Problem Solving Strategies

Triet Vo Nguyen Minh

01:07

So now I will redeclare my question: If mu is the coefficient of friction between 2 objects, and one object has an arbitrarily large mass (like the floor or the earth), the maximum static friction is F_max=mu*N. So far so good.
Assuming there are 2 stacked objects with the same mass, and the bottom one's friction to the floor is 0. When we drag the upper object to a side with a force of F_max,
the object still has a positive net acceleration to the ground, but with the same acceleration as the bottom object. The equation F_ext-F_max=0 still holds but 0 is not the net force of the upper object.

John Rennie

04:52

@TrietVoNguyenMinh Do you mean like this?

Triet Vo Nguyen Minh

Yes.

John Rennie

The way to approach questions like this is to work out the acceleration of both masses due to the applied force i.e. the total mass is m+M so if the two masses accelerate together then the acceleration is a = Fext/(m+M).

OK so far?

sonicsid

05:13

Wait, he says F_ext = F_max. Then if the object moves, mustn't it be moving with a constant velocity, and no acceleration?

And then wouldn't "the object still has a positive net acceleration to the ground" be untrue? Sure, the object may be in motion wrt the ground, but it won't be accelerated motion.

(assuming the coefficient of static friction to be roughly equal to that of kinetic friction)

John Rennie

Typically with questions like this you'll be asked something like "What is the maximum force before the two blocks start slipping over each other?"

So the approach I describe is targeted at understand this sort of question.

Triet Vo Nguyen Minh

Well, that friction is m's friction to M, that means m is stationary to M.
Newton's 3rd law states that if M exerts friction to m, m must also exert a force to M, friction between M and ground does not exist, so M has acceleration.

John Rennie

Yes, that is how we analyse the motion.

Suppose the blocks do not slip then they move together as a single unit, and in that case the acceleration of both blocks is given by the equation above:

a = Fext/(m+M)

Yes?

Triet Vo Nguyen Minh

Yes, but...
This is an example for me to understand friction, so what I need is to analyse the forces.

Apart from F_ext and F_f, object m must still have a force on it so that it has an acceleration to the ground, right?

John Rennie

So if the bottom block accelerates at a = Fext/(m+M) then the force on the bottom block must be given by F = Ma. So the force on the bottom block must be:

Fb = Fext M/(m+M)

And that force comes from the frictional force between the two blocks. Yes?

Triet Vo Nguyen Minh

05:29

That is true

What I am concerned about is the upper block.

John Rennie

And the maximum force that the top block can apply on the bottom block is equal to the static friction i.e. Fmax = μN = μmg.

So now by comparing Fmax and Fb we can tell if the two blocks will slide over each other or not.

So to summarise:
- we apply a force Fext to the top block
- the top block applies a force Fb to the bottom block through friction
- the maximum value of Fb is Fmax = μmg

And the net force on the top block is Fext - Fb

Triet Vo Nguyen Minh

Does that mean the net force on the top block is 0? I have stated that F_ext=F_max.

John Rennie

Are you defining Fmax = μmg? i.e. we apply a force Fext = μmg to the top block?

Triet Vo Nguyen Minh

Yes, that's true

If we drag the object on a floor with friction coefficient mu, it stays still. OK.

sonicsid

"it"?

John Rennie

05:37

@TrietVoNguyenMinh OK, since you are not exceeding the frictional force between the blocks the two blocks will move together without sliding over each other. Yes?

@TrietVoNguyenMinh I thought we were saying the friction with the floor was zero?

Triet Vo Nguyen Minh

@JohnRennie What I said is if the bottom object was a floor, sorry for not explicitly saying.

The floor has infinite mass so any force on it will not accelerate it

John Rennie

I'm lost now. Are we now talking about a single block sliding on a rough floor?

Triet Vo Nguyen Minh

(Why does this happen all the time, I always make others lost)
... To say in the 2 blocks context is M=infty

sonicsid

@JohnRennie I think he's trying to compare between the "one object on another object" case and "one object on rough floor" case.

John Rennie

Ok, so in that case the bottom block doesn't move at all and the top block moves only if Fext > μmg.

Triet Vo Nguyen Minh

05:43

Yes.

That says the top block doesn't move at all, because F_ext = F_max = mu*mg.

John Rennie

Yes, if Fext = μmg then the net force on the top block is zero so it has an acceleration of zero.

Triet Vo Nguyen Minh

But if the mass of the bottom block is finite, to the ground the top block would be accelerating

Which means either F_ext and F_f are not equal or there is another force apart from F_ext and F_f

John Rennie

If we make the mass of the bottom block finite (call it M) then in effect we have a single object with a mass equal to m + M, and we are applying the force μmg to the combined object. Yes?

Triet Vo Nguyen Minh

Yes but I want to resolve this the force analysis way

Hold on while I make a figure

John Rennie

OK, I'll go and make a coffee

sonicsid

05:53

"When we apply μmg on the upper object, the net force on it is 0 since it gets balanced by the friction between its and the lower object's surface. Because of Newton's Third Law, a force of μmg must also apply on the lower object. As the net force on the lower object is non zero, it must get accelerated. As the upper object does not move with respect to the lower object, it also moves with the same non zero acceleration with respect to the ground as the lower object.

How does this happen if the net force on the upper object is zero?"

Is this what you're trying to ask?

Triet Vo Nguyen Minh

@sonicsid Yes, that is an extremely accurate description of what I am asking.

Great I don't have enough reputation to upload images

John Rennie

@TrietVoNguyenMinh Is this the diagram you had in mind?

Ff is the frictional force between the two blocks.

Note that the maximum value of Ff is μmg but this is the maximum value. In general Ff < μmg.

Triet Vo Nguyen Minh

imgur.com/a/XFdSCSq

Lllt

@JohnRennie Why??

John Rennie

@TrietVoNguyenMinh F_f is smaller than F_ext

Triet Vo Nguyen Minh

06:07

There goes my image.
So even if F_ext = F_max, F_f is still not equal to F_max.

John Rennie

Yes

We can calculate F_f and show this is true.

sonicsid

Is F_f smaller than F_ext because its aim is to prevent relative motion between the two surfaces - and not merely to oppose F_ext?

John Rennie

@sonicsid F_f is related to the acceleration of the lower block by F_f = Ma. Yes?

(We assume the two blocks accelerate together with an acceleration equal to a)

sonicsid

right

and a = F_ext/(M+m)

John Rennie

And Fext is related to the cceleration of the two blocks together by Fext = (m+M)a. Yes?

@sonicsid Yes

So Fext - F_f = ma

sonicsid

06:16

Got it. Thank you.

John Rennie

:-)

Triet Vo Nguyen Minh

Does this make a different F_max value, and why?

Lllt

@JohnRennie We can have two situations here , either block moves relative to each other or block move together

In first situation kinetic friction accelerates the lower block and in second situation static friction

and as you proved above in second case the static friction will always be less than limiting friction

John Rennie

@Lllt Yes. The way to analyse the motion is to start by assuming the blocks do not slide, then find the force between the blocks.

Now compare the force between the blocks with the static friction. If the force is less than the static friction we know our assumption was valid and the blocks do not slide over each other.

However if we find the force is greater than the static friction we know our assumption was wrong and the blocks do slide.

Lllt

Yes

It becomes extremely confusing when we do opposite

John Rennie

06:25

If the blocks do slide then the force on the lower block is just the static friction, and the (net) force on the upper block is F_ext minus the static friction.

Lllt

Sir please see this

@JohnRennie sir when a q charge is inside a cubical surface then flux through cube is $\frac{q}{\epsilon_0}$, when it is outside it , then flux is zero
What will be the flux through a cubical surface if charge is on it ( for example at center of one of its face)

How will you do this?

John Rennie

@Lllt If the charge is on a surface then the flux through that surface is undefined. However there is a way we can approach problems like this.

Suppose we take a second identical cube and join it to the first cube at the face where the charge is.

Lllt

@JohnRennie Why flux through surface is defined

It should be zero I think?

Because the electric field is everywhere tangential to surface and area vector normal to it so surface integral evaluates to zero

John Rennie

Like that

Lllt

06:43

Sir??

John Rennie

Hi

Lllt

Hello sir

@JohnRennie Are you there sir??

John Rennie

Yes

Lllt

Why flux through surface is defined
It should be zero I think?
Because the electric field is everywhere tangential to surface and area vector normal to it so surface integral evaluates to zero

@JohnRennie We apply gauss law for cuboid and then use symmetry

John Rennie

In this case the flux through the side of the cube will be zero, and we can show it using the diagram above. But in general if you have a charge on a Gaussian surface it can cause problms with the calculation so this situation is best avoided.

Lllt

06:51

@JohnRennie Oh But will you see my method and check whether it is correct or not?

We know that a point charge is something like a sphere with radius tending towards zero, Yes?

John Rennie

The problem is that the flux changes discontinuously a the charge crosses the surface and that means it is not defined when the charge is exactly on the surface.

By saying it's zero you are effectively just choosing to define it in that way, but I'm not sure how rigorous this is. I would not use this in a JEE question.

Lllt

Ok sir , we can prove this by the keeping two observers on either side

sonicsid

I can see why (b) is correct. What about (c)?

Adil Mohammed

is this a HC verma question?

i think i remember asking here about it

John Rennie

I would have said only (a) is correct. The field is dU/dx, where U is the potential, and the question says U decreases uniformly, which I take to mean dU/dx is constant.

In that case dU/dx = E = 20V/cm is constant everywhere between x = -1 cm and x = +1 cm.

sonicsid

07:14

@AdilMohammed yep

@JohnRennie I thought so too, but (b) and (c) are given as the correct answers :/

John Rennie

Aha, wait, the component of the field E_x must be equal to 20V/cm but the other components E_y and E_z can have any value.

2

Because those two components do not cause the potential to change as we move along the x axis.

sonicsid

https://i.sstatic.net/ZpZrN.jpg

Found this online. Seems fine except for the part where it says costheta approximately equals one, which seems unnecessary.

Yes! Thank you.

John Rennie

:-)

sonicsid

Can the situation in part (c) be considered similar to an object being projected from the ground?

the electric field would be analogous to g, the charge to the mass, and so on

John Rennie

Yes. The situation is just like an object being thrown up from the ground at 60°

In both cases only the vertical component of the velocity changes.

sonicsid

07:22

Thrown up from the ground at 30°, you mean?

John Rennie

Oops, yes, 60° to the vertical so 30° to the ground.

sonicsid

Haha yeah. I'll check my calculations again then. Thanks for confirming.

John Rennie

:-)

Adil Mohammed

07:33

@JohnRennie can you check if you can see this link?

physics.stackexchange.com/questions/647421/…

John Rennie

The question was deleted. I can see it because you can see deleted questions if your rep is higher than 20K.

Adil Mohammed

i remember seeing a formula called the "General Gauss's law" where they talk about charge on the surface

John Rennie

There are probably ways to approach having a charge on the Gaussian surface, but you don't need them for the JEE.

Lllt

@JohnRennie @AdilMohammed when I said charge on surface I meant a small sphere with infinitesimal radius with center lying on surface

John Rennie

And I'm not sure I see the point anyway. We can always draw a Gaussian surface anywhere we want, so why make life difficult by drawing the surface through a charge?

Adil Mohammed

07:37

you are right

Lllt

@JohnRennie Yes that's easy!!

Sir , is there something like flux coming out of a charge

John Rennie

With an electric field we can define a property called the divergence.

Lllt

Ok

John Rennie

If we consider some small volume dV then integrate the divergence over this volume it tells us how much flux originates in the volume.

If you shrink the volume dV down to a point then the divergence of the field at that point tells how much flux originates from that point.

You can do this for a point charge.

Lllt

Divergence is therefore flux per unit volume at a point

John Rennie

07:41

Yes

Lllt

Ok

John Rennie

Though that isn't how you calculate it mathematically.

If you have an electric field E = (Ex, Ey, Ez) then the divergence is:

∇.E = dEx/dx + dEy/dy + dEz/dz

It is written as ∇.E because it's kind of like a dot product between the two vectors (d/dx, d/dy, d/dz) and (Ex, Ey, Ez).

Adil Mohammed

@Lllt this is the generalised one, but I am sure someone already made an answer but I cant find it

@JohnRennie what exactly is ∇?

John Rennie

Del

Del, or nabla, is an operator used in mathematics (particularly in vector calculus) as a vector differential operator, usually represented by the nabla symbol ∇. When applied to a function defined on a one-dimensional domain, it denotes the standard derivative of the function as defined in calculus. When applied to a field (a function defined on a multi-dimensional domain), it may denote any one of three operators depending on the way it is applied: the gradient or (locally) steepest slope of a scalar field (or sometimes of a vector field, as in the Navier–Stokes equations); the divergence of...

Vector calculus is a lot of fun, but you don't need it for the JEE.

Adil Mohammed

:(

John Rennie

07:51

If you go on to do engineering or physics at university you'll learn lots about vector calculus!

Adil Mohammed

i was actually planning to start multivariable calculus on MITcourseware cause it looked interesting

John Rennie

It's fascinating :-)

All I would say is be careful you don't get too interested in it and spend so much time on it that it affects your JEE preparation.

Adil Mohammed

yes you are right

i think if V is a function of (x,y,z) then to find E we have to partial differentiation right

i think thats how i learnt the basics

John Rennie

Yes, E = ∇V

Which written out in full is E = (dV/dx, dV/dy. dV/dz)

sonicsid

Umm, isn't E = -∇V?

John Rennie

08:06

Oops, yes, I have a tendency to forget the minus sign :-)

sonicsid

And I have a tendency to doubt what I have written in my notes :-)

Ashish Ahuja