Moment & Forces: Find Other Forces Present

[adjacent]

Homework Statement

Find the other forces present.
Assume the triangle to be weightless

Homework Equations

τ=Force x perpendicular distance

The Attempt at a Solution

The force,10N provides a 50Nm Torque. (10 x 5)
The triangle is in equilibrium so,anticlockwise moment is equal to clockwise moment
50Nm=2 x F
F=25N at p

But within the two meters,not only p touches but other parts of the triangle(in contact with the wall) touches too.
Should I account for that?

As the net force should be zero,pivot point exerts a force of 25N to the left.and 10N upwards.
Is this correct?

 

[Doc Al]

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[adjacent]

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I think it's okay now

 

[Doc Al]

But within the two meters,not only p touches but other parts of the triangle(in contact with the wall) touches too.
Should I account for that?

It's not clear exactly how the triangle is attached to the wall or how it makes contact. So I would just consider points O and P as making contact.

Can we assume the only fixed pivot is at point O and that the triangle just rests against the wall? (And that point P cannot exert an upward force?)

As the net force should be zero,pivot point exerts a force of 25N to the left.and 10N upwards.
Is this correct?

Looks reasonable to me.

 

[adjacent]

Can we assume the only fixed pivot is at point O and that the triangle just rests against the wall? (And that point P cannot exert an upward force?)

Yes.
And all the parts of line Op touches the wall

 

[adjacent]

I want the answer to this question:

But within the two meters,not only p touches but other parts of the triangle(in contact with the wall) touches too.
Should I account for that?

 

[Tanya Sharma]

But within the two meters,not only p touches but other parts of the triangle(in contact with the wall) touches too.
Should I account for that?

If the triangle presses against the wall,then you also need to consider Normal force from the wall .

 

[adjacent]

If the triangle presses against the wall,then you also need to consider Normal force from the wall .

How?Here only moment equation can be used.

 

[Tanya Sharma]

How?Here only moment equation can be used.

What do you mean by the above statement? When writing force and torque equations for the triangle ,you need to consider all the forces acting on the triangle .

The forces on the triangle are

1)Applied force -10 N downwards
2)Hinge force
3)Normal force from the wall

 

[adjacent]

What do you mean by the above statement? When writing force and torque equations for the triangle ,you need to consider all the forces acting on the triangle .

The forces on the triangle are

1)Applied force -10 N downwards
2)Hinge force
3)Normal force from the wall

I'm worried about the Normal force from the wall.How do I know it?I can calculate Normal force at point p. (50Nm=2*F)
But there are infinite number of normal forces acting by the wall.(In the distance 2m,it's 25N.In the distance 1m ,its 50.In the distance 0.0000001 meter,It's is ... and so on.Do I need to add all these?

 

[Tanya Sharma]

I'm worried about the Normal force from the wall.How do I know it?I can calculate Normal force at point p. (50Nm=2*F)
But there are infinite number of normal forces acting by the wall.(In the distance 2m,it's 25N.In the distance 1m ,its 50.In the distance 0.0000001 meter,It's is ... and so on.Do I need to add all these?

When two objects are in contact,there exists a contact force between them.The component of this force parallel to the surfaces in contact is called friction .The component of force perpendicular to the surfaces is called Normal force .

In this problem,there is no friction force,but there exists a normal force.

Even though there are infinite number of points on the surfaces in contact ,there is only one normal force ,which you may consider to be the net force which the bodies exert on each other ,perpendicular to the surfaces.

Well..now the question is where is the point of application of this Normal force ?

@Doc Al :Would you mind chipping in ? I hope I am not talking nonsense and misguiding the OP

 

[adjacent]

Well..now the question is where is the point of application of this Normal force ?

Yes.That was my question.

 

[Doc Al]

Even though there are infinite number of points on the surfaces in contact ,there is only one normal force ,which you may consider to be the net force which the bodies exert on each other ,perpendicular to the surfaces.

I agree with that.

Well..now the question is where is the point of application of this Normal force ?

There's the rub. To find the effective point of application of the normal force.

@Doc Al :Would you mind chipping in ? I hope I am not talking nonsense and misguiding the OP

No, you're fine. (Sorry, I was AFK for most of the day.)

I'll have to ponder just how to find that point of application.

 

[Tanya Sharma]

Thanks Doc...

If the normal force acts at a distance 'x' from O then we have the relation (N)(x) = 50 . But this gives us several possibilities .

 

[adjacent]

Thanks Doc...

If the normal force acts at a distance 'x' from O then we have the relation (N)(x) = 50 . But this gives us several possibilities .

And $0<x\leq2$

Eating my head.Help please

 

[Doc Al]

Is this a textbook problem? If so, please state the textbook and problem number. (If I have the book, I can check the context of the question.)

 

[adjacent]

Is this a textbook problem? If so, please state the textbook and problem number. (If I have the book, I can check the context of the question.)

This is not a textbook question.I made it.

 

[PhanthomJay]

You are getting into all sorts of deformation and indeterminate complexities if you consider any normal forces from the wall acting between O and P. The vertical member is not glued to the wall anyway. In addition to the applied force, the only other forces acting on the triangle occur ay joints O and P. Solve for them. There is no normal force from the wall between those points.

 

[adjacent]

You are getting into all sorts of deformation and indeterminate complexities if you consider any normal forces from the wall acting between O and P. The vertical member is not glued to the wall anyway. In addition to the applied force, the only other forces acting on the triangle occur ay joints O and P. Solve for them. There is no normal force from the wall between those points.

You are wrong.p is not a joint,it is just a point in the wall.O is the joint or pivot.Did you read the posts?

 

[Doc Al]

You are getting into all sorts of deformation and indeterminate complexities if you consider any normal forces from the wall acting between O and P.

Exactly. I don't think there is any simple way to address those forces, which are very dependent on exactly how the surfaces makes contact.

 

[PhanthomJay]

Also, if you look at the "joint " where the force P is applied, and draw a free body diagram of that joint, there will be a 25 N force in compression on the horizontal member of the triangle. Now if you look at "point" p, and assume a rigid wall, there is no place for that 25 N force to go except into the wall at that point. There will thus be no other normal forces from the wall between p and the hinge at O. You could remove or notch that wall between p and O and the results would be the same.

 

[adjacent]

Exactly. I don't think there is any simple way to address those forces, which are very dependent on exactly how the surfaces makes contact.

The whole surface meets the wall.I don't care whether it is simple or complex.