IIT-JEE problems - Page 11

Jaskanwar Singh · 01-02-2012, 02:02 PM

thanks hjpotter92.

a doubt -
suppose there is a rough table surface. on it i placed a block, and on top of that block, placed another smaller block. everything is rough.
now suppose i pushed the base block but it didn't move because of static friction. will only the table oppose the base block or the above placed block too?

thetechfreak · 01-02-2012, 02:12 PM

The force is applied on the base block hence it will have tendency to move. But it doesnt as friction opposes it. Hence friction works only on base block.

If the base block would have moved then there would have been relative motion on blocks. And friction opposes relative motion. Hence only in this case friction acts on both block. When there is relative motion the friction will speed up the top block and slows down base block.

Jaskanwar Singh · 01-02-2012, 02:44 PM

thetechfreak, when the base block moves, friction acts on base block in backward direction and for upper block in forward direction.
So why doesnt this friction come into play when base block ' tries' to move?

nims11 · 01-02-2012, 03:13 PM

^^ because then it doesn't move. the base block is trying is to move with respect to the ground. The upper blocks are there as they are and donot intend any relative motion unless the base blocks starts to move with respect to ground and thus 'tries' to move wrt the upper block

thetechfreak · 01-02-2012, 03:42 PM

Quote:

Originally Posted by Jaskanwar Singh

thetechfreak, when the base block moves, friction acts on base block in backward direction and for upper block in forward direction.
So why doesnt this friction come into play when base block ' tries' to move?

There will be friction ONLY WHEN base block moves. There won't be any friction no matter how hard it tries to move. There has to be relative motion

hjpotter92 · 02-02-2012, 07:08 PM

(some kind of static/tensile/dynamic etc...)Friction = (mu)*(Normal Reaction)

The normal reaction is because of combined weight of all blocks.

So, it's like saying that "actually, only the table is preventing base to move" but in reality, "the blocks/boxes above base are also contributing."

Jaskanwar Singh · 02-02-2012, 08:47 PM

thanks nims, thetechfreak, hjpotter.

Quote:

Originally Posted by thetechfreak

There will be friction ONLY WHEN base block moves. There won't be any friction no matter how hard it tries to move. There has to be relative motion

friction opposes both relative and impending motion

relative motion - kinetic friction
impending motion - static friction

Quote:

Originally Posted by hjpotter92

(some kind of static/tensile/dynamic etc...)Friction = (mu)*(Normal Reaction)

The normal reaction is because of combined weight of all blocks.

So, it's like saying that "actually, only the table is preventing base to move" but in reality, "the blocks/boxes above base are also contributing."

but then if we would have taken friction between blocks, it would have been mu times N due to upper block.

thetechfreak · 02-02-2012, 09:46 PM

I mean friction will act only when there is relative motion. Was saying about that case when there is RM.

hjpotter92 · 03-02-2012, 09:10 AM

Quote:

Originally Posted by Jaskanwar Singh

thanks nims, thetechfreak, hjpotter.

but then if we would have taken friction between blocks, it would have been mu times N due to upper block.

Nope, there are cases. First we take all the boxes as one body, and table as another. Next, we calculate internal forces between two boxes repeatedly till we reach the top two (or bottom two) boxes.

Jaskanwar Singh · 08-02-2012, 08:42 PM

^ok.

question - The potential energy function along the positive x axis is given by U(x) = -ax + b/x. a and b are constants. If it is known that the system has only one stable equilibrium configuration, the possible values of a and b are?

hjpotter92 · 09-02-2012, 06:12 PM

Quote:

Originally Posted by Jaskanwar Singh

^ok.

question - The potential energy function along the positive x axis is given by U(x) = -ax + b/x. a and b are constants. If it is known that the system has only one stable equilibrium configuration, the possible values of a and b are?

Are you sure it is (-ax)?

Jaskanwar Singh · 09-02-2012, 06:54 PM

^yes.

pramudit · 09-02-2012, 07:30 PM

a=b....?
"just a tukka"...

Jaskanwar Singh · 09-02-2012, 08:43 PM

^no

AcceleratorX · 11-02-2012, 07:13 AM

Quote:

Originally Posted by Jaskanwar Singh

^ok.

question - The potential energy function along the positive x axis is given by U(x) = -ax + b/x. a and b are constants. If it is known that the system has only one stable equilibrium configuration, the possible values of a and b are?

I think the context needs to be defined because the same question can be applied to statistical mechanics, classical mechanics, energy transfer and thermodynamics.

Which context is this?

If I had to garner a guess I'd be looking at the graph of this function and seeing local maxima and minima. Am I thinking along the right lines? Basically in the graph of the derivative we are looking for constancy and this is only achieved if b=0 or if b is variable, which it is not. As for a, I'm not exactly sure, the derivative returns a constant anyway and so a should be any value?

(Far removed from JEE days but it's always good to know a little physics

)

Jaskanwar Singh · 12-02-2012, 07:25 PM

^this is from chapter work, power energy

and yes it uses maxima, minima concept.

nims11 · 12-02-2012, 07:47 PM

^^ the closest i came is b>0 and a<0.

what is the answer?

Jaskanwar Singh · 12-02-2012, 07:58 PM

^yeah nims, its like that only. but they have even given values, a = -1 and b = 2. i am not able to understand how these came.

BTW do you have class 11 NCERT part 2?

nims11 · 12-02-2012, 08:06 PM

yes, but at home.

Jaskanwar Singh · 12-02-2012, 08:10 PM

i had a doubt in bernoulli equation (fluid mechanics), you revised it?

nims11 · 12-02-2012, 08:28 PM

nope, but i remember it, what is the doubt?

Jaskanwar Singh · 12-02-2012, 09:34 PM

^in efflux question, why we take pressure at hole to be atmospheric pressure?
when we proved bernoulli equation, we used pressure due to liquid for work done at right side of tube, and thus work done was = - liquid pressure x deltaV.