Problem Solving Strategies

11:03 AM

@JohnRennie Could you tell me the reason?

@Abcd There are two forces acting on the block 2. There is an upwards force that stops it falling down, and a force pointing from right to left that is accelerating the block sideways. Yes?

Abcd

@JohnRennie upwards force = ?

Do you mean the frictional/Normal force?

John Rennie

All I'm saying is that if the block isn't falling down there must be an upwards force stopping it from falling down ...

I'm not saying anything about where that force comes from.

Abcd

Okay

John Rennie

If we add the two forces we get a resultant force something like what I've drawn.

Abcd

11:15 AM

ok

John Rennie

Now suppose we rotate our diagram so the resultant force is vertically up:

Abcd

Okay

John Rennie

I haven't changed any of the physics, I've just twisted my head round so I'm now looking at an angle.

Abcd

ok

John Rennie

In this diagram the force on the block acts vertically upwards - there is no sideways force. So if there were no friction the block would now slide to the right i.e. up the wedge.

Abcd

11:19 AM

@JohnRennie and what if the block doesn't remain stationary?

@JohnRennie would this always be the case? Will block always tend to slide up the wedge?

John Rennie

Remember, I've just twisted my head round until the net force on the block is vertical. If the wedge is moving I'm also running along beside it to keep everything I'm looking at stationary.

@Abcd if the wedge isn't accelerating the block will slide down it.

As we increase the acceleration we stop the block sliding down the wedge, and as we increase the acceleration even more we can make the block slide up the wedge.

Abcd

@JohnRennie Thinking. Increasing the acceleration = increasing the horizontal component of force?

John Rennie

Incidentally, I'm not saying this is the way to the calculation. In fact this would be a bad way to do the calculation because it would be working in a non-inertial frame. I'm just trying to give you an intuition for how the system behaves.

@Abcd yes

Abcd

7 mins ago, by John Rennie

John Rennie

Well, increasing the horizontal component of the force we are applying to the wedge.

Abcd

11:24 AM

But look, horizontal component is pointing down the wedge.

John Rennie

Wwell, it's pointing from right to left, yes. But then it would have to be wouldn't it, because the block is accelerating from right to left.

But remember that we are now in a non-inertial frame.

Actually, given you haven't the experience of actually constructing these models I wonder if this is a good approach ...

I might be assuming a level of intuition that isn't there.

Abcd

11:42 AM

@JohnRennie I think I have got it.

@JohnRennie Last question for the day if you are free...?

Abcd

12:03 PM

never mind ...

Avnish Kabaj

12:15 PM

@Abcd

Abcd

@AvnishKabaj uploading

@AvnishKabaj Do you think Tension will act on the wedge toooooo?

Avnish Kabaj

@Abcd yes

The reaction from the pulley

That one is from irodov

Abcd

@AvnishKabaj How?

@AvnishKabaj You are done with irodov?

Avnish Kabaj

@Abcd action reaction pair

@Abcd no

It's far better that you google for a solution

Abcd

@AvnishKabaj how much are you done with

@AvnishKabaj Even you had to google this solution?

Avnish Kabaj

12:18 PM

@Abcd no

I'm not at my house right now

Abcd

@AvnishKabaj Wow!

@AvnishKabaj you must have been taught such a problem, right?

Avnish Kabaj

@Abcd don't remember

Maybe

Abcd

@AvnishKabaj but reaction will be on the thread?

Not on the wedge...

2 mins ago, by Abcd

@AvnishKabaj how much are you done with

Avnish Kabaj

@Abcd I stopped doing irodov when I reached rotation

Abcd

@AvnishKabaj Why?

Avnish Kabaj

12:21 PM

It's a waste of time

Not a waste of time

It's good

But DC Pandey cherry picks relevant questions from irodov and krotov and all

Abcd

@JohnRennie If you could please explain how tension acts on the wedge.

...or @Blue If you are free for 5 minutes.

Anonymous

Tension will act along the rope, on the wedge. One towards the right, away from the wedge and another towards bottom-left.

Anonymous

Consider pulley+wedge as one system to simplify the problem

Abcd

@Blue Oh that wayy! Got it! Thanks!

Anonymous

:)

Nehal Samee

1:10 PM

@JohnRennie ... If a body starts from rest in a frictionless plane and stops at a distance , then what would be the work done ..

..**the body is pulled by a rope by a person at a distance ...

Joe Stavitsky

1:54 PM

could someone please give a hint about F_3 in 4.4 and 4.5? i.imgur.com/pzSXN5a.png

Specifically, what to do with o.5 ft

Joe Stavitsky

2:48 PM

Could someone please explain why the directions of the moments made by F_a and F_b are said to be opposite in the textbook solutions? i.imgur.com/pzSXN5a.png

Abcd

@Tanuj What is the weight at the middle of a loop of radius $R$

For a body which moves with constant velocity $v$ in the loop

Tanuj

@Abcd I don't get what you're asking

Abcd

@Tanuj An aircraft loops the loop of radius $R$ with a const velocity $v$. Find the weight of the flyer of mass $m$ at the middle of the loop.

Tanuj

oh okay

Abcd

I feel it should be $m\dfrac{v^2}{R}$

What do you think @Tanuj???

It shouldn't take much time

Tanuj

3:03 PM

@Abcd I feel it should be what you're saying , but yea , can't say for sure what exactly would be happening

sorry @Abcd

Abcd

@Tanuj Dude, weighing machine records the normal reaction force.

That's how I got my answer to previous parts of the problem:

find the weight at the lower part

and at the uppermost part of the loop.

But I don't get the right answer for this ... using normal reaction

Tanuj

@Abcd what's the answer ?

Abcd

@Tanuj $\sqrt{(\dfrac {mv^2}{R})^2+ m^2 g^2 }$

Tanuj

hmm let me think

@Abcd seems straightforward. What's your doubt ?

Abcd

@Tanuj \ ... I thought weight = normal reaction force.

Because that's what is recorded by weighing machine

Tanuj

3:08 PM

Lol

Abcd

@Tanuj its nothing to laugh about, that's what it is and that's how I got my first two answers... But don't know why answer has done vector addition in this part...

1

Q: Which force does a weighing scale measure?

I have come across several answers on the internet, which address the question "Does a weighing scale measure mass or weight?" but I assure you that this is not one such question. My doubt is this; while solving problems involving free body diagrams of weighing machines, when asked to find the ...

Tanuj

@Abcd I'm not laughing at that dude . You asked the question , then on being asked what your doubt was , you said you got it and leave the question. That's what I found amusing.

Abcd

@Tanuj Oh, I see. Sorry. Btw I edited :p

Tanuj

yea , all good.

Abcd

@Tanuj so what is "weight" in the question...

"weight" is an ill defined term.

Tanuj

3:12 PM

I'm too not sure why would someone do that , but thinking logically , that answer seems much more correct.

Abcd

@Tanuj which answer, mine or his?

Tanuj

thinking of weight as "net force on the body " . hmm

his

Abcd

@Tanuj I don't see any logic in his answer though....

Tanuj

Look , the aeroplane must experience that centripetal force , right ? But how can $mg$ be not taken into account ? I mean no other force balances it.

Abcd

@Tanuj but that doesn't make it a part of the "weight"

Tanuj

3:15 PM

I guess it does .

Abcd

@Tanuj The accelerator of the aircraft balances $mg$

@Tanuj again depends on how you define weight...

Tanuj

@Abcd I knew you'd say that. Haha , I'll say we'll not assume it as its not given in question to consider that

Or atleast the one who gave the answer didn't assume it.

Abcd

@Tanuj Duh. The velocity cannot be constant -_- if net force acts on it in vertical direction.

Joe Stavitsky

Could someone please explain why the directions of the moments made by F_a and F_b are said to be opposite in the textbook solutions? i.imgur.com/pzSXN5a.png

Tanuj

@Abcd hmm , is it given that it is a vertical circle ? Well of course it would be .

Abcd

3:18 PM

@Tanuj lol

Anyway, think about it ttyl! Bye !

Tanuj

bye

Abcd

3:48 PM

@JohnRennie There seems to be a contradiction at a place on the basis of definition.

@Tanuj Are you there? Can I share that with you?

Tanuj

@Abcd sure

Gotta go rn , but will def have a look when I'm back.

Abcd

Okay...

Joe Stavitsky

5:42 PM

Can someone give me some hints here please? i.imgur.com/pHFQIQH.pngWhat I have so far is M1=2.5*50*cos25, M2=3.25*50*cos30,

Abcd

6:57 PM

@AvnishKabaj @GaurangTandon @Tanuj Please see !

JD_PM

10:37 PM

@Abcd as far as I am concerned right answers are both a and b. α does not remain positive on account of it has to come back to a state of rest (that means decrease in acceleration, that means change in sign).

Abcd

@JD_PM $c$ is also correct

JD_PM

As the mass just covers 1m, with kinematic laws in our hands we can comprehend it is impossible for the mass to reach 2

Abcd

@JD_PM Just out of curiosity, are you a JEE aspirant?

@JD_PM which kinematic laws are you talking about

JD_PM

If b) is correct, how can be c correct ?

Abcd

@JD_PM b is incorrect

JD_PM