Tension

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)

Regards,

Team,

TopperLearning.

 

 

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

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