibrium
NEXT STEPS
  • Study the principles of static equilibrium in physics
  • Learn how to derive torque equations for various configurations of beams
  • Explore the role of tension forces in static systems
  • Investigate the effects of different angles on torque calculations
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and static equilibrium, as well as educators looking for examples of torque problems in real-world applications.

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Homework Statement



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Homework Equations



torque = r*Fsin(x)

The Attempt at a Solution


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I really need help on walking through this problem...didn't really teach it in class...I tried watching a video and you're supposed to sum the forces * some perpendicular distance to the pivot? Yet what if the force is already perpendicular?
 
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Check your force diagram for missing and incorrect forces acting on the beam.

T2 is the tension force acting on the mass m (eg not acting on the end of the beam).

Your equation has force T1 acting at what distance from the pivot? Let's say the beam has length L so d = Lcos(60). Then T1 acts at h where h= Lsin(60) from the pivot.

The net torque on the beam should sum to zero because it's not rotating.

The net forces in x and y on the beam should also sum to zero because it's not translating.
 
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CWatters said:
Check your force diagram for missing and incorrect forces acting on the beam.

T2 is the tension force acting on the mass m (eg not acting on the end of the beam).

Your equation has force T1 acting at what distance from the pivot? Let's say the beam has length L so d = Lcos(60). Then T1 acts at h where h= Lsin(60) from the pivot.

The net torque on the beam should sum to zero because it's not rotating.

The net forces in x and y on the beam should also sum to zero because it's not translating.

How does this look?
2mepzk1.jpg


Also, how should I approach c? Is it just the weight of the beam?
 
What are the forces acting on the weight of mass M ?
 
There is a force on the beam at the pivot.
 
CWatters said:
There is a force on the beam at the pivot.
Could you please explicate?
 
Tanya Sharma said:
What are the forces acting on the weight of mass M ?
Gravity and the tension of cable 2.
 
mintsnapple said:
Gravity and the tension of cable 2.

Good...So considering that the weight is in equilibrium ,what is the relation between them ? You may refer tension as T2 and weight Mg .
 
Tanya Sharma said:
Good...So considering that the weight is in equilibrium ,what is the relation between them ? You may refer tension as T2 and weight Mg .

There is also a vertical force from the weight of the beam no? Which is M/2*G*sin60.

Also, how would I find the forces on the beam by the pivot?
 
  • #10
mintsnapple said:
There is also a vertical force from the weight of the beam no? Which is M/2*G*sin60.

No..that is not correct.The weight of the beam is acting on the beam ,not on the weight .There are only two forces acting on the weight ;T2 and Mg

So,what is the relation between them ?

mintsnapple said:
Also, how would I find the forces on the beam by the pivot?

Let us call the force exerted by pivot as P .The horizontal component as Px and vertical component as Py .

Can you write the torque equation for the beam with respect to the pivot ?

Note :Please type your work.Do not paste an image.
 
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  • #11
Tanya Sharma said:
No..that is not correct.The weight of the beam is acting on the beam ,not on the weight .There are only two forces acting on the weight ;T2 and Mg

So,what is the relation between them ?



Let us call the force exerted by pivot as P .The horizontal component as Px and vertical component as Py .

Can you write the torque equation for the beam with respect to the pivot ?

Note :Please type your work.Do not paste an image.

They are equal.

torque = px*sin60 + py*cos60 = 0?
 
  • #12
mintsnapple said:
They are equal.

Yes. So T2=Mg

mintsnapple said:
torque = px*sin60 + py*cos60 = 0?

Simply wrong .

The pivot force does not feature in the torque equation of the beam about the pivot . It would have played a role if we were writing torque equation about the center of mass .

List all the forces acting on the beam .
 
  • #13
Tanya Sharma said:
Yes. So T2=Mg



Simply wrong .

The pivot force does not feature in the torque equation of the beam about the pivot . It would have played a role if we were writing torque equation about the center of mass .

List all the forces acting on the beam .

Just gravity?
 
  • #14
mintsnapple said:
Just gravity?

It is only one of the forces acting on the beam .What about T1 and T2 ? Do you think any of them is acting on the beam ?
 
Last edited:
  • #15
Tanya Sharma said:
It is only one of the forces acting on the beam .What about T1 and T2 ? Do you think any of them is acting on the beam ?

I apologize for my lack of knowledge and thank you so much for helping me.

I suppose T2 does pull the beam in the vertical direction and T1 pulls it in horizontal direction?
 
  • #16
Yes...

So,basically there are four forces acting on the beam ; T1,T2,Mg/2 and P .

Do you agree ?
 
Last edited:

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