# Oxford PAT 2014, Question 19

## 48 thoughts on “Oxford PAT 2014, Question 19”

1. Sam says:

Hi there,
I have a question about the friction acting on the mass on the table. Why do we not include the coefficient of static friction in the equation when acceleration is greater than zero. I get that dynamic friction should be considered whilst it is actually moving but surely to begin any kind of movement you would also need to overcome the static friction too? Hence to achieve an equation for the overall acceleration of the mass I do not see how the static friction can not be included?

Thanks,
Sam

1. Sam says:

I think I may have actually justified to myself why the static friction coefficient is not used in this case. However just to clarify, if I had a box that I was pushing with a total force F and the box had a static and dynamic friction, would I have to factor in both of these to find the acceleration of the box?

Total force = F
Mass = M
Dynamic = Mud
Static = Mus

Ma= F-MusMg-MudMg

Would you consider both types of friction like this, since you would initially have to overcome static before you even consider dynamic???

1. No there is never an equation that involves the sum of both μs and μd. Have a look at https://en.wikipedia.org/wiki/Friction, which has some practical details including a graph of static and dynamic frictional forces. When the mass is stationary (i.e. the velocity and acceleration with respect to the surface it is touching is zero) the frictional force on it is mgμs, and when the mass is moving (with respect to the surface it is touching) the frictional force on it is mgμd.

2. In part c) why have you labelled the centripetal acceleration w^2r as pointing outwards? I labelled it as pointing inwards and got the same values for rmin and rmax as you but both multiplied by minus 1.
Can you please explain why the acceleration is outward?

1. I haven’t labelled the centripetal acceleration as pointing outwards — I’ve labelled it pointing inwards in both cases. But at rmax the mass itself accelerates outwards. For an explanation of why, look at the earlier comments.

3. haksaw says:

Hi, I’ve been through the comments, but I’m still a bit confused to why the coefficient of static friction is used. Surely if you want an object to actually move, the force must be at least greater the coefficient of dynamic friction, else, in the case where the coefficient of dynamic friction is greater than the coefficient of static friction, and tension is between the coefficient of dynamic friction and static friction, the block won’t actually ever move. I feel the use of it is justified in b) or wherever it asks for the condition for acceleration, since even if the coefficient of dynamic friction was greater than the coefficient of static friction, the block would accelerate infinitesimally , but I don’t see why it is used in c) where it implies constant “moving”. Sorry if I have misunderstood anything, it’d be nice to get clarification on this.
Thanks!

1. haksaw says:

Also this blog has been really helpful! Its a pain that I discovered it 2 days before my actual test date. 😦

2. > Surely if you want an object to actually move, the force must be at least greater the coefficient of dynamic friction

No, the coefficient of dynamic friction only applies when the object is actually moving. At the point where it starts moving this is because the static friction has been overcome — normally the dynamic friction is less than the static friction.

1. haksaw says:

Thanks for the prompt reply. I’d still be inclined to think that the object wouldn’t really be moving if m1gUd<T<m1gUs, since the moment T> m1gUs, it will undergo an acceleration. However that acceleration won’t last long at all and no longer exists as soon as any tiny velocity is gained, since immediately m1gUd>T which implies that any slight change in velocity due to a slight acceleration is essentially counteracted and can be regarded to be insignificant, as you would immediately have a deceleration? I’m probably over-complicating this a bit much :s

1. I think the question is just assuming that μd is less than μs, because this is the standard case that always applies apart from in esoteric situations. Maybe I’m wrong in this case, but I doubt it as that would be the normal way of working.

1. haksaw says:

Thanks for all the help! Didn’t really know much about how the frictions compare. I do feel that a lot of PAT questions can be quite vague and open to different interpretations, hope tomorrow goes alright. Do you have any tips for last minute brushing up?

4. Shan says:

Hi, I have a question: what significance does the last part in question (c) have? The part that mentions about m1 not moving radially for r(min)<r<r(max).
Thank you!

1. Look at the comment and answer below and see if that helps.

5. ian says:

Hi Oxford PAT,

I am just wondering, why is it at r max, the friction acts to the left, but at r min, friction acts to the right? thanks!

1. Normally I would not recommend using your physical intuition, because it’s so easy to mislead yourself, but I think circular motion is one thing that it is worth relating back to your own personal experience. Imagine m1 is you and m2 is a big heavy weight that you are holding up with a rope, and you are standing on a roundabout. When you are near the centre, the effect of circular motion is very small and if the roundabout is quite slippery you will be dragged to the centre, so whatever friction is acting on you is directed outwards because it is working against motion. When you are a long way from the centre the effect of circular motion is very big and if the roundabout is quite slippery and the sack isn’t heavy enough you won’t be pulled hard enough to stay moving in a circle and you will drift away from the centre, so whatever friction is acting on you is directed inwards because it is always working against motion.

The more formally-acceptable answer would be to look at the equation (2) in both cases, having replaced T with_m2g_ (using equation (1)). It should be clear that m1, m2, the magnitude of μs, g, and ω are all constants and the only value that is changing is r. So there are two values of r that would satisfy the equation: one smaller one where friction is subtracted from T and one bigger one where friction is added to T.

I think that to get a good handle on this kind of question you need to be able to feel comfortable with both the intuitive and the formal approaches. Obviously the mathematical approach is the one that matters but sometimes the intuition can be a good pointer to help you to generate the necessary equation.

1. I see. I figured the mathematical approach this morning, but the intuition one definitely also helps me. Thanks!

2. Bulbasaur10 says:

Why would you be dragged into the centre if ‘the effect of circular motion’ is small? I’m sorry if this is a stupid question I’m just having trouble thinking this through. From what I can see, said effect is due to the centripetal force on m1 – the sum of friction and tension acting on it, so as tension is constant, friction must be smaller but I am finding it hard to tell which direction it acts in…

1. When you are close to the centre, r is small, so the centripetal acceleration ω²r is also small. You would be dragged towards the centre because you are pulled towards it by the rope by the gravitational force acting on the mass m2. Maybe it’s worth having a look at https://oxfordpat.wordpress.com/circular-motion-and-centripetal-acceleration/ and https://oxfordpat.wordpress.com/oxford-pat-2012-question-17/ to help yourself understand the principles behind circular motion problems. After that you could revisit this one…

1. Bulbasaur10 says:

I’m probably just picturing this wrong but I’d just like to ask a bit more. Thank you for the prompt response though! This site is very useful.

I am a tad confused about why a smaller centripetal acceleration closer to the centre would result in the mass being pulled further inward… Am I picturing the situation wrong? Is the pulley at the centre like I’m imagining? Again I am very sorry if this is a stupid question.

1. There are no stupid questions! You’re right to be confused — circular motion is a confusing thing until you have got it worked out in your head. Together, the links I posted higher up bring out the two key points you need to remember about circular motion.

Firstly, just because of the geometry of the situation, when any object is moving with angular velocity ω a distance r from the centre of the circle, it is accelerating towards the centre with acceleration ω²r.

Secondly, when you are considering the forces on a mass in circular motion, you need to remember that F = ma, so the resultant force on the mass corresponds centripetal acceleration multiplied by the mass.

For example, if I am tethered to the centre of a circle by a rope and rotating about the centre with angular velocity ω a distance r from the centre of the circle, then if my mass is m, because F = ma, there must be a tension in the rope which equals mω²r. As r increases, the tension in the rope increases.

Maybe it’s worth you trying to look over the answer to this question, bearing in mind these two principles.

1. Bulbasaur10 says:

Wait so the further in you go, the less the tension is, meaning the string can no longer support M2 as its weight is greater than the tension, so M2 is ‘pulled along with it? So the fact that the centripetal force is smaller actually leads to the mass moving inwards and means there is friction which now acts against this?

Thank you so much for your help.

1. Well, kind of. In my comment above I was describing a simpler example in which a mass is tethered to the centre of the circle, so don’t get too hung up on that.

One other way of trying to aid your understanding is to compare the diagrams in section (b) and (the first part of) section (c), and notice that they are absolutely identical, apart from the centripetal acceleration of m1 in (c).

In general, when you are analyzing any problem with constant-angular-velocity circular motion, all you need to do is treat the problem as though there were no circular motion, and then add in the centripetal acceleration. Then use F = ma. For everything to stay still (apart from the fact that its all rotating, of course), the resultant force inwards minus the centripetal acceleration times the mass (m₁ω²r in this case) must be zero. If it’s greater than zero, everything will start moving towards the centre; if it’s less than zero, everything will start moving away from the centre.

In this particular problem the question is a bit more subtle because you need to remember that firctional forces always act to resist motion, so the direction of the friction is different in the two cases in section (c) that you are asked to consider.

6. Rickardo says:

(c) is it because the object is not radially that we are solving the question at equillibirum so we can equation the horizontal forces and vertical
because initally i was equation T=m2A and mromega^2-T-Udmg=M1a for my second equation
thanks also posted about the circuit question on the 2015 question that you said youd get back to once you were home

1. I think basically yes you are right — because the object is not moving radially, the acceleration is zero so all the acceleration terms disappear from the equations.

The form of your equations looks about right, but I’m not sure what some of your symbols mean in the comment above: what is A, and what is Ud?

1. Rickardo says:

sorry both the A should have been “g” and Ud was the dynamic co-efficent of friction
lIke you said i also think because the object is not moving radially (which i guess means along the radius tot he centre) we instead take the equations as it’s just about to move, so my Ud whould have been Us (static friction) and i equation the horizontal and vertical

Could you please help me with the 2012 mechanics question as i am completly lost on the last part. Is there anywhere else i could talk to you easier like twitter or email?

Thank you for all your help. You really help people like me who have not much other help to go with.

1. Rickardo says:

i equate the horizontal and vertical* is what i meant to say

1. I think you are basically doing it the right way, but I’m not sure I like the language ‘equating the horizontal and vertical’ — we’re just getting two equations for T and then using them to generate another equation without T in it. Also, check that you have got your signs right in the second equation: F = ma, so Tension – Friction = mass * centripetal acceleration so T – m1gUs = mw^2r, so mw^2r – T + m1gUs = 0, but you have got a minus for the m1gUs term.

2. If you have a problem with the 2012 question then make a comment on that page.

7. claire says:

What does rotating about the vertical axis actually mean? I thought it means m1 is now on an inclined plane, but that doesn’t seem right…

1. Hi Claire — the question says “rotating about the vertical axis going through the middle of the table”. This just means that the axis of rotation is vertical and the centre of the rotation is the middle of the table — so the table stays horizontal but starts to rotate about its central point.

8. Harry says:

On the very last part of the last question, it says m1 will not be moving radially
What does this mean?
and what is the difference between moving horizontally and radially- are they the same?
and how is m1 moving in this case?

1. Moving radially means moving along the radius (i.e. moving along the line that radiates from the axis about which the table is rotating). It is not the same as moving horizontally, because motion perpendicular to the radius is also possible in the horizontal plane and, in fact, the mass m1 is moving horizontally (but not radially) because the table is rotating.

1. Harry says:

ok thank you 🙂

9. Clemens says:

Hi Oxford PAT

In part (b), when including friction into your calculations, why does your expression for tension include μd for dynamic friction only? Arguably, the tension can exist even when acceleration=0. In which case, wouldn’t the static friction also exist in such a case. Thus, shouldn’t the μ in the expression be general (i.e. can either be μs or μd).

Thanks

1. Hi Clemens. I’m not 100% sure what you mean here, so maybe you can clarify your question.

Just to clarify what I have written: I’ve said that the frictional force is μsm1g (when the mass m1 is stationary) or μdm1g when a > 0 (i.e. when the mass is moving — of course the mass has no initial velocity and so it can only start moving when a > 0). I can’t see how this is different from what you are suggesting.

1. Clemens says:

Oh I apologize. I did not realize that was the working you were going for. If that is the case, a few follow up questions just to understand your working better.

Hi Clemens — sorry for the late reply, I posted a response to this a couple of days ago but it has vanished 😦

1) In your workings for (b), you took into account the two cases when a=0 and when a>0, then constructed the free-body equations. Why isn’t it possible instead to construct general free-body equations for a and T first then apply the two cases (a=0,a>0) to find the conditions. I did exactly that yet got differing equations to you.

Yes it’s true that you can consider the general case where the frictional force actually is equal to μm1g, and then substitute in the values μs or μd depending on whether the mass is static or moving. But, it’s important to remember that the actual frictional force might be less than μm1g, which brings us to …

2) I noticed that when a=0, your equation simplifies to μsm1 ≥m2. I’m a bit confused as to how the LHS can be more than the RHS. Cause arguably, if the frictional force is higher than the tension downwards, you would get a net force to the right of m1. This would mean that the m1 will move to the right, which doesn’t make physical sense. Thus, shouldn’t the simplified equation be μsm1=m2?

Thanks again for the help!

You’re right that a=0 is true if and only if μsm1 ≥ m2. But μsm1g is the maximum frictional force, not the actual force acting on the mass m1. When a=0, the actual frictional force acting on m1 is m2g. I’m afraid my labelling of the diagram is a bit lazy because I’ve shown a force arrow labelled with μsm1g, so I hope these further details help.

10. There is a coefficient of friction that applies to a mass in contact with the table. When the mass is stationary on the table, the coefficient is mu(s), and when it is moving over the table the coefficient is mu(d). But, whether it is static or dynamic, the frictional only applies to a mass that is pressing against the table. The mass m2 doesn’t touch the table so no frictional force applies to it.

1. Yip Shing says:

Hi Oxford PAT

In part (b), why is the frictional force only μdm1g, instead of (μsm1g+μdm1g)N when a>0? Is it because when the m1 block is moving, the frictional force is only μdm1g? Thanks

1. Hi — I put your later corrections into this question. Yes you are right, the frictional force is only μdm1g when the block is moving. This is what μs and μd are defined to be: the coefficient of friction is μs in the stationary case, and μd in the dynamic case. See a link like this for more details: http://www.roymech.co.uk/Useful_Tables/Tribology/co_of_frict.htm

11. Yip Shing says:

For part b), why there is no mu(d)m2g acting on the vertical mass(m2) assuming the mass is accelerating downward?
Thank you

1. Hi, thanks for your question. In the wording of part (b), the question asks us to consider friction for the table but not for the pulley, so the only source of frictional force is from the contact between a mass and the table. The mass m2 is not in contact with the table so we won’t get any frictional forces involving m2.

If there were some friction acting in the pulley then it would increase as m2 increased, but modelling the friction in a pulley is more complicated than modelling the friction due to a horizontal plane, like the table.

1. Yip Shing says:

I thought the frictional force mu(d)m2g is depending whether the mass is at rest? So when m2 is moving, there will be a frictional force mu(d)m2g acting on it?
Thank you!

12. Harry says:

i have read the link. it was helpful but I don’t understand how the tension and friction can pull it towards the axis when they are acting in the opposite direction to this motion?

1. Here’s a list of statements that explain why the mass behaves as it does. Have a look at each one and try to satisfy yourself that each stage of the argument is true.
1. The mass is rotating about the axis and its centripetal acceleration must be proportional to its distance from the axis, so the further the mass gets from the axis, the bigger its acceleration has to be if it is going to keep going round in a circle.
2. The tension and friction are pulling the mass towards the axis, and they are related to the centripetal acceleration by the equation F = m a.
3. But there is a limit to the size of the tension-plus-friction force: the tension is a constant (it is m2 g) and the friction can’t get any bigger than m1 g mu(s).
4. So there must be a corresponding limit to the size of the acceleration towards the axis.
5. But (remember point (1)) as the mass gets further from the axis the acceleration has to get bigger, so, once the mass gets to a certain distance from the axis, it can’t accelerate towards the axis enough to keep moving in a circle round the axis.
6. Once the mass has got to this point it is still accelerating towards the axis, but not quite enough to keep going in a circle, so its distance from the axis slowly starts to increase.
7. If you think about the table immediately underneath the mass, this is still moving in a circle and so it is accelerating towards the axis faster than the mass is, so the mass will start to slide across the table.

13. Hi Harry, thanks for your question.

For part b, first we work out that T = m2 g, second we observe that the mass m1 won’t move unless T > m1 g mu(s). The system won’t move unless the mass m1 moves, so, substituting m2 for T, the system won’t move unless m2 g > m1 g mu(s), i.e. when m2 > m1 mu(s).

In part c, consider the forces on the mass m1 and apply F = ma. This gives you Tension + Friction = m1 w^2 r. As r increases the frictional force increases up to the point where we overcome the friction. At that point the mass m1 starts moving away from the hole, and this means that it pulls the mass m2 upwards. Hope this helps.

1. Harry says:

actually for part b, you have written T < m1 g mu(s) not greater than and for part c how can acceleration occur in the opposite direction to motion- i.e when m2 is coming up m1 moves to the right but tension and friction and the resultant m1 w^2 r are all facing the left hand side?

1. _actually for part b, you have written T < m1 g mu(s) not greater than _

No I haven’t — I’ve written a (acceleration) = 0 when T <= mu(s) m1 g, i.e. the mass m1 won’t move unless T > m1 m mu(s).

for part c how can acceleration occur in the opposite direction to motion- i.e when m2 is coming up m1 moves to the right but tension and friction and the resultant m1 w^2 r are all facing the left hand side?

Because m1 is moving in a circle about the vertical axis that passes though m2, so it is accelerating towards that axis; it is being pulled towards the axis by the tension in the string plus the friction. This is actually quite a subtle circular motion question. Maybe you should take a look at some of the material on circular motion here in order to get your head around circular motion and then come back to it.

14. Harry says:

for part b why is m2 greater than mu(s)*m1 because if downwards is the positive direction then the inequality sign doesn’t flip? and for part c, why is m2 going to accelerate upwards at r max?