Re: A problem

[Surendranath <bsreddy@HD1.VSNL.NET.IN>]

> > Force of normal reactioon due to the surface would not act through the
> > center of the bottom surface. It would act closer towards the front edge.
> >
> > Surendranath. B.
>
>
> This would indeed create a net 'stopping' torque since it would then
> exceed the opposing torqe due to gravity.
>
> Can we easily see HOW MUCH closer to the the front edge the normal
> force is? _*I*_ can't!

Angular momentum about the specified axis is mva/2( a is the side of the
cube, v is the instantaneous velocity) and its rate of change is -kmga/2 (k
coefficient of  friction). Torques( about the specified axis) of mg is mgx
and of N is -mg(x+d) {d is the distance between lines of action of mg and N
and x is the instatnataneous postion}
Net torque -mgd =-kmga/2  gives d=ka/2;
distance from front edge is a/2-d;


Surendranath. B.

Surendranath Reddy. B.
http://surendranath.tripod.com
http://members.nbci.com/Surendranath