- #36
thevinciz
- 49
- 1
haruspex said:
Thanks I really made a mistake, I have deleted already.
haruspex said: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.
haruspex said: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.
Why?thevinciz said:so ODB=OBD=45°
It is correct that ODB = OBD which makes triangle DOB isosceles; @haruspex questioned that the angles are 45o, which is not necessarily true. Knowing one of these angles means knowing all of them because their sum must be equal to 180o. 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 said:OD=OB (because they are radii) so ODB=OBD=45°. Then we will have θ=45°.
haruspex said:Why?
Compare DAB with DOB. There is a theorem you should know.
kuruman said:It is correct that ODB = OBD which makes triangle DOB isosceles; @haruspex questioned that the angles are 45o, which is not necessarily true. Knowing one of these angles means knowing all of them because their sum must be equal to 180o. 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##?
Are you asking me or are you telling me? What do you think?thevinciz said:Is it θ?
kuruman said:Are you asking me or are you telling me? What do you think?
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.thevinciz said:Actually it was a answer to your question.
Going out into details of all the forces can be avoided using my hint in post #35.kuruman said: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.
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.haruspex said:Going out into details of all the forces can be avoided using my hint in post #35.