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# M2 Rigid bodies help watch

1. The diagram shows a uniform rod AB of weight 2w and lenth 2a which is smoothly hinged to a fixed support at A. A light inextensible string is attached to a smooth ring which is free to slide on the rod. The string passes over a smooth pulley at C, a distance 3a vertically above A and carries on its end a particle of weight w. The rod makes an angle theta with the vertical.

The system is in equilibrium with the reaction and tension perpendicular to the rod.

a). show that tan theta is 3/2
b). find the magnitude of the reaction at A

please help, I cant resolve the forces correctly. Diagram is attached, the picture is correct, only Ive added the forces that are acting in my opinion in orange.
Attached Images

2. (Original post by F=ma is strong)
The diagram shows a uniform rod AB of weight 2w and lenth 2a which is smoothly hinged to a fixed support at A. A light inextensible string is attached to a smooth ring which is free to slide on the rod. The string passes over a smooth pulley at C, a distance 3a vertically above A and carries on its end a particle of weight w. The rod makes an angle theta with the vertical.

The system is in equilibrium with the reaction and tension perpendicular to the rod.

a). show that tan theta is 3/2
b). find the magnitude of the reaction at A

please help, I cant resolve the forces correctly. Diagram is attached, the picture is correct, only Ive added the forces that are acting in my opinion in orange.
i am currently studying M2 also. The Reaction force at A i am not sure about if its going the right direction, because shouldn't it be the reaction of the hinge (on a wall ?). this is a really hard question. sorry I can tbe more help i have had a go but no idea.
i am currently studying M2 also. The Reaction force at A i am not sure about if its going the right direction, because shouldn't it be the reaction of the hinge (on a wall ?). this is a really hard question.
for hinges, the reaction is always at an angle, so you have to split it into horizontal and vertical components. Im starting to think the reaction at the ring acts in the opposite direction because what is stopping the rod at b from going up? the ring has no mass so the reaction ,ust acts downwards relative to the rod to keep it in equilibirum
4. (Original post by F=ma is strong)
for hinges, the reaction is always at an angle, so you have to split it into horizontal and vertical components. Im starting to think the reaction at the ring acts in the opposite direction because what is stopping the rod at b from going up? the ring has no mass so the reaction ,ust acts downwards relative to the rod to keep it in equilibirum
ok cracked the first half woop woop
give me one moment just need to swap gadgets to add a picture of my basic diagram
ok cracked the first half woop woop
give me one moment just need to swap gadgets to add a picture of my basic diagram
ok thank you
6. from this we know that t= w because the string is the same either side of the pulley (M1)
There is no reaction of the ring at B so no force. Taking moments at A we get
Hen draw a triangle and bam !!!
Tan theta =3/2

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7. Taking the moments at A means you can ignore any forces caused by the hinge and make sure that your forces are at right angles to the rod hope that helps

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Taking the moments at A means you can ignore any forces caused by the hinge and make sure that your forces are at right angles to the rod hope that helps

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thanks for that, the only thing Im not sure about is your third picture because you have a hypotenuse of 3 with adjacents and opposites of 2 and 3. This doesnt obey pythagoras
9. (Original post by F=ma is strong)
thanks for that, the only thing Im not sure about is your third picture because you have a hypotenuse of 3 with adjacents and opposites of 2 and 3. This doesnt obey pythagoras
Oh dear in my excitement to have actually understood M2 I missed that now I have no idea sorry

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10. Which exam board is this ? Is this an exam question or from a textbook ?

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Oh dear in my excitement to have actually understood M2 I missed that now I have no idea sorry

Posted from TSR Mobile
no worries. I think you may have to consider the rod and ring as two seperate systems in equilibirum. Therefore the weight on the string does not affect the rod, it's the tension and reaction of the ring. If we can resolve the forces correctly, I think we've got it but thats the trickiest part
12. (Original post by F=ma is strong)
no worries. I think you may have to consider the rod and ring as two seperate systems in equilibirum. Therefore the weight on the string does not affect the rod, it's the tension and reaction of the ring. If we can resolve the forces correctly, I think we've got it but thats the trickiest part
edexcel oxford -oxbox textbook, one of the stretch questions haha
13. (Original post by F=ma is strong)
edexcel oxford -oxbox textbook, one of the stretch questions haha
haha i can tell only the Heinemann textbooks seem to have solution banks
haha i can tell only the Heinemann textbooks seem to have solution banks
yeah so frustrating!

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Updated: February 14, 2016
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