# Solving Gravitational Torque Homework Equations

• mintsnapple
In summary: But I'm not sure if that is the only way.I apologize for my lack of knowledge and thank you so much for helping me.
mintsnapple

## Homework Equations

torque = r*Fsin(x)

## 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?

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.

1 person
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?

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.

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?

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 ?

1 person
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 ?

They are equal.

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

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 .

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?

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:
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?

Yes...

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

Do you agree ?

Last edited:

## 1. What is gravitational torque?

Gravitational torque is a force that causes rotation around a fixed point due to the gravitational pull of an object. It is the product of the force of gravity and the distance between the object and the fixed point.

## 2. How do you calculate gravitational torque?

The formula for calculating gravitational torque is T = r x F, where T is the torque, r is the distance between the object and the fixed point, and F is the force of gravity. This formula is also known as the cross product of the two vectors.

## 3. What are the units of gravitational torque?

The units of gravitational torque are newton-meters (Nm) in the SI (metric) system and foot-pounds (ft-lb) in the English system. Both units represent the product of force and distance.

## 4. How does gravitational torque affect objects?

Gravitational torque causes objects to rotate around a fixed point. The direction of the rotation depends on the direction of the force and the direction of the distance vector between the object and the fixed point.

## 5. What factors affect gravitational torque?

The magnitude and direction of the force of gravity, as well as the distance between the object and the fixed point, are the main factors that affect gravitational torque. The angle between the force and the distance vector also plays a role in determining the magnitude and direction of the torque.

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