Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    Hi
    Ive been stuck on this question for days! its an edexcel M1 question from chapter 4 mixed exercises. Its question 21 (and 22 is also similar)
    I would be so thankful if anyone could help me answer it - im so stuck, it depresses me.
    The question is:

    21) A light inextensible string passes over a smooth peg, and it attached at one end to a particle of mass m kg and at the other end to a ring also of mass m kg. The ring is threaded on a rough vertical wire. The system is in limiting equilibrium with part of the string between the ring and the peg making an angle of 60degrees with the vertical wire.
    Calculate the coefficient of friction between the ring and the wire, giving your answer to three significant figures.

    (answer at the back of the book = 0.577)

    How do i answer this question?
    Offline

    4
    ReputationRep:
    Have you drawn a diagram to represent the situation and have you drawn all the forces etc?
    • Thread Starter
    Offline

    0
    ReputationRep:
    Yes i have.

    Im so depressed right now - these type of questions irritate me.

    what do i do next?

    ive uploaded what the situation looks like in the book
    Attached Images
     
    Offline

    14
    ReputationRep:
    Why not use the books CD which has worked answers?
    • Thread Starter
    Offline

    0
    ReputationRep:
    I dont have the CD
    thats the problem.
    the school gave me the book and took the CD out
    Offline

    12
    ReputationRep:
    (Original post by kirino1)
    I dont have the CD
    thats the problem.
    the school gave me the book and took the CD out
    Torrent it
    • Thread Starter
    Offline

    0
    ReputationRep:
    tried and failed.
    many times.
    Offline

    12
    ReputationRep:
    http://thepiratebay.org/torrent/5833...el_M1_LiveText

    Used this 1
    • Thread Starter
    Offline

    0
    ReputationRep:
    wont work on my comp
    • Thread Starter
    Offline

    0
    ReputationRep:
    anyone. i would really appreciate help on the question
    Offline

    12
    ReputationRep:
    (Original post by kirino1)
    anyone. i would really appreciate help on the question
    Im in the process of doin it
    Offline

    1
    ReputationRep:
    Never seen a question like this before. I was on AQA though.
    I've had a play around with it and I've got the right answer.

    The forces acting on the ring are the friction which is pointing horizontally left, the reaction force directly upwards, the weight directly downwards and the tension (=mg since both tension forces must be equal) upwards at the 60 degree angle.

    The Reaction force is -mgcos60 + mg (resolve forces vertically)
    Therefore the friction is Z(-mgcos60 + mg)

    Friction is also equal to Tsin60 = mgsin60. (resolve horizontally)

    Therefore mgsin60 = mg(cos60Z + Z)
    sin60 = Z(cos60 + 1)
    Z = ((sin60)/cos60+1)
    =0.577

    EDIT: I've called the coef of friction Z since the greek letter I used came up as quesiton marks
    Offline

    0
    ReputationRep:
    T in string = mg
    mgcos60=Fr (solving vertically)
    R=mgsin60 (solving horizontally)

    (MU)=mgcos60/mgsin60
    Offline

    12
    ReputationRep:
    (Original post by kirino1)
    anyone. i would really appreciate help on the question
    Ok so I've some how got the answer..here's how

    First you wanna split the middle part into two triangles.
    The horizontal component of on the ring will therefore be mcos30.
    SInce this is perpendicular to the plane this is the reaction force.
    Now you wanna find the friction thingy on the vertical string not the string on the pully. If the system was not in rest the ring would move up the vertical string (logic).
    Therefore friction acts downwards of the vertical string and force acts up. Now you want the force that acts up the string. This is found from resolving the forces on the top triangle. If you resolve it the verticale force is msin30. m being the mass of that black particle. Since the ring is stationary, mewR=msin30. (mew is coefficient of friction). But you know R is mcos30 as explained above. so Mewmcos30=msin30.
    Cancel the m out from both sides. gives sin30=Mewcos30. Mew=sin30/cos30

    Which gives 0.577.


    Hope it helps and tanks for posting this got me thinkin.
    Excuse spelling mistakes please lol
    Offline

    0
    ReputationRep:
    (Original post by Insanity514)
    Ok so I've some how got the answer..here's how

    First you wanna split the middle part into two triangles.
    The horizontal component of on the ring will therefore be mcos30.
    SInce this is perpendicular to the plane this is the reaction force.
    Now you wanna find the friction thingy on the vertical string not the string on the pully. If the system was not in rest the ring would move up the vertical string (logic).
    Therefore friction acts downwards of the vertical string and force acts up. Now you want the force that acts up the string. This is found from resolving the forces on the top triangle. If you resolve it the verticale force is msin30. m being the mass of that black particle. Since the ring is stationary, mewR=msin30. (mew is coefficient of friction). But you know R is mcos30 as explained above. so Mewmcos30=msin30.
    Cancel the m out from both sides. gives sin30=Mewcos30. Mew=sin30/cos30

    Which gives 0.577.


    Hope it helps and tanks for posting this got me thinkin.
    Excuse spelling mistakes please lol
    Why m though, shouldn't the tension be mg
    • Thread Starter
    Offline

    0
    ReputationRep:
    Thanks for the help guys!
    Offline

    12
    ReputationRep:
    (Original post by gunmetalpanda)
    Why m though, shouldn't the tension be mg
    My bad sorry mg.
    Offline

    0
    ReputationRep:
    I am confused after looking through the solution bank on the livetext CD-rom. Why is the frictional force pointing vertically upwards on the wire and the normal reaction force pointing horizontally left ?
    Shouldn't the normal reaction force pointing vertically upwards and the frictional force horizontally left?

    Can anyone help me on this? Many thanks.
    Offline

    14
    ReputationRep:
    (Original post by pearlypearl)
    I am confused after looking through the solution bank on the livetext CD-rom. Why is the frictional force pointing vertically upwards on the wire and the normal reaction force pointing horizontally left ?
    Shouldn't the normal reaction force pointing vertically upwards and the frictional force horizontally left?

    Can anyone help me on this? Many thanks.
    Ollie got the right answer by assuming what you did. I'm not sure about your examining board but there were occassional errors in my live text CD and it sounds like you did edexcel like me because of the live text cd you have.
    Offline

    0
    ReputationRep:
    (Original post by Ollie901)
    Never seen a question like this before. I was on AQA though.
    I've had a play around with it and I've got the right answer.

    The forces acting on the ring are the friction which is pointing horizontally left, the reaction force directly upwards, the weight directly downwards and the tension (=mg since both tension forces must be equal) upwards at the 60 degree angle.

    The Reaction force is -mgcos60 + mg (resolve forces vertically)
    Therefore the friction is Z(-mgcos60 + mg)

    Friction is also equal to Tsin60 = mgsin60. (resolve horizontally)

    Therefore mgsin60 = mg(cos60Z + Z)
    sin60 = Z(cos60 + 1)
    Z = ((sin60)/cos60+1)
    =0.577

    EDIT: I've called the coef of friction Z since the greek letter I used came up as quesiton marks


    How did you rearrange Z(-mgcos60+mg) to get mg(cos60Z+Z) ?
    Shouldn't it be mgZ(1-cos60) ?
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: September 28, 2011
Poll
Do you think parents should charge rent?
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.