# Homework Help: Gravitational torque

1. Mar 11, 2014

### mintsnapple

1. The problem statement, all variables and given/known data

2. Relevant equations

torque = r*Fsin(x)

3. The attempt at a solution

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?

2. Mar 12, 2014

### CWatters

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? Lets 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.

3. Mar 12, 2014

### mintsnapple

How does this look?

Also, how should I approach c? Is it just the weight of the beam?

4. Mar 12, 2014

### Tanya Sharma

What are the forces acting on the weight of mass M ?

5. Mar 12, 2014

### CWatters

There is a force on the beam at the pivot.

6. Mar 12, 2014

### mintsnapple

Could you please explicate?

7. Mar 12, 2014

### mintsnapple

Gravity and the tension of cable 2.

8. Mar 12, 2014

### Tanya Sharma

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

9. Mar 12, 2014

### mintsnapple

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. Mar 12, 2014

### Tanya Sharma

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.

11. Mar 12, 2014

### mintsnapple

They are equal.

torque = px*sin60 + py*cos60 = 0?

12. Mar 12, 2014

### Tanya Sharma

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 .

13. Mar 12, 2014

### mintsnapple

Just gravity?

14. Mar 12, 2014

### Tanya Sharma

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: Mar 12, 2014
15. Mar 12, 2014

### mintsnapple

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. Mar 12, 2014

### Tanya Sharma

Yes...

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

Do you agree ?

Last edited: Mar 12, 2014
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