Finding Equilibrium of Forces: A Shortcut Method

[thevinciz]

We have already established that BD is perpendicular to the rod. This means that triangle ABD is a right triangle with hypotenuse AD. How is AD related to the diameter of the circle? You may not be able to see what's going on unless you draw a good diagram as I have repeatedly suggested. The one provided to you appears to be deliberately misshapen perhaps so that nothing is given away. Try redrawing it, making the circle as round as possible and the perpendiculars as "perpendicular" as possible.

Oh I see, thanks you very much, I'll try draw it.

 

[thevinciz]

We have already established that BD is perpendicular to the rod. This means that triangle ABD is a right triangle with hypotenuse AD. How is AD related to the diameter of the circle? You may not be able to see what's going on unless you draw a good diagram as I have repeatedly suggested. The one provided to you appears to be deliberately misshapen perhaps so that nothing is given away. Try redrawing it, making the circle as round as possible and the perpendiculars as "perpendicular" as possible.

I have drawn it, AD is a diameter?

 

[kuruman]

Yes! Based on that knowledge, what kind of triangle is BOD? What are its 3 angles in relation to $\theta$?

 

[thevinciz]

Yes! Based on that knowledge, what kind of triangle is BOD? What are its 3 angles in relation to $\theta$?

But I still don't know how to find AE to answer question5, could you give me more hint?

 

[haruspex]

Those 3 angles are 60°

No, it is not necessarily equilateral. BD need not equal the radius.
It would help to find the answer to my question in post #27.
If exactly three forces act on a body in equilibrium, what can you say about the lines of action of the three forces? Consider moments.

 

[thevinciz]

No, it is not necessarily equilateral. BD need not equal the radius.
It would help to find the answer to my question in post #27.
If exactly three forces act on a body in equilibrium, what can you say about the lines of action of the three forces? Consider moments.

Thanks I really made a mistake, I have deleted already.

 

[thevinciz]

No, it is not necessarily equilateral. BD need not equal the radius.
It would help to find the answer to my question in post #27.
If exactly three forces act on a body in equilibrium, what can you say about the lines of action of the three forces? Consider moments.

I will think with your method, but could you check my method please.

OD=OB (because they are radii) so ODB=OBD=45°. Then we will have θ=45°. Therefore question 3 and 4 are impossible.

 

[haruspex]

so ODB=OBD=45°

Why?
Compare DAB with DOB. There is a theorem you should know.

 

[kuruman]

OD=OB (because they are radii) so ODB=OBD=45°. Then we will have θ=45°.

It is correct that ODB = OBD which makes triangle DOB isosceles; @haruspex questioned that the angles are 45^o, which is not necessarily true. Knowing one of these angles means knowing all of them because their sum must be equal to 180^o. If you do not know or do not remember the theorem that @haruspex mentioned, do this. Draw a line perpendicular to diameter CB at point B. What is the angle formed by this perpendicular and the normal force at B in relation to angle ABC? Remember your answers to parts (1) and (2). What is this angle's relation to $\theta$?

 

[thevinciz]

Why?
Compare DAB with DOB. There is a theorem you should know.

Do you mean DOB = 2(DAB)?

 

[thevinciz]

It is correct that ODB = OBD which makes triangle DOB isosceles; @haruspex questioned that the angles are 45^o, which is not necessarily true. Knowing one of these angles means knowing all of them because their sum must be equal to 180^o. If you do not know or do not remember the theorem that @haruspex mentioned, do this. Draw a line perpendicular to diameter CB at point B. What is the angle formed by this perpendicular and the normal force at B in relation to angle ABC? Remember your answers to parts (1) and (2). What is this angle's relation to $\theta$?

Is it θ?

 

[kuruman]

Is it θ?

Are you asking me or are you telling me? What do you think?

 

[thevinciz]

Are you asking me or are you telling me? What do you think?

Actually it was a answer to your question.

 

[thevinciz]

For question4 (θ = 75/2), I still don’t see how impossible it is.

 

[kuruman]

Actually it was a answer to your question.

Great! Yes, it is $\theta$. Now you should be able to label all relevant angles as $\theta$, $2\theta$, $90^o-\theta$ or $90^o-2\theta$. Having done this, 1. Write all forces in component form.
2. Choose a convenient origin (I chose point A), write the position vectors for the force at B and the force at G.
3. Find the torque generated by each force and set the sum of torques equal to zero. This should give you a relation between the magnitude of the force at B, the weight and the ratio L/D.
4. Use the fact that the sum of all the horizontal and vertical components of forces are separately equal to zero to find an equation relating L/D to θ. You also do this by noting that since the forces add to zero, they must form a triangle which is not necessarily right, isosceles or equilateral. If you label the internal angles correctly, you should be able to use the law of sines to get the relation between L/D and θ in fewer steps.
5. Once you have the desired expression, substitute the values of 3-5 and see whether you get a reasonable value for L/D.

 

[haruspex]

Great! Yes, it is $\theta$. Now you should be able to label all relevant angles as $\theta$, $2\theta$, $90^o-\theta$ or $90^o-2\theta$. Having done this, 1. Write all forces in component form.
2. Choose a convenient origin (I chose point A), write the position vectors for the force at B and the force at G.
3. Find the torque generated by each force and set the sum of torques equal to zero. This should give you a relation between the magnitude of the force at B, the weight and the ratio L/D.
4. Use the fact that the sum of all the horizontal and vertical components of forces are separately equal to zero to find an equation relating L/D to θ. You also do this by noting that since the forces add to zero, they must form a triangle which is not necessarily right, isosceles or equilateral. If you label the internal angles correctly, you should be able to use the law of sines to get the relation between L/D and θ in fewer steps.
5. Once you have the desired expression, substitute the values of 3-5 and see whether you get a reasonable value for L/D.

Going out into details of all the forces can be avoided using my hint in post #35.

 

[kuruman]

Going out into details of all the forces can be avoided using my hint in post #35.

Yes. I thought OP may be more comfortable with the "brute force" method. Nevertheless, I did mention a shortcut under item 4, post #45 that works well.