@Blue Please help me, I've got a new laptop and while I'm trying to acess my new lectures the player says " Device count exceeded". How do I deregister my previous laptop?

@SwapnilDas what site hosts the lectures i.e. what web site do you go to when you want to view them?

@JohnRennie that's why friction in incline is dependedent on mgcos (theta)

Because N is equal to it

Yes

My fbd was wrong as @DarkKnight said.I figured out I had to resolve the force along the incline to justify the motion of block which helped me realize I can't just resolve forces randomly

Will you be here for some time

@JohnRennie but I have doubt.Does the force mgsin(theta) responsible for both vertical and horizontal motion ? If I resolve it then it seems to

Gravity is directly downwards so it has no horizontal component. So any horizontal motion is due to the normal force. Both gravity and the normal force have non-zero vertical components, so they both contribute to the vertical motion.

@JohnRennie but for proving N=mgcos(theta) ,I equalised N with mgcos(theta).{There is no motion perpendicular to incline}.So here the force mgsin{theta) will be responsible for vertical and horizontal motion or else our motion will not be justified, right?

I can't help feeling you're making this more complicated than it needs to be.

It's certainly true that without the normal force there would be no sideways motion. For example suppose you took the wedge away. In that case the object falls straight down due to gravity. Put the wedge back and now we have a normal force and the object now starts moving horizontally as well as vertically.

So what does mgsin (theta) do....where does it cause displacement of block

I have seen that sometimes Normal gets resolved and sometimes mg gets resolved. When do we know when to resolve

To be honest that comes with experience

Do enough of these problems and you get used to which force you have to work with.

But consider this:

Suppose you're trying to work out the acceleration of the object along the incline. What is the component of the normal force in this direction?

Zero

So the acceleration in this direction cannot be due to the normal force. Yes?

Yes

So if you're being asked about the acceleration along the incline, which is what question will usually ask, you take the component of the gravitational force along the incline.

Okay...I am gonna try some questions now

Usually wedge constraints proceed by resolving normal forces ........

@harambe I'll be here for a few hours if you want to post any questions

Thanks....I have a simple doubt

imgur.com/Lx0h9BM

I understand that the string is light so every object will have same acceleration

What happens if assume anyone has greater acceleration

Will string become slack

If the acceleration of the left block is less than the acceleration of the other two then yes the string will become slack. However it should be obvious that this can't happen.

Okay

@JohnRennie I have a doubt

@harambe Yes?

@JohnRennie imgur.com/tmRIOqY

Sorry for the delay...my phone misfunctions a lot....don't know why but it's screen looks kinda broken so that's why

What is a crystalline solid?

Solid arranged with regular repeating units as much as 1000 units I guess

Why is c wrong @Abcd @AvnishKabaj this was in our test

@Jasmine it's not canceling there

The force

Kids are getting 290 in the paper

@AvnishKabaj C is a constant right?

But all I'm getting is depression

Oh they are awesome I mean wow is there score

@Jasmine multiply and see

Net force is not 0

ihat.k hat is 0 right

@AvnishKabaj I.k is 0 right

@AvnishKabaj please say

Oh god man..

Ohhhh

@Jasmine you got it?

@Blue thanks :)