Recent content by nirmalya1basu

I have solved the problem in the following way. When the weight W is raised a distance y, F(raising) acts through a distance [{2*r1/(r1 - r2)} * y], and F(friction) acts on the chain through a distance [{2*(r1 + r2)/(r1 - r2)} * y]. Then, we have F(raising) * [{2*r1/(r1 - r2)} * y] - F(friction)...

I have tried to solve the problem in the following way. When the weight W is raised a distance y, then Useful work done = Actual work done by the user - Work done by friction Therefore, W*y = F(raising) * y - F(friction) * y or, W = F(raising) - F(friction) or, F(raising) - F(friction) = W --->...

Sir, I have understood your point. I will try to solve the problem using this concept.

Sir, it does not appear to me that Manual F times R = W because the following answers to the problem are given: Force needed to raise W = W*R / [n*(R - 1) + 1] Force needed to lower W = W / [n*(R - 1) + 1]

It appears to me that if the weight is raised a distance y, then the useful work done = W*y Actual work done by the user = F(raising) * y Work done by friction = F(friction) * y Now, if we say that Useful work done = Actual work done by the user - Work done by friction, then we have W*y =...

Thank you very much, PeroK.

Thank you, Sir, for your reply. I understand that friction is relevant, but, I am unable to write equations involving friction that will help in solving the problem. What I have thought is the following: Work done by the lowering force = F(lowering) * {r2*x/(n - 1)} Work done by the raising...

Let the radius of the larger sheave in the upper pulley be r1, and the radius of the smaller sheave in this pulley be r2. Given, the larger sheave has n notches in its circumference and the smaller sheave, (n - 1) notches. Therefore, 2*pi*r1 corresponds to n notches, and 2*pi*r2 corresponds to...

Hi, I am Nirmalya. I am from Kolkata, West Bengal, India. I have a lot of interest in Physics.