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M2 Diagram for Reaction Help Needed

Here is the question and I'm on part c). I'm attaching my workings. The final answer is 4(root5)mg tan^-1 (1/2) to vertical.. I have no idea about where the reaction is and I would appreciate a diagram or some clues on how to get the correct diagram.

Also as far as part d) is concerned, if you lifted the ring wouldn't the 5m then tighten the string, pull the ring to the wire and then the ring wouldn't fall at all. Mechanics is a mystery to me.

Thanks
Reply 1
Original post by maggiehodgson
Here is the question and I'm on part c). I'm attaching my workings. The final answer is 4(root5)mg tan^-1 (1/2) to vertical.. I have no idea about where the reaction is and I would appreciate a diagram or some clues on how to get the correct diagram.

Also as far as part d) is concerned, if you lifted the ring wouldn't the 5m then tighten the string, pull the ring to the wire and then the ring wouldn't fall at all. Mechanics is a mystery to me.

Thanks


I haven't looked at d yet but for c, breaking things into horizontal and vertical components is probably the clean way to do it. The pulley has to be in equilibrium, but it has the particle and ring pulling it down, and the ring pulling it horizontally. It needs a force which balances this out, so it will, using your diagram, be a force which has "up" and "right" components, acting at the pulley. Then you just need to use the particle force which is self-explanatory and some trig to get the components of the ring force.
(edited 8 years ago)
Reply 2
Original post by maggiehodgson
Here is the question and I'm on part c). I'm attaching my workings. The final answer is 4(root5)mg tan^-1 (1/2) to vertical.. I have no idea about where the reaction is and I would appreciate a diagram or some clues on how to get the correct diagram.

Also as far as part d) is concerned, if you lifted the ring wouldn't the 5m then tighten the string, pull the ring to the wire and then the ring wouldn't fall at all. Mechanics is a mystery to me.

Thanks


Think about the forces acting on the ring when it is at the same horizontal level as the pulley. The tension in the string can only be acting horizontally on the ring, while the weight of the ring pulls it straight down. So there is nothing to keep the ring up, no matter how big the tension is.
Original post by 1 8 13 20 42
I haven't looked at d yet but for c, breaking things into horizontal and vertical components is probably the clean way to do it. The pulley has to be in equilibrium, but it has the particle and ring pulling it down, and the ring pulling it horizontally. It needs a force which balances this out, so it will, using your diagram, be a force which has "up" and "right" components, acting at the pulley. Then you just need to use the particle force which is self-explanatory and some trig to get the components of the ring force.


Ah! So Part b) was a hint! Up = 8N. Right= 4 N hence root(64+16).

I would not have thought of that.

Thank you. Now to trying to get to grips with part d).
Original post by 1 8 13 20 42
Think about the forces acting on the ring when it is at the same horizontal level as the pulley. The tension in the string can only be acting horizontally on the ring, while the weight of the ring pulls it straight down. So there is nothing to keep the ring up, no matter how big the tension is.


I've tried it and here is my attempt.

Looking forward to more clues.
Reply 5
Original post by maggiehodgson
I've tried it and here is my attempt.

Looking forward to more clues.


I haven't done this type of dynamics question in which I assume the tension of the string is constantly changing and am not really sure how to go about it with the methods I know. I may be being stupid or missing something obvious but eh, haven't done mechanics in a while. I hope somebody else more experienced can assist. :s-smilie:
Original post by 1 8 13 20 42
I haven't done this type of dynamics question in which I assume the tension of the string is constantly changing and am not really sure how to go about it with the methods I know. I may be being stupid or missing something obvious but eh, haven't done mechanics in a while. I hope somebody else more experienced can assist. :s-smilie:



Thanks. I have done all the questions in the chapter before this one and there's nothing like it alll. Let's hope someone else picks up the thread.
Original post by 1 8 13 20 42
I haven't done this type of dynamics question in which I assume the tension of the string is constantly changing and am not really sure how to go about it with the methods I know. I may be being stupid or missing something obvious but eh, haven't done mechanics in a while. I hope somebody else more experienced can assist. :s-smilie:



I just thought I'd let you know that your idea was absolutely, perfectly correct. Ghostwalker answered my next post and it confirmed your thoughts. The answer that I posted to you was nearly there, I just used the wrong hypotenuse. I had 9d/8 rather than 9d/8 + d.

So, once again, thanks.
Reply 8
Original post by maggiehodgson
I just thought I'd let you know that your idea was absolutely, perfectly correct. Ghostwalker answered my next post and it confirmed your thoughts. The answer that I posted to you was nearly there, I just used the wrong hypotenuse. I had 9d/8 rather than 9d/8 + d.

So, once again, thanks.


I didn't really have much of an idea; it was more "I don't know how to handle a question with changing tension". :redface: I did think of using the inextensible nature of the string to get distances like you seem to have done but didn't make much of an attempt (I was going out anyway to be fair). You're welcome for the other part anyhow. :tongue:

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