@JohnRennie: Hi sir. Good morning :-)

@GuruVishnu hi :-)

Are you free now sir? I'm having a doubt about coefficient of friction.

Yes I'm free. In fact I finished work early today so I'm free for the rest of the morning :-)

Ok sir :-)

Here a ball is rotating between a rough wall and a smooth inclined plane.

The following is the free body diagram of the ball:

Yes, that looks good to me.

Ok sir. On equating the forces in the vertical direction, I get:

$$\mu N_1+N_2\cos\theta=W$$

Similarly for the horizontal direction:

$$N_1=N_2\sin\theta$$

From these two, we get:

$$N_1=\frac{W}{\mu+\cot\theta}$$

Or in other words, the normal contact force exerted by the wall depends on the coefficient of friction. As $\mu$ increases, $N_1$ decreases and vice versa.

I haven't checked the algebra, but that makes sense.

We know that the coefficient of friction is independent of the normal contact force. But here the normal contact force depends on coefficient of friction. What is the reason for this?

The ball is pressed against the wall by its own weight. Yes?

Yes sir.

But the effect of the frictional force is to reduce the weight of the ball because it applies an upwards force that counteracts the downwards force due to gravity.

But if the (net) weight of the ball is reduced then the force with which it presses against the wall is also reduced.

Ok sir. I understand the apparent weight is less than the true weight due to the upward frictional force. But I think here we need to compare the degree of roughness of the wall and the normal contact force.

Or in other words, how can the wall repel away the ball when the roughness (coefficient of friction) increases?

The wall doesn't repel the ball.

Suppose the friction was very high then as soon as the rotating ball touched the wall it would fly upwards so the (net) weight would be zero. If the ball is not in contact with the slope then there is nothing pressing it against the wall so the normal force is zero.

Fine sir. I get the idea. So is it incorrect to say $N_1$ depends on $\mu$ when in general $\mu$ doesn't depend on the normal contact force?

$N_1$ and $\mu$ are related so they both depend on each other.

Although normally $\mu$ is a constant because it's a property of the surface.

So I guess I'd probably say $N_1$ depends on $\mu$ (as well as the mass of the ball and the slope angle).

Ok sir. Thank you :-)

I think the problem with my comparison of the normal force and coefficient of friction is here there are other factors disturbing these two quantities where under normal conditions only these two are present.

@JohnRennie: Hi sir :-)

Are you free now for a question?

I'm just answering another question, but I won't be long so post the question and I'll look at it as soon as I'm free.

Ok sir.

DCP-04-067-10

Question:

> Two blocks of mass 4 kg and 2 kg are connected by a heavy string and placed on rough horizontal plane. The 2 kg block is pulled with a constant force F. The coefficient of friction between the blocks and the ground is 0.5. What is the value of F so that tension in the string is constant throughout the motion of the blocks? (g=10 m/s^2)

My opinion:

If the string has mass, I don't think for any non-zero value of the applied force, the tension remains uniform throughout and it decreases with increase in distance from the 2 kg block. Could you tell whether I'm missing any details or is the situation described in the question impossible?

Let me draw a diagram:

@GuruVishnu Does that look right?

@JohnRennie Yes sir. That's how I also drew on my notebook.

When the string is massless, I understand the tension is uniform throughout but I don't think the same is true when the string has mass.

The whole system accelerates uniformly at an acceleration given by $a = F_t/M_t$ where the total force $F_t = F - 30$ and the total mass $M_t = 6 + m$.

Yes sir.

$$ a = \frac{F - 30}{6 + m} $$

Now starting at the left end with the 4kg block the total force on the block is $F_t = F_b - 20 = ma = 4a$

Ok sir.

As we move towards right, the tension in the string increases progressively as it needs to accelerate an increased mass on the left.

So $F_b = 20 + 4a$

Likewise $F_a = 20 + (4 + m)a$

Ok sir.

Hmm, yes, They can't be equal unless $m=0$.

Ok sir. Then I was correct while thinking this :-)

Thank you.

I wonder what the question means then?

Don't know. All the four options are incorrect then.

30N would mean the blocks don't move, so the tension in the string would be a steady 20N

Maybe that's what they mean i.e. the answer is (B)

In fact that has to be right doesn't it?

Yes sir. I'm able to understand your reason. So the tension in the heavy string is constant when there's zero acceleration and non zero external force. Am I right?

I hope we need to assume the string doesn't bend down due to gravity.

I know to find the value of different tension in different regions of the string - it's simple. I think this question is not clearly stated. So shall we move on to the next question sir?

Yes, if we equate $F_a$ and $F_b$ we get $20 + 4a = 20 + (4+m)a$

And for non-zero $m$ that has to mean $a = 0$

@JohnRennie Ok sir. Then it's possible only when the net applied force is less than or equal to the limiting friction.

Yes

Fine sir. Now it seems the question is very simple. Are there any other cases (beyond this problem) when the tension is uniform for non-zero accelerations?

I think we've thought more than what is needed for this one :-)

No, our equation only has two solutions $a=0$ and $m=0$.

Ok sir. But here we've equated only $F_a$ and $F_b$. Does it guarantee, the tension at all other points be the same as of this value?

I can't think of any way for the tension to be equal at the ends but not in the middle of the string ...

Ok sir.

Anyway, can you please ping me once you're free? I'm having an another doubt.

@JohnRennie: Hi sir.

May I ask one last question?

@GuruVishnu hi, yes. I'm doing a few bits and pieces, but post the question and I'll look at it as soon as I'm free.

Ok sir. Thank you.

_____________________________________________________________________________________

DCP-04-071-50

> A frame is rotating in a circle with varying speed $v=(2t-4)~\rm{m/s}$ where $t$ is in second. An object is viewed from this frame. The pseudo force:

I'm very familiar with the application of pseudo forces like centrifugal and Coriolis force. But, I felt this question lacked some details and chose "Data is insufficient", but the answer says the pseudo force is "minimum at 2 s" i.e., option (b). Could you please provide any hints? I think without the data on object's position relative to the frame's axis of rotation, it's not possible to find the centrifugal force on the object.

In the rotating frame the object appears to be rotating at $v = -(2t - 4) m/s$

Ok sir.

So if the distance to the object is $r$ then there appears to be a centripetal force $F = v^2/r = (2t - 4)^2/r$

OK so far?

So to find the minima/maxima we differentiate $dF/dt = 4(2t-4)$

Ok sir. So there's an extrema at $t=2$.

So we get a maximum/minimum at $t=2$.

Differentiate again to find out if it's a minimum of maximum $d^2F/dt^2 = +8$

So it's a minimum.

Ok sir. I understood this. But why isn't it "zero at 2 s"?

Hmm ...

It is zero at $t = 2$ ...

Yes sir.

Further, this question is from "only single correct answer" category and not "multiple correct answers" type, so we need to choose only one. I guess this question was mistyped in this section.

Or are we missing something else?

I think you're right. Both B and C are clearly correct.

Ok sir. Maybe the formula you provided for $F$ is only taking the centrifugal force into account. So if we want to justify (b), then we could say there might be some non-zero Coriolis force.

Does this reason look good?

But again another question, how do we know whether the Coriolis force is constant, or attaining maximum/minimum? Also the circular motion is also not uniform.

Oh, hang on.

There are two components to the (pseudo) acceleration, radial and tangential, and we have only calculated the radial component.

Ah. Fine. I hope it's constant however.

The tangential component is a constant $a = 2 m/s^2$.

Fine sir. So (b) is correct and the pseudo force is non-zero. Is this tangential component - the Coriolis force?

So we were correct that there is a minimum at $t = 2$, but the acceleration at that minimum is $a = \sqrt{a_r^2 + a_t^2} = \sqrt{0^2 + 2^2} = 2 m/s^2$.

So C is wrong.

Got it sir :-)

The Coriolis force is a different phenomenon.

Ok sir. So here, we're assuming the object under inspection is not moving in that rotating frame, right? There's no mention about this in the question however.

@GuruVishnu hi, sorry, I was on the phone.

The Coriolis force is seen when the radial distance is changing, and that isn't happenung here.

@JohnRennie No problem sir :-)

@JohnRennie Fine. Thank you for the clarification sir.

:-)

@JohnRennie: Hi sir. Are you having your lunch or preparing it now?

@GuruVishnu lunch is finished :-)

Ok sir. I'd like to ask what was your lunch even though I can't eat that now?

I had Dosa for my dinner. I think this name will be new for you.

I know a student who lives in Kochi, and before that Chennai, so I know what dosa are :-)

Ah. Fine :-)

Lunch today was sandwiches made with chutney and cheese and mayonnaise garnished with black pepper.

Hm. Chutney - is that an international food (side dish to be precise)? And I've to Google what is "mayonnaise"?

"It is a stable emulsion of oil, egg yolk, and an acid, either vinegar or lemon juice."

I see why a Colloid scientist loves to eat "mayonnaise" :-)

@GuruVishnu chutney is Indian isn't it? I have the idea it entered British cuisine from India.

@JohnRennie Yes sir. We have chutneys here. But only today I came to know you're also familiar with that. There are many types of it :-)

Which type did you have? Like coconut, tomato, etc.?

My father spent several years in India during the second world war, and he loved Indian food and used to cook it for us.

Though he was stationed in what is now Pakistan so the cooking he knows is Punjabi.

That's great sir!

This was the chutney I used in my sandwiches

:-)

I thought you made it yourselves.

No, I'm too lazy to make that much effort :-)

I'm sure some people do make their own chutneys, but not me.

Haha :-)

There is a restaurant in Chester that does Keralan food, though it's closed at the moment because of the lockdown.

I see. I haven't been to UK. But I've also seen some Indian restaurants in countries like South Korea, and USA.

And I've also seen other countries' restaurants here :-)

@JohnRennie: Are you free now for a small doubt sir?

Yes.

Here we pull the top part of the string with a speed $v$.

I'm having a great confusion whether, the speed of the block is $v\cos\theta$ or $v/\cos\theta$. Could you please help me out, sir?

Remember what I said about this type of problem. Write an equation that links th block position to the string length, then differentiate both sides.

Let's simplify the diagram. Give me a moment and I'll draw it.

Ok sir.

This is what I did:

The speed of the block is the derivative of $h\cos\theta$. Assuming $\theta$ to be constant as we're interested in the instantaneous speed, I get:

$$\frac{dx}{dt}=\cos\theta \frac{dh}{dt}$$

Or,

$$\frac{dx}{dt}=v\cos\theta$$

Ah, you beat me to it :-)

Hooray :-)

You cannot assume $\theta$ to be constant

Ok sir. I thought in a very small amount of time, the variation in $\theta$ is tiny to be neglected.

I would write $L^2 = h^2 + x^2$ so differentiating gives:

$$ 2L \frac{dL}{dt} = 2x\frac{dx}{dt} $$

Ok sir. Then that gives $$\frac{dx}{dt}=\frac{v}{\cos\theta}$$

And this is the correct answer.

$$ \frac{dx}{dt} = \frac{L}{x} \frac{dL}{dt} = \frac{L}{x} v $$

Dammit, you beat me again :-)

See how easy it is using this approach!

:-) But not in considering $\theta$ as a constant.

@JohnRennie Yes sir. The problem was with considering $\theta$ as constant. If it changes with time, I might have to dealt with the angular speed term $d\theta/dt$. But I agree using Pythagoras theorem is easier.

What happens if you write $x = L\cos\theta$. Then we get $dx/dt = d/dt(L\cos\theta)$

Yes sir. And it gives: $$L(-\sin\theta)\frac{d\theta}{dt}+\frac{dL}{dt}\cos\theta$$

For the product we get $d/dt(L\cos\theta) = Ld(\theta/dt) + \cos\theta dL/dt$

But then you need to get an expression for $d\theta/dt$ ...

I think this is going to get messy.

@JohnRennie Yes sir. May I know whether $d(\theta/dt)$ is a short hand notation of $(-\sin\theta)\frac{d\theta}{dt}$?

No, it's a typo :-)

Ok sir :-)

Argh, it's too late for me to edit it. Oh well :-)

:-) No issues sir. I understood that however,

@JohnRennie Yes sir. I'm satisfied with your method. But could you please tell why is the component of one velocity along the another gives the correct result whereas the other way round gives the incorrect one?

If I didn't clearly state my doubt in the previous message, kindly let me know. I'll draw a diagram to show that and how it led to my confusion although it's clear now mathematically.

@GuruVishnu to be honest I find trying to do these questions by playing with components of velocity utterly confusing and I almost always get the wrong answer.

That's why I always do them using the method I've described.

So if you're asking me why the intuitively obvious approach gives the wrong answer I can't say.

@JohnRennie Ok sir. Will you wish to have a look at it? It's interesting however.

@JohnRennie Fine sir.

Maybe tomorrow.

Ok sir. No problem :-)

Thank you for your time and help.

Good bye :-)

Bye :-)