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# mechanics 1 question ....help me plz ..its tomorrow!! watch

1. Hi!! my m1 exam is going to be tomorrow { 15/01/2010 } .... i cantlet this one part just this part c) of the question make me loose 5 MRKS !! please those reading this do clarify how to get a hold of such questions .

Thanks alot for your time and help!!

P.S : do not ignore or mind my grammar & help is needed urgently. Thanks once again!

I don't get this that why is it ( the angle) divided by 2 ??? Please do explain from A to Z.
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2. Resultant on the pulley acts down between the two Ts, therefore when you resolve, you have to do [ T x cos ( (90-alpha) / 2 ) ] x 2. Ie, you have to resolve to half way, not the full angle
3. (Original post by 2710)
Resultant on the pulley acts down between the two Ts, therefore when you resolve, you have to do [ T x cos ( (90-alpha) / 2 ) ] x 2. Ie, you have to resolve to half way, not the full angle
If that is so then why do we multiply the answer by 2?

Its really nice of you to reply!
4. (Original post by bluefire)
Hi!! my m1 exam is going to be tomorrow { 15/01/2010 } .... i cantlet this one part just this part c) of the question make me loose 5 MRKS !! please those reading this do clarify how to get a hold of such questions .

Thanks alot for your time and help!!

P.S : do not ignore or mind my grammar & help is needed urgently. Thanks once again!

I don't get this that why is it ( the angle) divided by 2 ??? Please do explain from A to Z.

i guess this is the 2009 jan paper...right? its the one i did in my exam.
i know how to get it but its difficult to explain.
you have to use the parallelogram rule. i dont know how they have done it.
force acting on both strings is T which is 6g.
the equation that i get is F^2= (T^2) + (T^2) - 2 x T x T x cos(90 + a)

from here i get 105N. i used this method in the exam and got it right
5. (Original post by bluefire)
If that is so then why do we multiply the answer by 2?

Its really nice of you to reply!
You mulitply by 2, because there are 2 tensions acting on the pulley.

6. (Original post by kosy91)
i guess this is the 2009 jan paper...right? its the one i did in my exam.
i know how to get it but its difficult to explain.
you have to use the parallelogram rule. i dont know how they have done it.
force acting on both strings is T which is 6g.
the equation that i get is F^2= (T^2) + (T^2) - 2 x T x T x cos(90 + a)

from here i get 105N. i used this method in the exam and got it right
It seems quiet mathematical, you have used cosine rule. Shouldn't your tetha (Q) value be cos (90-a)?? as your angle between the two T values is 90-a?
7. Yh (90-a) It's hard to explain without diagrams.

check out this vid : http://www.youtube.com/watch?v=rC2IQ3R5im4
8. (Original post by 2710)
You mulitply by 2, because there are 2 tensions acting on the pulley.

Does this method prooves to be accurate ? Just asking to be 100% confident of it any how i seem to have more comfartable with your method !!! it seems simple multiplied by 2 cuz there are two T and angle after subtraction from 90 is divided by 2, ....right. thnks man!
9. (Original post by bluefire)
It seems quiet mathematical, you have used cosine rule. Shouldn't your tetha (Q) value be cos (90-a)?? as your angle between the two T values is 90-a?
nope...it is 90+ a. i have not used the angle shown on the diagram, rather the one opposite the hypotenuse of the traingle. i got it by 180-(90-a)
10. bluefire check out the vid, it's easier when you can view diagrams as guidance!!
11. (Original post by bluefire)
Does this method prooves to be accurate ? Just asking to be 100% confident of it any how i seem to have more comfartable with your method !!! it seems simple multiplied by 2 cuz there are two T and angle after subtraction from 90 is divided by 2, ....right. thnks man!
but how would you explain to the examiner what you did?
12. (Original post by Kameo)
Yh (90-a) It's hard to explain without diagrams.

check out this vid : http://www.youtube.com/watch?v=rC2IQ3R5im4
are you sure about this { i do think u are right } just asking cuz my friend kosky91 wrote 90+a that is why ...i would like him to clarify too!! thnks any way!
13. He just used another angle, just use the angle that you are at ease with.
14. (Original post by kosy91)
but how would you explain to the examiner what you did?
But why to explain ..is it neccessary to explain?? am i right over here or wrong ...examiners know all kinds of methods but i am not aware of what should be done thats why i need your help my friend!
15. yeah...well we are all right....OP should use the method he is comfortable with
16. Its probably sed alredy but its times the tension in the string
17. (Original post by kosy91)
nope...it is 90+ a. i have not used the angle shown on the diagram, rather the one opposite the hypotenuse of the traingle. i got it by 180-(90-a)
yeah 180-(90-a) thats right thats what i did!!! so now i get it thanks
18. (Original post by Kameo)
Yh (90-a) It's hard to explain without diagrams.

check out this vid : http://www.youtube.com/watch?v=rC2IQ3R5im4
19. maybe this could better explain my method

EDIT: its -2T^2 cos....
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20. thanks to u too kosy91 ! but i think i will stick to the resolving method kameo showed me !! its easier for me that way!! once again thanks alot1!!

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