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Difficult M2 Question, Help needed

Hello, this question is question 10 of exercise 2D in M2 Mechanics by Brian Jefferson. There is no image to go along with it which is causing the majority of my problems.

A uniform rod AB of length 2a and weight W is hinged to a horizontal ceiling at A and is suspended by a light in extensible string BC of length a connecting B to a point C on the ceiling such that angle ABC is 90 degrees. Show that the tension in the string is W Sqrt 5, and find the horizontal and vertical components of the reaction of the hinge on the rod.

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Reply 1
These are the answers in the back of the book but I am looking for working.

Spoiler

Original post by Sail last
Hello, this question is question 10 of exercise 2D in M2 Mechanics by Brian Jefferson. There is no image to go along with it which is causing the majority of my problems.

A uniform rod AB of length 2a and weight W is hinged to a horizontal ceiling at A and is suspended by a light in extensible string BC of length a connecting B to a point C on the ceiling such that angle ABC is 90 degrees. Show that the tension in the string is W Sqrt 5, and find the horizontal and vertical components of the reaction of the hinge on the rod.


We can't post solutions ..

Try taking moments about A and post what you get.
you could take moments about A;

let the tension in the string = t

t*2a = W*aCosµ where µ is the angle BCA

from the information given we can say that tanµ = 1/2....
Reply 4
Drawing (3).png
First is this the correct drawing.

We can't post solutions ..
Reply 5
Original post by the bear
you could take moments about A;

let the tension in the string = t

t*2a = W*aCosµ where µ is the angle BCA

from the information given we can say that tanµ = 1/2....


So there is no horizontal and vertical reaction at C?
Original post by Sail last
So there is no horizontal and vertical reaction at C?


the moments are for the rod...
Original post by Sail last
Drawing (3).png
First is this the correct drawing.

We can't post solutions ..


Label A, B and C; I'd use a ruler in future.

You need to draw a perpendicular to the rod at the mid point and label the angle between the force W and this perpendicular - you need this for taking moments. If you are putting the forces at C you shoud have a T ..

Take moments about A for the rod..
(edited 8 years ago)
Reply 8
Original post by the bear
the moments are for the rod...


:smile: I have been going around in circles for hours cause of this!
Wow that was pretty silly of me, I was doing moments along the ceiling and everything.

I think I can do the question pretty easily now thanks
Reply 9
Original post by Muttley79
Label A, B and C; I'd use a ruler in future.

You need to draw a perpendicular to the rod at the mid point and label the angle between the force W and this perpendicular - you need this for taking moments. If you are putting the forces at C you shoud have a T ..

Take moments about A for the rod..


so moments about A
2T = W cosa

but from the pic cos a is 2/root 5

so 2T = 2/root 5 W
so root 5 T = W

not T = root 5 W
:s-smilie:

maybe I spoke too soon...
Original post by Sail last
so moments about A
2T = W cosa

but from the pic cos a is 2/root 5

so 2T = 2/root 5 W
so root 5 T = W

not T = root 5 W
:s-smilie:

maybe I spoke too soon...


That is what I got also. Been sat staring at the paper for 20 minutes.
Reply 11
Original post by 16Characters....
That is what I got also. Been sat staring at the paper for 20 minutes.


I'm soooo tired of this question, I have been doing it for hours. I just want to move on but I cant let myself until I understand whats going wrong.
Original post by Sail last
I'm soooo tired of this question, I have been doing it for hours. I just want to move on but I cant let myself until I understand whats going wrong.


My suggestion is to carry on to the next part of the question, you can get the answers to that question using T = W/root 5 it seems.
Reply 13
Original post by 16Characters....
My suggestion is to carry on to the next part of the question, you can get the answers to that question using T = W/root 5 it seems.


No I don't think it works because lets say we resolve it all vertically:

Tcosa + R (Horizontal reaction) = W

but T = root 5 W and cos a = 2/root 5 so:

2W + R = W
W = -R
nonsense
Original post by Sail last
No I don't think it works because lets say we resolve it all vertically:

Tcosa + R (Horizontal reaction) = W

but T = root 5 W and cos a = 2/root 5 so:

2W + R = W
W = -R
nonsense


I said carry on using our answer, that T = W /root 5 not T = W root 5.
Reply 15
Original post by 16Characters....
I said carry on using our answer, that T = W /root 5 not T = W root 5.


:angry: our result works to get the answer in the back of the book.

SO I have been wasting my time with a misprint...







:leaf::leaf::leaf::charm::leaf::leaf::leaf:
Original post by Sail last
No I don't think it works because lets say we resolve it all vertically:

Tcosa + R (Horizontal reaction) = W

but T = root 5 W and cos a = 2/root 5 so:

2W + R = W
W = -R
nonsense


Are you sure the answer is not (W root 5)/5
Original post by Sail last
No I don't think it works because lets say we resolve it all vertically:

Tcosa + R (Horizontal reaction) = W

but T = root 5 W and cos a = 2/root 5 so:

2W + R = W
W = -R
nonsense


If you are resolving vertically then R has no component and should be ignored - it;s Y the vertical component you need.

Then resolve horizontally...
Reply 18
Original post by Muttley79
If you are resolving vertically then R has no component and should be ignored - it;s Y the vertical component you need.

Then resolve horizontally...


mistype I meant vertical.

So previoulsy we found that T root5 = W not W root5 = T like it says in the question.

If you use that to do the next part then resolving Vertically (sorry) gives you W over root 5 times 2/root 5 (this is cos a) + R = W.

This gives you R = W - 2W/5 = 3W/5

So the tension is W/root 5 not W root 5

Spoiler

(edited 8 years ago)
Original post by Sail last
mistype I meant vertical.

So previoulsy we found that T root5 = W not W root5 = T like it says in the question.

If you use that to do the next part then resolving Vertically (sorry) gives you W over root 5 times 2/root 5 (this is cos a) + R = W.

This gives you R = W - 2W/5 = 3W/5

So the tension is W/root 5 not W root 5

Spoiler



You can put the final answer but you asked us to explain the solution which isn't allowed.

I also got the right answer when I resolved horizontally - did you check the answer was not what I posted above?

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