CBSE Class 11-science Answered
Tension
Asked by | 06 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)
Regards,
Team,
TopperLearning.
Answered by | 09 Jan, 2010, 03:17: PM
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