Recent content by kirakun

Yes changing frames of reference helped in finding the solution. Thanks everyone.

Yes. Resolving the centripetal along the surface and equating with the frictional force and the weight (In the same direction) gives the required equation...

Isn't the centripetal force a resultant force, so it cannot be resolved?

What I did was: Let W = weight of vehicle N = Normal reaction at surface acting on vehicle F = Frictional force developed and y = coefficient of friction. x = Angle of inclination Resolving weight W perpendicular to the surface we have N = W cos x The frictional force F =...

Hello, I was browsing a derivation for the relationship between the curvature of horizontal road curves and the angle of elevation/banking. The derivation is shown on page 4 here: http://www.cdeep.iitb.ac.in/nptel/Civil%20Engineering/Transportation%20Engg%201/14-Ltexhtml/nptel_ceTEI_L14.pdf...

Thank you!

Hmm the answer seems right there but its not clicking. Any more hints. The work done on the hand by mass = average force x distance = (mg/2 x mg/k) ? Stored energy = Tension x distance = (mg x mg/k) ? (Which is obviously not good since we know beforehand that the stored energy = work done on...

There would be 3 forces: 1. Force exerted by the wire on the mass (upwards), T 2. Force exerted by the hand on the mass (upwards), F 3. And the weight (downwards), mg Resultant force = ma The equation should be this: mg - T - F = ma ?

This relates to the extension of a uniform cross-section, homogenous, ideal wire which extends within the proportional limit (Hooke's law). From my understanding, only half of the gravitational potential energy lost when the mass is lowered on the initially un-extended wire, is stored in the...

Thank you!

Oh god yes, i forgot the 9 in the denominator. Well that partial fraction makes the integrand now easier. I should complete the denominators to the square and use arctan integral?

Alright. But it should give the same answer as in my edited post

It will be of the form \frac {At+B}{1+t^2} + \frac {Ct+D}{7+4t^2} right? Consequently For the numerator, t^2 = (At+B)(7+4t^2) + (Ct+D)(1+t^2) (4A+C)t^3 + (4B+D) t^2 + (7A+C)t + (7B+D) = t^2 Comparing coefficients of t^2 and comparing constants we get the 2 equations respectively; 4B...

What method do u think would be best to tackle this one? well i chose tan x because with the denominator containing terms squared, i thought i would end up with an arctan case, but the t^2 in the numerator after simplifying makes it harder to achieve the form if not impossible. Also when t =...

Homework Statement Well this is part of an integration process, namely: \int \frac {sin^2x}{4+3cos^2x}dx Homework Equations My attempt involved using a u-substitution, namely t = tan x The Attempt at a Solution Using t = tan x, sin^2 x = \frac {t^2}{1+t^2} and cos^2 x = \frac...