Problem Solving Strategies

And if we start applying a small force then steadily increase the force we find the block remains static as long as the force is less than or equal to 15N, but above 15N the block slides. Yes?

Srijan M.T

One sec

It doesn’t say equal to sir.

Only less than 15N

which is like 14.999N.

But never 15

John Rennie

8:29 AM

OK ...

> And if we start applying a small force then steadily increase the force we find the block remains static as long as the force is less than 15N, but at or above 15N the block slides. Yes?

Srijan M.T

:one sec

@JohnRennie Yes sir

John Rennie

So that means the maximum static friction is 14.9999999999999 (let's just call it 15).

Yes?

Srijan M.T

:ok

Got it.

John Rennie

:-)

Srijan M.T

:1st one shall we do now.

@JohnRennie

John Rennie

8:46 AM

Which one is that?

Srijan M.T

In the adjoining figure, the coefficient of friction between wedge (of mass M) and block (of mass m) is μ. Find the minimum horizontal force F required to keep the block stationary with respect to wedge.[![enter image description here][1]][1]

1) For this Q , total acc of $M+m$= $F/M+m. $

2) Considering inertial frame of reference , for mass M and m.FBD is as follows :[![enter image description here][2]][2]

I haven’t marked N1 or N2 but can assume them according to FBD.

For mass M ,

1) μN1 on left means the friction from surface.

This one sir

John Rennie

Neither of your figures show a wedge ...

Srijan M.T

K. I’ll put them again sir.

John Rennie

Srijan M.T

Yes sir.

John Rennie

8:51 AM

Those are the two figures, but I don't understand how they relate to the question.

Can you post a picture of the question from your book showing the diagram?

Srijan M.T

K sir.

There is no FBD.I drew that

John Rennie

Ah OK, I understand now.

OK, there is a downwards force mg on the small block. Yes?

Srijan M.T

Normal by the mass M.

So , two forces are present on mass m

John Rennie

Yes. I've labelled the normal force N.

Srijan M.T

Sir , why is mew N in upwards direction ?

John Rennie

9:01 AM

And there is a frictional force that I've shown by the blue arrow. Yes?

Srijan M.T

Yes but how ?

😅 Didn’t get why upward ?

John Rennie

@SrijanM.T it's just conventional to use the Greek letter μ for the friction coefficient.

Srijan M.T

No. I meant upwards

John Rennie

I'm drawing the forces acting on the small mass.

The frictional force stops the small mass sliding downwards. Yes?

Srijan M.T

Yes

Ohk.

I was thinking direction of normal in right means the mass is pushed forward

John Rennie

9:04 AM

So the frictional force must be in the opposite direction to the gravitational force. Yes?

Srijan M.T

so friction is back

@JohnRennie Yes

John Rennie

The whole system is accelerating to right due to the applied force F, and that means the small mass is accelerating to the right.

So the normal force acting on the small mass, N, points horizontally right. Yes?

Srijan M.T

Yes

John Rennie

That's why the forces are in the directions I have drawn.

Srijan M.T

Sir , I got why direction is up because is supposed to slide down or up. So , it’s motion is up or down

John Rennie

9:06 AM

Can you do the question now or shall we go through it?

Srijan M.T

The major mistake you solved sir. I’ll try and if I don’t get it. Then I’ll aks

John Rennie

OK :-)

Srijan M.T

Sir , one Q. Different one.

I had been wondering about it for long.

John Rennie

Yes ... ?

Srijan M.T

Consider like more than 5 pulleys

I’ll draw. One sec

Sth like this,

So , we solve it like writing T first

Now. When we write those equation.

If I say 4 goes up. Then I say M goes down and 2M goes up and 3M goes down and 4M goes up and then write equation according to that. Right ?

No friction

John Rennie

9:11 AM

No, you can't assume anything about the movements of the masses.

Srijan M.T

My sir always solves like that sir.

Like if T is up

T-4g=4a

M-2T=M*a1

Like this.

John Rennie

You would have to write a separate equation for the force on each mass, and you'd get a whole load of simultaneous equations that you'd need to solve. It would all become rather messy.

Srijan M.T

@JohnRennie Yes.K. So , we consider all at rest

What my sir did right above , it is right ?

Because when pulleys gets complicated , you can’t tell which is going up or down. Right

So , My Q was that can I write the equation in such a way that all go up

John Rennie

You need to write an equation like:

Srijan M.T

First write T and then - 3M,-2M

John Rennie

9:14 AM

2 mins ago, by Srijan M.T

M-2T=M*a1

for each mass, so that gives you five equations.

Srijan M.T

Ok. All having Mass first ?

John Rennie

Then you get another equation linking the accelerations of the masses because the length of the string is fixed.

Srijan M.T

Yes. That will be 1/2a for all

except 4kg

John Rennie

So you get six simultaneous equations that you can solve for the five accelerations and the tension.

@SrijanM.T No, you cannot assume anything about the different accelerations of the different masses. You need to grind through the calculation and it will give you the accelerations.

Srijan M.T

Ok sir . But will they all have masses written first

@JohnRennie k

@JohnRennie Let us take A1,a2,a3,a4

John Rennie

9:17 AM

I'm not going to do this problem. It is long and messy and a pain, and you'll never get a problem that complicated in the JEE.

Srijan M.T

Sir. The main problem is considering how to write equation my only Q

It can be less also(pulleys)

But the thing was of considering pulleys to go up or down

That changes the equation right

Thats all

Only take till mass M and 4

Then ,

4 hours later…

Srijan M.T

1:14 PM

@JohnRennie Sir , I solved it. I wish to know if from NIF(From frame of reference of mass M). Then , will there be a pseudo force on mass M.

5 hours later…

Anonymous

5:50 PM

I have a question.

Suppose you have a planck of infinite length kept on a frictionless horizontal surface, and a small block kept on that planck with friction coefficient between the plank and block being non-zero.

Now suppose you give a sharp impulse to the block and it now has some velocity v (w.r.t. ground) and starts sliding on the planck.

So there will be kinetic friction which will decelerate the block and accelerate the planck (in the ground reference frame).

Anonymous

6:03 PM

Now at some instant the block and the plank will have the same velocity (w.r.t. ground) [correct me if I am wrong]. So there is no relative motion between planck and block at this instant. Does this mean that the kinetic friction will stop acting from this instant, and the corresponding acceleration and deceleration will just disappear from that very moment?

And can we say for sure that there will also be no static friction at that instant or at any time after it, as there is no tendency of slipping? And the block and the planck will just move together with a constant velocity?

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May '2112

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Problem Solving Strategies

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