M1 Exercise 4D HELP
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?
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?
Scroll to see replies
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
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
(edited 13 years ago)
Why not use the books CD which has worked answers?
Original post by kirino1
I dont have the CD 
thats the problem.
the school gave me the book and took the CD out
thats the problem.
the school gave me the book and took the CD out
Torrent it
Original post by kirino1
anyone. i would really appreciate help on the question
Im in the process of doin it
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
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
(edited 13 years ago)
T in string = mg
mgcos60=Fr (solving vertically)
R=mgsin60 (solving horizontally)
(MU)=mgcos60/mgsin60
mgcos60=Fr (solving vertically)
R=mgsin60 (solving horizontally)
(MU)=mgcos60/mgsin60
(edited 13 years ago)
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
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
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
Original post by gunmetalpanda
Why m though, shouldn't the tension be mg
My bad sorry mg.
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.
Shouldn't the normal reaction force pointing vertically upwards and the frictional force horizontally left?
Can anyone help me on this? Many thanks.
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.
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.
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
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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