Asked by  | 6th Jan, 2010, 12:03: AM

Expert Answer:

Since F1 pulls one end and F2 other end of the rod, the net force F1 - F2,

and the net acceleration a = (F1 - F2)/M

Now at a distance x from F1, the mass uptil x = Mx/L

and remaining mass = M(L-x)/L

Since all of the rod is moving with an acceleration a,

The net force on rod section upto distance x,

F1 - P = (Mx/L)a .... where p is the contact force, a is the acceleration and Mx/L the mass.

Similarly for remaining section,

P - F2 = (M(L-x)/L)a.

Taking the ratio,

(F1 - P)/(P - F2) = (L-x)/x

Solving this for P the contact force i.e. tension at distance x,

P = F2(L-x)/L + F1(x/L)






Answered by  | 9th Jan, 2010, 03:17: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.