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# M1 Exercise 4D HELP watch

1. 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?
2. Have you drawn a diagram to represent the situation and have you drawn all the forces etc?
3. 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

4. Why not use the books CD which has worked answers?
5. I dont have the CD
thats the problem.
the school gave me the book and took the CD out
6. (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
7. tried and failed.
many times.
8. wont work on my comp
9. anyone. i would really appreciate help on the question
10. (Original post by kirino1)
anyone. i would really appreciate help on the question
Im in the process of doin it
11. 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
12. T in string = mg
mgcos60=Fr (solving vertically)
R=mgsin60 (solving horizontally)

(MU)=mgcos60/mgsin60
13. (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.
14. (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.
Why m though, shouldn't the tension be mg
15. Thanks for the help guys!
16. (Original post by gunmetalpanda)
Why m though, shouldn't the tension be mg
17. 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.
18. (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.
19. (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) ?

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