The forum discussion revolves around solving a physics problem involving equilibrium of forces on a rod positioned on a hemisphere. Participants analyze the geometry of the situation, particularly focusing on angles and forces acting on points A, B, C, D, and E. Key insights include the identification of normal forces and the necessity of drawing accurate diagrams to visualize the problem. The conclusion emphasizes that understanding the relationships between angles and forces is crucial for solving the problem effectively.
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Normal to what? What does that tell you about the geometry?thevinciz said:
mfb said:
kuruman said:
Right, but exactly what is your reasoning?thevinciz said:
haruspex said:
You are right that BAD=ABC, but I do not see how you deduced that. Might be something simple I am missing.thevinciz said:
haruspex said:
mfb said:
Ah yes, I see. Except, BD is not perpendicular to the tangent. It is perpendicular to the rod.thevinciz said:
For 3, try to draw the diagram specific to that case.thevinciz said:
haruspex said:
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.thevinciz said:
I believe thevinciz did that in determining that OA and OB are radii.kuruman said:
Indeed, but what about OD?haruspex said:
Yes OD is interesting, but please see my edit to post #19.kuruman said:
I read the edit. I meant the intersection of AD and CB. Note that the normal force at A is centripetal ...haruspex said:
I was hoping you meant that. Thanks for confirming.kuruman said:
Yes.thevinciz said:
You've lost me. Can you explain using the names of the points? Name some more points if necessary.thevinciz said:
haruspex said:
Can you just say where it puts D in relation to the rod?thevinciz said:
haruspex said:
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 said:
