Finding Equilibrium of Forces: A Shortcut Method

[thevinciz]

Homework Statement

The questions are in the comment box. Sorry!
https://www.img.in.th/image/V9K8Rg

Homework Equations

The Attempt at a Solution

1. I think α = 90 because BD is normal force.

and I don't know anymore

PS. sorry for bad english

 

[thevinciz]

 

[mfb]

Looks like we tried to fix the images at the same time.

1. I think α = 90 because BD is normal force.

Normal to what? What does that tell you about the geometry?

 

[thevinciz]

Looks like we tried to fix the images at the same time.

Normal to what? What does that tell you about the geometry?

I think it is a normal force (from hemisphere) acting on the bar, so ABD which is angle of normal force should be 90 degree.

 

[mfb]

I agree.

 

[kuruman]

Can you answer (2) now? Hint: Draw a good diagram, to scale if you can.

 

[thevinciz]

Can you answer (2) now? Hint: Draw a good diagram, to scale if you can.

If I call a point (located at middle of BC) O, I think OA and OB are radius. Because OA and OB have same length so BAD = ABC. Is this correct?

 

[haruspex]

If I call a point (located at middle of BC) O, I think OA and OB are radius. Because OA and OB have same length so BAD = ABC. Is this correct?

Right, but exactly what is your reasoning?

 

[thevinciz]

Right, but exactly what is your reasoning?

I'm sorry I don't understand what you mean

 

[haruspex]

I'm sorry I don't understand what you mean

You are right that BAD=ABC, but I do not see how you deduced that. Might be something simple I am missing.

 

[mfb]

OA and OB having the same length tells you that BAO and ABO=ABC are the same. There is a step missing.

 

[thevinciz]

You are right that BAD=ABC, but I do not see how you deduced that. Might be something simple I am missing.

AD and BD are normal force which are perpendicular to a tangent line of hemisphere. Therefore AO and BO are radius of the hemisphere.

 

[thevinciz]

OA and OB having the same length tells you that BAO and ABO=ABC are the same. There is a step missing.

I just explain it above this comment, could you check it please?

sorry for bad English

 

[thevinciz]

Could you guys help me the question 3-5 please

 

[haruspex]

AD and BD are normal force which are perpendicular to a tangent line of hemisphere. Therefore AO and BO are radius of the hemisphere.

Ah yes, I see. Except, BD is not perpendicular to the tangent. It is perpendicular to the rod.

 

[haruspex]

Could you guys help me the question 3-5 please

For 3, try to draw the diagram specific to that case.
Where is A now?

 

[thevinciz]

Ah yes, I see. Except, BD is not perpendicular to the tangent. It is perpendicular to the rod.

Thanks , I made a mistake

 

[kuruman]

Could you guys help me the question 3-5 please

The key to this problem is this. Consider the point of intersection of segments AD and CE. What's special about it? Then either draw to-scale diagrams specific to each case as @haruspex suggested or derive a general expression for the ratio of the length to diameter in terms of angle $\theta$ and see what it is for each angle.

 

[haruspex]

Consider the point of intersection of segments AD and CE.

I believe thevinciz did that in determining that OA and OB are radii.
Ah, I read your post as intersection of AD and CB. You really did mean CE?

 

[kuruman]

I believe thevinciz did that in determining that OA and OB are radii.

Indeed, but what about OD?

 

[haruspex]

Indeed, but what about OD?

Yes OD is interesting, but please see my edit to post #19.

 

[kuruman]

I believe thevinciz did that in determining that OA and OB are radii.
Ah, I read your post as intersection of AD and CB. You really did mean CE?

I read the edit. I meant the intersection of AD and CB. Note that the normal force at A is centripetal ...

 

[thevinciz]

For the question3, I have found that if θ=45° (β=45° too because β=θ) this configuration is impossible.

Because we will have AOB = 90° and the nearby angle 90° (the angle of above triangle that is located on AD) but the above triangle has 90° angle already, so the other angle will be 0°. Therefore it is impossible.

Is this correct. Sorry for bad English again. I'm trying my best

 

[haruspex]

I meant the intersection of AD and CB

I was hoping you meant that. Thanks for confirming.

 

[haruspex]

we will have AOB = 90°

Yes.

the angle of above triangle that is located on AD

You've lost me. Can you explain using the names of the points? Name some more points if necessary.

 

[thevinciz]

Yes.

You've lost me. Can you explain using the names of the points? Name some more points if necessary.

I'm trying to insert a picture, sorry.

 

[haruspex]

I'm trying to insert a picture, sorry.

Can you just say where it puts D in relation to the rod?

 

[thevinciz]

Can you just say where it puts D in relation to the rod?

I don't know

 

[thevinciz]

 

[kuruman]

I don't know

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.

 

[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.