Recent content by Alif Yasa

I already got the answer for the case a but at the case b, i don't know how to sum all of the torques. I'm thinking of using integral, but i don't know where to put the pivot point for the torque. I don't know how to search for the pivot point. Can anybody give me hints so i could find the pivot...

Thanks !

So, if i make the angle between vertical axis and PC is y, and the equilibrium angle x The net torques are mg sin(x)(R cos(x+y)+R+L)-mg cos(x)(R sin(x+y)) =0 sin(x)(R+L) = R sin(y) and from the forces tan(y)=u sin(y)=u/sqrt(1+u^2) then i get sin(x) = (Ru)/((R+L)sqrt(1+u^2)) Is that correct ?

Oh, so is it like Mg cos(x)R-Mg sin(x)(R+L)=0 and tan(x) = R/(R+L) Also, from the force equilibrium, i got tan(x)=1/u cos(x)=u/sqrt(1+u^2) The answer from the book is x=arcsin((Ru)/((R+L)sqrt(1+u^2)) I can get it if i multiply tan(x) from the torque with cos(x) from the force, but why it...

it is something like this

I find that the torque by the mass is T=Mg sin(x) *(R+L) where x is the equilibrium angle

-I tried to draw the forces on the hoop when it is in the equilibrium state. I know there are friction and normal force on the contact point of the shaft and the hoop -I also put the weight force to the M object -But when i used the torque equilibrium, where the pivot is the contact point of the...