maths - m1 edexcel Light Inextensible String Smooth Peg

mufc802 · 3 Jan 2012 at 16:41

There are two questions that I am unsure on.
1)

A light inextensible string passes over a smooth peg, and is 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 as shown in the diagram. The system is in limiting equilibrium with the part of the string between the ring and the peg making an angle of 60° with the vertical wire.
Calculate the coefficient of friction between the ring and the wire, giving your answer to three significant figures.

2)

A light inextensible string passes over a smooth peg, and is attached at one end to a particle of mass 3m kg and at the other end to a ring of mass 2m kg. The ring is threaded on a rough vertical wire as shown in the diagram. The system is in limiting equilibrium with the part of the string between the ring and the peg making an angle of 30° with the vertical wire.
Calculate the coefficient of friction between the ring and the wire, giving your answer to three significant figures.

The wording seems almost exact except for the angle and mass yet in the solution it shows the first answer with the force pointing up, but on the second one it shows the force pointing down. I get the same answer anyway it is just negative but I want to know how you know whether the force is pointing up or down. Here are the diagrams for both..
1)

2)

Killerkebab · 3 Jan 2012 at 16:56

I think the second diagram is just wrong (FN should be upwards pointing) - unless I'm totally missing something, why would friction act in the same direction as weight?

MuttsNutts · 3 Jan 2012 at 17:02

WTF?

Looks to me the complex version of the hangman hanging post and noose lol

RomanNose · 3 Jan 2012 at 17:07

Is this an exam question or out of a textbook? They usually say which direction it is traveling in as it's quite ambiguous.
To answer your question, you can do it through a series of steps to see which way it will go.
In the first diagram, the particle on the left, has lower resistive forces. This is because the tension is acting at an angle, so the particle on the left has a higher overall force downwards than the particle on the right

In the second diagram, 3mg> 2mg, so the 2mg is going to move upwards. So friction is going to act in the same direction as the weight.
There is still some confusion, not a well written question imo.

mufc802 · 3 Jan 2012 at 17:19

Yeah I had to calculate the tension which was 3mg, then find out what the frictional force was which is Tcos30 + 2mg which is 3mgcos30 + 2mg which is greater than the downwards force 2mg, so the friction must point down to keep the particle from going up. On the first one the force is less than mg, so it must point up to prevent the particle moving down. They are two questions from the book and aren't exam questions lucky, I just wanted to know how to solve it anyway.

Koom · 3 Jan 2012 at 17:19

If you tell me which year and which paper it was from I can find the mark scheme for you.

mufc802 · 3 Jan 2012 at 17:21

Koom said:
If you tell me which year and which paper it was from I can find the mark scheme for you.

Thanks but it's from the book, that diagram is part of the solution. It's not in any exam papers.

Smithy · 3 Jan 2012 at 17:27

Killerkebab said:
I think the second diagram is just wrong (FN should be upwards pointing) - unless I'm totally missing something, why would friction act in the same direction as weight?

Because the resulting vertical vector of the ring is pointing up (derived from the tension force of the string attached to the ring). Since the ring is in equilibrium, we know that there must be a counter force acting against the upward force, hence a downward frictional force.

Smithy · 3 Jan 2012 at 17:31

OP, I don't understand. Surely you answered your own question when you worked out the answer. You must have assumed the actual direction of the frictional forces when you worked out the resulting vertical component force of the ring?

mufc802 · 3 Jan 2012 at 18:04

Smithy said:
OP, I don't understand. Surely you answered your own question when you worked out the answer. You must have assumed the actual direction of the frictional forces when you worked out the resulting vertical component force of the ring?

Yeah I ended up with a negative coefficient of friction so I had the frictional force pointing the wrong way. I was just asking how you can find out the direction before working out the final answer.

Liampope · 3 Jan 2012 at 18:49

The diagrams are correct - you can quickly see that by comparing the weight of the mass of the rings with the upwards components of the tensons. e.g. in 1) you have mg down vs 0.5mg up (mg.cos60), so the ring wants to move down, hence the balancing friction will act up. In 2) you have 2mg down vs 2.6mg up (3mg.cos30), so the ring wants to move up, hence the balancing friction will act down. You can't have a negative coefficient of friction, of course.