Hey there! Sign in to join this conversationNew here? Join for free

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

    • Thread Starter
    Offline

    0
    ReputationRep:
    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!:yes:

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

    17
    ReputationRep:
    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
    • Thread Starter
    Offline

    0
    ReputationRep:
    (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!
    • PS Helper
    Offline

    0
    ReputationRep:
    PS Helper
    (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
    Offline

    17
    ReputationRep:
    (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.

    Glad to help
    • Thread Starter
    Offline

    0
    ReputationRep:
    (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?
    Offline

    2
    ReputationRep:
    Yh (90-a) It's hard to explain without diagrams.

    check out this vid : http://www.youtube.com/watch?v=rC2IQ3R5im4
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by 2710)
    You mulitply by 2, because there are 2 tensions acting on the pulley.

    Glad to help
    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!
    • PS Helper
    Offline

    0
    ReputationRep:
    PS Helper
    (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)
    Offline

    2
    ReputationRep:
    bluefire check out the vid, it's easier when you can view diagrams as guidance!!
    • PS Helper
    Offline

    0
    ReputationRep:
    PS Helper
    (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?
    • Thread Starter
    Offline

    0
    ReputationRep:
    (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!
    Offline

    2
    ReputationRep:
    He just used another angle, just use the angle that you are at ease with.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (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!
    • PS Helper
    Offline

    0
    ReputationRep:
    PS Helper
    yeah...well we are all right....OP should use the method he is comfortable with
    Offline

    0
    ReputationRep:
    Its probably sed alredy but its times the tension in the string
    • Thread Starter
    Offline

    0
    ReputationRep:
    (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
    • Thread Starter
    Offline

    0
    ReputationRep:
    (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
    Your video was extremely helpful thanks alot!!c: C;
    • PS Helper
    Offline

    0
    ReputationRep:
    PS Helper
    maybe this could better explain my method


    EDIT: its -2T^2 cos....
    Attached Images
     
    • Thread Starter
    Offline

    0
    ReputationRep:
    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!!
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.