Recent content by saket

yeah right.. and what we do normally use is, Newtonian assumption (for dry friction): µk & fk doesn't depend upon surface area, or speed.. (µk doesn't depend upon even mass) ..but then it is just an assumption.. for we to be able to work easily, and possibly it matches agreeably (although not...

If I am not mistaken following are the findings which some standard textbooks have quoted: µk, for rigid bodies, depends on the two materials in contact and is independent of mass, m. Therefore, in a given problem fk is proportional to normal reaction (as fk = µkN). But also, some have quoted...

Something amiss in the question.. or, there shouldn't be (2) part.

I think your statements have flaws. Have you directly quoted the question or have you tried to put it up in your own words? And, when you say, "i got the wrong answer" .. you should be knowing the correct answer? Anyways, as far as classical Kinematics go: Kinetic Energy = m*(v^2)/2...

Method I: Ideal Fluid, by definition, can not tolerate any shear force. Use this fact. Assume the centre of the water surface to be origin. Let P(x,y) be a point on the water surface. Consider forces on it. Impose the condition that tangential force will be zero. I hope you can proceed now.

well.. 11.2 * 1000 = 11200 and not 112000! what is the combined mass?? 105kg? or, 2,3105kg, or what?? what is the answer?

(b) seems fine.. (assuming constant acceleration) why don't you try for (a) in a similar fashion!

Okay, although not provided in the question, I am assuming that you are supposed to find coefficient of kintic friction. (Your attempt at the solution tells me that.) You erred in this statement: Note that, torque produced by F is not the one which is producing deceleration in the disc (It...

Consider a stick of length L. Where will its centre of mass be? Mid-point, right? This is at a distance of L/2 from either end. Similarly in the given problem, we have water column of height, say h. So, centre of mass will be at a height of h/2 .. isn't it? {You have studied centre of mass...

Hmm.. gone to sleep? Hope, you will solve it in your dreams! Anyways, if in case you are not able to, to make you start your day happily, here is the soltion: (Both methods are almost same.) Method I: Let bottom of the cylinders be our reference line for height. Let h1 = 1.56 m, h2 = 0.854 m...

No, you won't need integral.

Wait, I am a bit confused about the figure. First thing, your wedge is a right-triangular prism, of some thickness, t. The triangular face has sides L and h. (None of these are hypotenuse of the triangle.) Now let us consider this triangular face on the plane of the paper (such that...

If that is the case, then don't you think, your following formula doesn't hold here: (Why?) Simply because, it is a thin rod (of thickness dx) and length -- that can be obtained using similar triangles (or trigonometry). So, you should use moment of inertia for a thin rod (about its centre)...

yep, is it okay? (Attachment is still in pending for approval. :( )

No no, you don't need to go for Bernoulli. Note that as liquid is enetring/leaving the cylinder, mass of the liquid in it doesn't remain constant. So, the formula you have used for change in potential energy is wrong. Instead, use \DeltaPE = (mgh)final - (mgh)initial, for each of the cylinders.