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# OCR M3: Rigid Objects Question 4 Watch

1. Hi, I've drawn a diagram and derived that
0.2 * R = sin alpha * X + 0.5 cos alpha * W. where R is the normal reaction force where the drum meets the rod, W is the weight alpha is the angle the rod makes with the ground. I did this by taking moments at the point where the rod touches the ground. I also know that a vertical force destroys symmetry, I am just stuck at this point. Thanks for the help.

This is the question:

Two uniform rods AB and BC, each 1 metre and W newtons in weight, are connected at P by a smooth pin joint. The rods rest symmetrically in equilibrium in contact with a smooth cylindrical drum of radius 0.8 metes with its axis fixed and horizontal. Explain why the force from the pin on each rod is horizontal, and calculate the angle between the rods.
2. Edit: they don't actually touch the ground ( the rods)
3. I think
5. (Original post by HashBrowns)
Hi, I've drawn a diagram and derived that
0.2 * R = sin alpha * X + 0.5 cos alpha * W. where R is the normal reaction force where the drum meets the rod, W is the weight alpha is the angle the rod makes with the ground. I did this by taking moments at the point where the rod touches the ground. I also know that a vertical force destroys symmetry, I am just stuck at this point. Thanks for the help.
OK, how did you get that equation first of all, as it doesn't look right.

PS, the more info the better - I don't want to play 20 questions.
6. I presume you knew why forces I was referring to? I also mentioned how I derived it
7. (Original post by ghostwalker)
OK, how did you get that equation first of all, as it doesn't look right.

PS, the more info the better - I don't want to play 20 questions.
8. (Original post by HashBrowns)
I presume you knew why forces I was referring to? I also mentioned how I derived it
Yes, sorry you did say how you'd derived it.

Though in your later post, you realised they don't actually touch the ground - which is true, they don't.

Also where does the 0.2 come from?
9. The length of the rod is 1m, and the radius is 0.8m , from the diagram the distance to R from the bottom of the rod is 1-0.8m.
10. (Original post by ghostwalker)
Yes, sorry you did say how you'd derived it.

Though in your later post, you realised they don't actually touch the ground - which is true, they don't.

Also where does the 0.2 come from?
11. (Original post by HashBrowns)
The length of the rod is 1m, and the radius is 0.8m , from the diagram the distance to R from the bottom of the rod is 1-0.8m.
Here's my drawing.
Attached Images

12. (Original post by ghostwalker)
Here's my drawing.
So if you construct two radii to the points you should see why. I think
13. Sorry I am wrong!
14. (Original post by HashBrowns)
Sorry I am wrong!
OK.

I'd resolve vertically initially, so you can find R in terms of W and the angle.

Then take moments about the pin.

PS: Don't worry about keep quoting me; I normally only look at this forum, and I check the threads I've posted on already.
15. Ok so do I take moments about the pin for separate rods or for the whole system. I ask because the question is asked before the topic of whole systems is mentioned - it is mentioned later.
16. (Original post by HashBrowns)
Ok so do I take moments about the pin for separate rods or for the whole system. I ask because the question is asked before the topic of whole systems is mentioned - it is mentioned later.
Just take moments about the pin for one of the rods.
17. So x * R = 0.5 cos a * W and R * cos a = W
18. (Original post by HashBrowns)
So x * R = 0.5 cos a * W and R * cos a = W
OK, so we now need to get rid of "x".

Can you see a triangle involving x, that would allow you to express it in terms of a, and some constant?
19. I used the sine rule for the triangle with two radii as sides two show the length of the longer side is 1.6 cos a. From there I used another triangle to show x = 0.8. Is this wrong?
20. It may help if you show me your solution.

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